Definition (Convex subset).
A subset (C) of a vector space (V) over (\mathbb{R}) is called convex if for every pair of points (x,y\in C) and every real number (t) with (0\le t\le 1), the point [ tx+(1-t)y ] also belongs to (C). Geometrically, this means that the line segment joining any two points of (C) lies completely inside (C).

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Let (P_1,P_2,\ldots,P_n) be points of a vector space (V). Define [ S=\left{t_1P_1+t_2P_2+\cdots+t_nP_n:\ t_i\ge 0,\ \sum_{i=1}^n t_i=1\right}. ] We have to prove that (S) is convex.

Let (A,B\in S). Then there exist non-negative real numbers (a_i,b_i) such that [ A=a_1P_1+a_2P_2+\cdots+a_nP_n,\qquad \sum_{i=1}^n a_i=1, ] and [ B=b_1P_1+b_2P_2+\cdots+b_nP_n,\qquad \sum_{i=1}^n b_i=1. ]

Now take any (\lambda\in[0,1]). Consider the point [ \lambda A+(1-\lambda)B. ] Substituting the expressions for (A) and (B), we get [ \lambda A+(1-\lambda)B =\lambda\sum_{i=1}^n a_iP_i+(1-\lambda)\sum_{i=1}^n b_iP_i. ] Hence [ \lambda A+(1-\lambda)B =\sum_{i=1}^n \bigl(\lambda a_i+(1-\lambda)b_i\bigr)P_i. ] Let [ c_i=\lambda a_i+(1-\lambda)b_i. ] Since (\lambda\ge 0), (1-\lambda\ge 0), (a_i\ge 0), and (b_i\ge 0), we have [ c_i\ge 0 ] for every (i=1,2,\ldots,n). Also, [ \sum_{i=1}^n c_i =\sum_{i=1}^n \bigl(\lambda a_i+(1-\lambda)b_i\bigr) =\lambda\sum_{i=1}^n a_i+(1-\lambda)\sum_{i=1}^n b_i =\lambda\cdot 1+(1-\lambda)\cdot 1=1. ] Thus [ \lambda A+(1-\lambda)B =c_1P_1+c_2P_2+\cdots+c_nP_n ] with (c_i\ge 0) and (\sum c_i=1). Therefore, [ \lambda A+(1-\lambda)B\in S. ]

Since (A,B\in S) and (\lambda\in[0,1]) were arbitrary, the line segment joining any two points of (S) lies entirely in (S). Hence (S) is a convex subset of (V).

[ \boxed{\text{Therefore, } S \text{ is convex.}} ]

In fact, (S) is precisely the convex hull of the points (P_1,P_2,\ldots,P_n).

Definition: Convex Subset of a Vector Space

Let (V) be a vector space over the field of real numbers (\mathbb{R}). A non-empty subset (C \subseteq V) is called a convex subset if for every pair of points (x, y \in C) and for every real number (t) with (0 \le t \le 1),

[ tx + (1-t)y \in C. ]

Equivalently, (C) is convex if the line segment joining any two points of (C) lies entirely in (C).

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Given

Let (P_1, P_2, \dots, P_n) be points of a vector space (V). Let

[ S = \left{ t_1P_1 + t_2P_2 + \dots + t_nP_n ;:; t_i \ge 0,; t_1+t_2+\dots+t_n=1 \right}. ]

We have to prove that (S) is convex.

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Proof

Take any two points (X, Y \in S). Then by definition of (S), there exist real numbers (a_i, b_i) such that

[ X = a_1P_1 + a_2P_2 + \dots + a_nP_n, ]

[ Y = b_1P_1 + b_2P_2 + \dots + b_nP_n, ]

where

[ a_i \ge 0,\quad b_i \ge 0, ]

and

[ a_1+a_2+\dots+a_n=1,\qquad b_1+b_2+\dots+b_n=1. ]

Let (\lambda) be any real number such that (0 \le \lambda \le 1). Consider the point

[ Z = \lambda X + (1-\lambda)Y. ]

Substituting the expressions for (X) and (Y), we get

[ Z = \lambda(a_1P_1+\dots+a_nP_n) + (1-\lambda)(b_1P_1+\dots+b_nP_n). ]

Rearranging the terms,

[ Z = [\lambda a_1 + (1-\lambda)b_1]P_1 + \dots + [\lambda a_n + (1-\lambda)b_n]P_n. ]

Let

[ c_i = \lambda a_i + (1-\lambda)b_i,\qquad i=1,2,\dots,n. ]

Then

[ Z = c_1P_1 + c_2P_2 + \dots + c_nP_n. ]

Now, since (a_i \ge 0), (b_i \ge 0), (\lambda \ge 0), and (1-\lambda \ge 0), we have

[ c_i = \lambda a_i + (1-\lambda)b_i \ge 0. ]

Also,

[ \sum_{i=1}^n c_i = \sum_{i=1}^n [\lambda a_i + (1-\lambda)b_i] = \lambda \sum_{i=1}^n a_i + (1-\lambda)\sum_{i=1}^n b_i. ]

But

[ \sum_{i=1}^n a_i = 1,\qquad \sum_{i=1}^n b_i = 1. ]

Therefore,

[ \sum_{i=1}^n c_i = \lambda(1) + (1-\lambda)(1) = \lambda + 1 - \lambda = 1. ]

Thus,

[ c_i \ge 0,\qquad c_1+c_2+\dots+c_n=1. ]

Hence (Z = c_1P_1 + c_2P_2 + \dots + c_nP_n) is a linear combination of (P_1, P_2, \dots, P_n) with non-negative coefficients whose sum is (1). Therefore,

[ Z \in S. ]

So, for any (X, Y \in S) and any (\lambda \in [0,1]),

[ \lambda X + (1-\lambda)Y \in S. ]

Therefore, by the definition of a convex set, (S) is convex.

[ \boxed{\text{Hence } S \text{ is a convex subset of } V.} ]

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Note

The set (S) is precisely the convex hull of the points (P_1, P_2, \dots, P_n), i.e. the smallest convex set containing all these points.

Option 1: Equivalence Relation

Definition.
A relation (R) on a non-empty set (A) is called an equivalence relation if it satisfies the following three properties:

1. Reflexive: (aRa) for every (a \in A).
2. Symmetric: If (aRb), then (bRa) for all (a,b \in A).
3. Transitive: If (aRb) and (bRc), then (aRc) for all (a,b,c \in A).

Proof that equality ((=)) is an equivalence relation.
Let (A) be any non-empty set and define the relation (R) on (A) by
[ aRb \iff a=b. ]

Reflexive:
For any (a \in A), we have (a=a). Hence (aRa).
So (R) is reflexive.

Symmetric:
Suppose (aRb). Then (a=b). Therefore (b=a), which gives (bRa).
So (R) is symmetric.

Transitive:
Suppose (aRb) and (bRc). Then (a=b) and (b=c). Hence (a=c), so (aRc).
So (R) is transitive.

Since equality is reflexive, symmetric and transitive, it is an equivalence relation. (\square)

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Option 2: Bijective Map

Definition.
A function (f: A \to B) is called:

• Injective (one-one) if (f(a_1)=f(a_2) \Rightarrow a_1=a_2) for all (a_1,a_2 \in A).
• Surjective (onto) if for every (b \in B), there exists (a \in A) such that (f(a)=b).
• Bijective if it is both injective and surjective.

A function (g: B \to A) is called the inverse of (f) if
[ g \circ f = \mathrm{id}_A \quad \text{and} \quad f \circ g = \mathrm{id}_B. ]

Theorem.
A function (f: A \to B) has an inverse if and only if it is bijective.

Proof.

((\Rightarrow)) Suppose (f) has an inverse (g: B \to A).
Then (g \circ f = \mathrm{id}_A) and (f \circ g = \mathrm{id}_B).

Injectivity: Let (f(a_1)=f(a_2)). Applying (g),
[ g(f(a_1))=g(f(a_2)) \Rightarrow a_1=a_2. ] So (f) is injective.

Surjectivity: Let (b \in B). Put (a=g(b)). Then
[ f(a)=f(g(b))=(f \circ g)(b)=b. ] So every (b \in B) has a preimage. Thus (f) is surjective.

Hence (f) is bijective.

((\Leftarrow)) Suppose (f) is bijective.
Define (g: B \to A) by
[ g(b)=a, \quad \text{where } f(a)=b. ] This is well-defined because:

• (f) is surjective, so for every (b \in B) there exists at least one such (a).
• (f) is injective, so this (a) is unique.

Now for any (a \in A), let (f(a)=b). Then by definition (g(b)=a), so
[ g(f(a))=a. ] Thus (g \circ f = \mathrm{id}_A).

For any (b \in B), let (g(b)=a). Then (f(a)=b), so
[ f(g(b))=b. ] Thus (f \circ g = \mathrm{id}_B).

Therefore (g) is an inverse of (f). Hence (f) has an inverse if and only if it is bijective. (\square)

Given system: $x+2y-z=3,\quad 3x-y+2z=1,\quad 2x-2y+3z=2$

Write in matrix form:

$$\begin{bmatrix} 1 & 2 & -1\\ 3 & -1 & 2\\ 2 & -2 & 3 \end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix} = \begin{bmatrix} 3\\1\\2 \end{bmatrix}$$

Test for consistency:

$$D= \begin{vmatrix} 1 & 2 & -1\\ 3 & -1 & 2\\ 2 & -2 & 3 \end{vmatrix}$$ $$=1[(-1)(3)-2(-2)]-2[(3)(3)-2(2)]+(-1)[(3)(-2)-(-1)(2)]$$ $$=1[-3+4]-2[9-4]-1[-6+2] =1-10+4=-5$$

Since (D=-5\neq 0), the coefficient matrix has rank (3), equal to the number of unknowns.
Hence the system is consistent and has a unique solution.

Solve by Cramer’s rule:

$$D_x= \begin{vmatrix} 3 & 2 & -1\\ 1 & -1 & 2\\ 2 & -2 & 3 \end{vmatrix} =5$$ $$D_y= \begin{vmatrix} 1 & 3 & -1\\ 3 & 1 & 2\\ 2 & 2 & 3 \end{vmatrix} =-20$$ $$D_z= \begin{vmatrix} 1 & 2 & 3\\ 3 & -1 & 1\\ 2 & -2 & 2 \end{vmatrix} =-20$$

Therefore,

$$x=\frac{D_x}{D}=\frac{5}{-5}=-1$$ $$y=\frac{D_y}{D}=\frac{-20}{-5}=4$$ $$z=\frac{D_z}{D}=\frac{-20}{-5}=4$$

Hence the solution is:

$$\boxed{x=-1,\quad y=4,\quad z=4}$$

The system is consistent and has the unique solution ((-1,4,4)).

Answer (First alternative)

Scalar projection

Let $\vec a$ and $\vec b$ be two non-zero vectors and let $\theta$ be the angle between them. The scalar projection (or component) of $\vec a$ on $\vec b$ is the scalar

$\text{comp}_{\vec b}\,\vec a=\frac{\vec a\cdot\vec b}{|\vec b|}=|\vec a|\cos\theta .$

Geometrical interpretation. Drop a perpendicular from the terminal point $A$ of $\vec a$ onto the line containing $\vec b$, and let the foot of this perpendicular be $M$. Then the scalar projection is precisely the directed length of the segment $OM$ measured along the direction of $\vec b$. In other words, it is the length of the "shadow" (projection) of $\vec a$ on the line of $\vec b$, taken with a sign:

• $\text{comp}_{\vec b}\vec a>0$ when $\theta$ is acute (the shadow falls along $\vec b$),
• $\text{comp}_{\vec b}\vec a<0$ when $\theta$ is obtuse (the shadow falls opposite to $\vec b$),
• $\text{comp}_{\vec b}\vec a=0$ when $\vec a\perp\vec b$ (the shadow shrinks to a point).

Vector projection

The vector projection of $\vec a$ on $\vec b$ is the vector

$\text{proj}_{\vec b}\,\vec a=\left(\frac{\vec a\cdot\vec b}{|\vec b|^{2}}\right)\vec b=(\vec a\cdot\hat b)\,\hat b ,\qquad \hat b=\frac{\vec b}{|\vec b|}.$

Geometrical interpretation. It is the vector $\overrightarrow{OM}$ lying along the line of $\vec b$ whose length is the scalar projection $\text{comp}_{\vec b}\vec a$ and whose direction is that of $\vec b$ or opposite to it according as the scalar projection is positive or negative. Thus the vector projection is the component of $\vec a$ parallel to $\vec b$, and the remaining part

$\vec r=\vec a-\text{proj}_{\vec b}\,\vec a$

is the component of $\vec a$ perpendicular to $\vec b$, so that $\vec a=\text{proj}_{\vec b}\vec a+\vec r$ with $\vec r\cdot\vec b=0$. This is exactly the resolution of $\vec a$ into two perpendicular components along and normal to $\vec b$.

Note. The projection of $\vec a$ on $\vec b$ and of $\vec b$ on $\vec a$ are generally different, since $\text{comp}_{\vec b}\vec a=|\vec a|\cos\theta$ while $\text{comp}_{\vec a}\vec b=|\vec b|\cos\theta$.

Example. For $\vec a=2\hat i+3\hat j$ and $\vec b=\hat i+\hat j$: $\vec a\cdot\vec b=5,\quad |\vec b|=\sqrt2,\quad \text{comp}_{\vec b}\vec a=\frac{5}{\sqrt2},$ $\text{proj}_{\vec b}\vec a=\frac{5}{2}(\hat i+\hat j).$

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OR (Second alternative)

Theorem

A set of non-zero vectors $\{v_1,v_2,\dots,v_n\}$ of a vector space $V$ over a field $F$ is linearly dependent iff one of the vectors can be expressed as a linear combination of the others.

Proof

Condition necessary. Suppose $\{v_1,v_2,\dots,v_n\}$ is linearly dependent. Then there exist scalars $c_1,c_2,\dots,c_n\in F$, not all zero, such that

$c_1v_1+c_2v_2+\cdots+c_nv_n=0. \quad (1)$

Since not all the $c_i$ are zero, choose an index $k$ for which $c_k\neq 0$. Then from (1),

$c_kv_k=-(c_1v_1+\cdots+c_{k-1}v_{k-1}+c_{k+1}v_{k+1}+\cdots+c_nv_n),$

and dividing by $c_k\neq 0$,

$v_k=\sum_{i\neq k}\left(-\frac{c_i}{c_k}\right)v_i .$

Thus $v_k$ is a linear combination of the remaining vectors $v_1,\dots,v_{k-1},v_{k+1},\dots,v_n$.

Condition sufficient. Conversely, suppose one of the vectors, say $v_k$, is a linear combination of the others:

$v_k=a_1v_1+\cdots+a_{k-1}v_{k-1}+a_{k+1}v_{k+1}+\cdots+a_nv_n .$

Transposing all terms to the left,

$a_1v_1+\cdots+a_{k-1}v_{k-1}+(-1)v_k+a_{k+1}v_{k+1}+\cdots+a_nv_n=0.$

This is a linear relation among $v_1,\dots,v_n$ in which the coefficient of $v_k$ is $-1\neq 0$. Hence not all the coefficients are zero, and therefore the set $\{v_1,v_2,\dots,v_n\}$ is linearly dependent. $\blacksquare$

Remark. The hypothesis that the vectors are non-zero is not actually needed for the argument above; it is retained because in the usual treatment it guarantees that a single vector $v_1\neq 0$ forms an independent set, so that dependence can arise only when $n\ge 2$.

Note on notation. Since (G:W\to U), its matrix must use the basis (B') of (W) as the domain basis and (B'') of (U) as the codomain basis. Thus the correct statement is [ M_{B''}^{B'}(G),M_{B'}^{B}(F)=M_{B''}^{B}(G\circ F). ]

Let [ B={v_1,v_2,\dots,v_n},\qquad B'={w_1,w_2,\dots,w_m},\qquad B''={u_1,u_2,\dots,u_p}. ] Define [ M_{B'}^{B}(F)=(a_{ij}){m\times n},\qquad M{B''}^{B'}(G)=(c_{ki}){p\times m}. ] By definition of the matrix of a linear map, [ F(v_j)=\sum{i=1}^{m} a_{ij}w_i,\qquad j=1,2,\dots,n, ] and [ G(w_i)=\sum_{k=1}^{p} c_{ki}u_k,\qquad i=1,2,\dots,m. ]

Now compute the image of the basis vector (v_j) under (G\circ F): [ (G\circ F)(v_j)=G(F(v_j)) =G\left(\sum_{i=1}^{m}a_{ij}w_i\right). ] Using linearity of (G), [ (G\circ F)(v_j) =\sum_{i=1}^{m}a_{ij}G(w_i) =\sum_{i=1}^{m}a_{ij}\left(\sum_{k=1}^{p}c_{ki}u_k\right). ] Interchanging the order of summation, [ (G\circ F)(v_j) =\sum_{k=1}^{p}\left(\sum_{i=1}^{m}c_{ki}a_{ij}\right)u_k. ]

Thus the coefficient of (u_k) in the expression of ((G\circ F)(v_j)) with respect to the basis (B'') is [ \sum_{i=1}^{m}c_{ki}a_{ij}. ] But this is exactly the ((k,j))-entry of the matrix product [ M_{B''}^{B'}(G),M_{B'}^{B}(F). ] Therefore the matrix of (G\circ F) with respect to the bases (B) and (B'') is [ M_{B''}^{B}(G\circ F)

M_{B''}^{B'}(G),M_{B'}^{B}(F). ]

Hence, [ \boxed{M_{B''}^{B'}(G),M_{B'}^{B}(F)=M_{B''}^{B}(G\circ F)}. ]

So the matrix of a composite linear map is the product of the matrices of the individual maps, taken in the correct order.

Option 1: Eigenvalues, Eigenvectors and Linear Independence

Definitions

Let (V) be a vector space over a field (K), and let (A:V\to V) be a linear map.

A scalar (\lambda\in K) is called an eigenvalue of (A) if there exists a non-zero vector (v\in V) such that
[ A(v)=\lambda v. ] The non-zero vector (v) is called an eigenvector of (A) corresponding to the eigenvalue (\lambda).

Equivalently, (\lambda) is an eigenvalue of (A) if and only if
[ \ker(A-\lambda I)\neq {0}, ] where (I) is the identity map on (V). The subspace
[ E_\lambda=\ker(A-\lambda I)={v\in V:A(v)=\lambda v} ] is called the eigenspace corresponding to (\lambda).

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Theorem

Let (v_1,v_2,\ldots,v_n) be eigenvectors of (A) with corresponding eigenvalues
[ \lambda_1,\lambda_2,\ldots,\lambda_n ] such that (\lambda_i\neq \lambda_j) whenever (i\neq j). Then (v_1,v_2,\ldots,v_n) are linearly independent.

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Proof

We prove the result by induction on (n).

Base case: For (n=1), the eigenvector (v_1) is non-zero by definition. Hence ({v_1}) is linearly independent.

Inductive step: Assume the statement is true for any set of (n-1) eigenvectors corresponding to distinct eigenvalues. Now consider (n) eigenvectors (v_1,\ldots,v_n) with distinct eigenvalues (\lambda_1,\ldots,\lambda_n).

Suppose there exist scalars (c_1,c_2,\ldots,c_n\in K) such that
[ c_1v_1+c_2v_2+\cdots+c_nv_n=0. \quad (1) ]

Apply the linear map (A) to both sides of (1): [ c_1A(v_1)+c_2A(v_2)+\cdots+c_nA(v_n)=0. ] Since (A(v_i)=\lambda_i v_i), we get
[ c_1\lambda_1v_1+c_2\lambda_2v_2+\cdots+c_n\lambda_nv_n=0. \quad (2) ]

Multiply equation (1) by (\lambda_n): [ c_1\lambda_nv_1+c_2\lambda_nv_2+\cdots+c_{n-1}\lambda_nv_{n-1}+c_n\lambda_nv_n=0. \quad (3) ]

Subtract (3) from (2): [ c_1(\lambda_1-\lambda_n)v_1+c_2(\lambda_2-\lambda_n)v_2+\cdots+c_{n-1}(\lambda_{n-1}-\lambda_n)v_{n-1}=0. ]

By the induction hypothesis, the vectors (v_1,v_2,\ldots,v_{n-1}) are linearly independent. Therefore, [ c_i(\lambda_i-\lambda_n)=0 \quad \text{for } i=1,2,\ldots,n-1. ] Since the eigenvalues are distinct, (\lambda_i-\lambda_n\neq 0). Hence
[ c_i=0 \quad \text{for } i=1,2,\ldots,n-1. ]

Substituting these values into (1), we obtain
[ c_nv_n=0. ] But (v_n\neq 0), so (c_n=0).

Thus all the scalars (c_1,c_2,\ldots,c_n) are zero. Therefore (v_1,v_2,\ldots,v_n) are linearly independent. By induction, the theorem holds for all (n). (\square)

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OR

Option 2: Solve (x^3-12x-8=0) by Cardan’s Method

The given equation is
[ x^3-12x-8=0. ]

Compare it with the standard cubic
[ x^3+px+q=0. ] Here, [ p=-12,\qquad q=-8. ]

Put
[ x=u+v. ] Then [ (u+v)^3-12(u+v)-8=0. ] Expanding, [ u^3+v^3+3uv(u+v)-12(u+v)-8=0, ] so [ u^3+v^3+(3uv-12)(u+v)-8=0. ]

Choose (u) and (v) such that
[ 3uv-12=0 \quad \Rightarrow \quad uv=4. ] Then the equation reduces to
[ u^3+v^3=8. ]

Let
[ U=u^3,\qquad V=v^3. ] Then [ U+V=8,\qquad UV=(uv)^3=4^3=64. ] Hence (U) and (V) are the roots of
[ t^2-8t+64=0. ] Solving, [ t=\frac{8\pm\sqrt{64-256}}{2} =\frac{8\pm\sqrt{-192}}{2} =\frac{8\pm 8\sqrt{3},i}{2} =4\pm 4\sqrt{3},i. ]

Thus, [ u^3=4+4\sqrt{3},i=8\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)=8e^{i\pi/3}, ] [ v^3=4-4\sqrt{3},i=8\left(\cos\frac{\pi}{3}-i\sin\frac{\pi}{3}\right)=8e^{-i\pi/3}. ]

Taking cube roots, [ u=2e^{i(\pi/9+2k\pi/3)},\qquad v=2e^{-i(\pi/9+2k\pi/3)},\qquad k=0,1,2. ] Therefore, [ x=u+v=4\cos\left(\frac{\pi}{9}+\frac{2k\pi}{3}\right),\qquad k=0,1,2. ]

Hence the three roots are
[ x_1=4\cos\frac{\pi}{9}, ] [ x_2=4\cos\frac{7\pi}{9}, ] [ x_3=4\cos\frac{13\pi}{9}. ]

Numerically, [ x_1\approx 3.759,\qquad x_2\approx -3.064,\qquad x_3\approx -0.695. ]

Thus the required roots are
[ \boxed{x=4\cos\frac{\pi}{9},;4\cos\frac{7\pi}{9},;4\cos\frac{13\pi}{9}}. ]

Let
[ \vec l=(l_1,l_2,l_3),\quad \vec m=(m_1,m_2,m_3),\quad \vec n=(n_1,n_2,n_3) ] and
[ \vec a=(a_1,a_2,a_3),\quad \vec b=(b_1,b_2,b_3),\quad \vec c=(c_1,c_2,c_3). ]

Define the matrices [ L=\begin{pmatrix} l_1&l_2&l_3\ m_1&m_2&m_3\ n_1&n_2&n_3 \end{pmatrix}, \qquad A=\begin{pmatrix} a_1&a_2&a_3\ b_1&b_2&b_3\ c_1&c_2&c_3 \end{pmatrix}. ]

By the definition of the scalar triple product, [ [\vec l\ \vec m\ \vec n]=\vec l\cdot(\vec m\times \vec n)=\det L ] and [ [\vec a\ \vec b\ \vec c]=\vec a\cdot(\vec b\times \vec c)=\det A. ]

Now consider the product (LA^T). Since [ A^T=\begin{pmatrix} a_1&b_1&c_1\ a_2&b_2&c_2\ a_3&b_3&c_3 \end{pmatrix}, ] we get [ LA^T= \begin{pmatrix} \vec l\cdot\vec a&\vec l\cdot\vec b&\vec l\cdot\vec c\ \vec m\cdot\vec a&\vec m\cdot\vec b&\vec m\cdot\vec c\ \vec n\cdot\vec a&\vec n\cdot\vec b&\vec n\cdot\vec c \end{pmatrix}. ]

Taking determinants on both sides, [ \det(LA^T)= \begin{vmatrix} \vec l\cdot\vec a&\vec l\cdot\vec b&\vec l\cdot\vec c\ \vec m\cdot\vec a&\vec m\cdot\vec b&\vec m\cdot\vec c\ \vec n\cdot\vec a&\vec n\cdot\vec b&\vec n\cdot\vec c \end{vmatrix}. ]

But by the multiplicative property of determinants, [ \det(LA^T)=\det L\det A^T=\det L\det A. ] Hence [ \det L\det A= \begin{vmatrix} \vec l\cdot\vec a&\vec l\cdot\vec b&\vec l\cdot\vec c\ \vec m\cdot\vec a&\vec m\cdot\vec b&\vec m\cdot\vec c\ \vec n\cdot\vec a&\vec n\cdot\vec b&\vec n\cdot\vec c \end{vmatrix}. ]

Therefore, [ [\vec l\ \vec m\ \vec n][\vec a\ \vec b\ \vec c]

\begin{vmatrix} \vec l\cdot\vec a&\vec l\cdot\vec b&\vec l\cdot\vec c\ \vec m\cdot\vec a&\vec m\cdot\vec b&\vec m\cdot\vec c\ \vec n\cdot\vec a&\vec n\cdot\vec b&\vec n\cdot\vec c \end{vmatrix}. ]

Thus proved.

Question: Show that the necessary and sufficient condition for the vector function $\vec{f}$ of a scalar variable $t$ to have a constant direction is $\vec{f} \times \dfrac{d\vec{f}}{dt} = 0$.

Solution:

Let $\vec{f}(t)$ be a vector function of the scalar variable $t$.

Condition for constant direction: A vector function $\vec{f}(t)$ has a constant direction if its direction (unit vector) does not change with $t$. This means $\vec{f}(t)$ remains parallel to a fixed direction at all values of $t$, i.e., $\vec{f}$ and $\dfrac{d\vec{f}}{dt}$ are parallel (collinear) at every $t$.

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Part 1: Necessary condition

Suppose $\vec{f}(t)$ has a constant direction. Then we can write $\vec{f}(t) = \lambda(t)\,\hat{a}$ where $\hat{a}$ is a fixed unit vector (constant direction) and $\lambda(t)$ is a scalar function of $t$.

Differentiating with respect to $t$: $\frac{d\vec{f}}{dt} = \frac{d\lambda}{dt}\,\hat{a}$

Now, $\vec{f} \times \frac{d\vec{f}}{dt} = \big(\lambda\,\hat{a}\big) \times \left(\frac{d\lambda}{dt}\,\hat{a}\right) = \lambda\,\frac{d\lambda}{dt}\,(\hat{a} \times \hat{a}) = 0$

since $\hat{a} \times \hat{a} = 0$.

Hence, $\vec{f} \times \dfrac{d\vec{f}}{dt} = 0$ is necessary.

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Part 2: Sufficient condition

Conversely, suppose $\vec{f} \times \frac{d\vec{f}}{dt} = 0$

If $\vec{f} = \vec{0}$, the direction is not defined, so assume $\vec{f} \neq \vec{0}$. Then the above equation implies that $\vec{f}$ and $\dfrac{d\vec{f}}{dt}$ are parallel (collinear) vectors. Hence there exists a scalar $\mu(t)$ such that $\frac{d\vec{f}}{dt} = \mu(t)\,\vec{f}$

Let $\hat{u} = \dfrac{\vec{f}}{|\vec{f}|}$ be the unit vector along $\vec{f}$. Then $\vec{f} = |\vec{f}|\,\hat{u}$.

Differentiating: $\frac{d\vec{f}}{dt} = \frac{d|\vec{f}|}{dt}\,\hat{u} + |\vec{f}|\,\frac{d\hat{u}}{dt}$

But we also have $\dfrac{d\vec{f}}{dt} = \mu\,\vec{f} = \mu\,|\vec{f}|\,\hat{u}$.

Comparing the two expressions: $\frac{d|\vec{f}|}{dt}\,\hat{u} + |\vec{f}|\,\frac{d\hat{u}}{dt} = \mu\,|\vec{f}|\,\hat{u}$

Rearranging: $|\vec{f}|\,\frac{d\hat{u}}{dt} = \left(\mu\,|\vec{f}| - \frac{d|\vec{f}|}{dt}\right)\hat{u}$

Since $\hat{u} \cdot \dfrac{d\hat{u}}{dt} = 0$ (because $|\hat{u}| = 1$), taking the dot product with $\hat{u}$ on both sides: $|\vec{f}|\left(\hat{u} \cdot \frac{d\hat{u}}{dt}\right) = \left(\mu\,|\vec{f}| - \frac{d|\vec{f}|}{dt}\right)(\hat{u} \cdot \hat{u})$ $0 = \mu\,|\vec{f}| - \frac{d|\vec{f}|}{dt}$

Thus, $\frac{d\hat{u}}{dt} = 0$

which means $\hat{u}$ is a constant vector. Therefore, the direction of $\vec{f}(t)$ is constant.

Hence, $\vec{f} \times \dfrac{d\vec{f}}{dt} = 0$ is sufficient.

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Conclusion: Combining both parts, the necessary and sufficient condition for the vector function $\vec{f}$ of a scalar variable $t$ to have a constant direction is $\boxed{\vec{f} \times \frac{d\vec{f}}{dt} = 0}$

Let the position vector be
[ \vec r=x\hat i+y\hat j+z\hat k ] and let (\phi,\psi) be scalar point functions of (x,y,z), while (\vec a,\vec b) are vector point functions of (x,y,z).

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i) Prove that (\operatorname{grad}(\phi\pm\psi)=\operatorname{grad}\phi\pm\operatorname{grad}\psi)

By definition, [ \operatorname{grad}\phi=\nabla\phi =\hat i\frac{\partial\phi}{\partial x} +\hat j\frac{\partial\phi}{\partial y} +\hat k\frac{\partial\phi}{\partial z} ] and [ \operatorname{grad}\psi=\nabla\psi =\hat i\frac{\partial\psi}{\partial x} +\hat j\frac{\partial\psi}{\partial y} +\hat k\frac{\partial\psi}{\partial z}. ]

Now, [ \operatorname{grad}(\phi\pm\psi)=\nabla(\phi\pm\psi) ] [ =\hat i\frac{\partial}{\partial x}(\phi\pm\psi) +\hat j\frac{\partial}{\partial y}(\phi\pm\psi) +\hat k\frac{\partial}{\partial z}(\phi\pm\psi). ]

Since partial differentiation is linear, [ \frac{\partial}{\partial x}(\phi\pm\psi) =\frac{\partial\phi}{\partial x}\pm\frac{\partial\psi}{\partial x}, ] and similarly for (y) and (z). Therefore, [ \operatorname{grad}(\phi\pm\psi)

\hat i\left(\frac{\partial\phi}{\partial x}\pm\frac{\partial\psi}{\partial x}\right) + \hat j\left(\frac{\partial\phi}{\partial y}\pm\frac{\partial\psi}{\partial y}\right) + \hat k\left(\frac{\partial\phi}{\partial z}\pm\frac{\partial\psi}{\partial z}\right). ]

Separating the terms, [ \operatorname{grad}(\phi\pm\psi)

\left( \hat i\frac{\partial\phi}{\partial x} +\hat j\frac{\partial\phi}{\partial y} +\hat k\frac{\partial\phi}{\partial z} \right) \pm \left( \hat i\frac{\partial\psi}{\partial x} +\hat j\frac{\partial\psi}{\partial y} +\hat k\frac{\partial\psi}{\partial z} \right). ]

Hence, [ \boxed{\operatorname{grad}(\phi\pm\psi)=\operatorname{grad}\phi\pm\operatorname{grad}\psi} ]

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ii) Prove that (\operatorname{Curl}(\vec a\pm\vec b)=\operatorname{Curl}\vec a\pm\operatorname{Curl}\vec b)

Let [ \vec a=a_1\hat i+a_2\hat j+a_3\hat k, \qquad \vec b=b_1\hat i+b_2\hat j+b_3\hat k. ]

By definition, [ \operatorname{Curl}\vec a=\nabla\times\vec a

\begin{vmatrix} \hat i & \hat j & \hat k\[2mm] \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z}\[2mm] a_1 & a_2 & a_3 \end{vmatrix}. ]

Expanding, [ \operatorname{Curl}\vec a

\hat i\left(\frac{\partial a_3}{\partial y}-\frac{\partial a_2}{\partial z}\right) + \hat j\left(\frac{\partial a_1}{\partial z}-\frac{\partial a_3}{\partial x}\right) + \hat k\left(\frac{\partial a_2}{\partial x}-\frac{\partial a_1}{\partial y}\right). ]

Similarly, [ \operatorname{Curl}\vec b

\hat i\left(\frac{\partial b_3}{\partial y}-\frac{\partial b_2}{\partial z}\right) + \hat j\left(\frac{\partial b_1}{\partial z}-\frac{\partial b_3}{\partial x}\right) + \hat k\left(\frac{\partial b_2}{\partial x}-\frac{\partial b_1}{\partial y}\right). ]

Now, [ \vec a\pm\vec b

(a_1\pm b_1)\hat i+(a_2\pm b_2)\hat j+(a_3\pm b_3)\hat k. ]

Therefore, [ \operatorname{Curl}(\vec a\pm\vec b)

\nabla\times(\vec a\pm\vec b) ] [

\begin{vmatrix} \hat i & \hat j & \hat k\[2mm] \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z}\[2mm] a_1\pm b_1 & a_2\pm b_2 & a_3\pm b_3 \end{vmatrix}. ]

Expanding, [ \operatorname{Curl}(\vec a\pm\vec b)

\hat i\left[ \frac{\partial(a_3\pm b_3)}{\partial y}

\frac{\partial(a_2\pm b_2)}{\partial z} \right] ] [ + \hat j\left[ \frac{\partial(a_1\pm b_1)}{\partial z}

\frac{\partial(a_3\pm b_3)}{\partial x} \right] ] [ + \hat k\left[ \frac{\partial(a_2\pm b_2)}{\partial x}

\frac{\partial(a_1\pm b_1)}{\partial y} \right]. ]

Using linearity of partial derivatives, [ \frac{\partial(a_3\pm b_3)}{\partial y}

\frac{\partial a_3}{\partial y}\pm\frac{\partial b_3}{\partial y}, ] and similarly for all other terms. Hence, [ \operatorname{Curl}(\vec a\pm\vec b)

\hat i\left[ \left(\frac{\partial a_3}{\partial y}-\frac{\partial a_2}{\partial z}\right) \pm \left(\frac{\partial b_3}{\partial y}-\frac{\partial b_2}{\partial z}\right) \right] ] [ + \hat j\left[ \left(\frac{\partial a_1}{\partial z}-\frac{\partial a_3}{\partial x}\right) \pm \left(\frac{\partial b_1}{\partial z}-\frac{\partial b_3}{\partial x}\right) \right] ] [ + \hat k\left[ \left(\frac{\partial a_2}{\partial x}-\frac{\partial a_1}{\partial y}\right) \pm \left(\frac{\partial b_2}{\partial x}-\frac{\partial b_1}{\partial y}\right) \right]. ]

Separating the terms, [ \operatorname{Curl}(\vec a\pm\vec b)

\left[ \hat i\left(\frac{\partial a_3}{\partial y}-\frac{\partial a_2}{\partial z}\right) + \hat j\left(\frac{\partial a_1}{\partial z}-\frac{\partial a_3}{\partial x}\right) + \hat k\left(\frac{\partial a_2}{\partial x}-\frac{\partial a_1}{\partial y}\right) \right] ] [ \pm \left[ \hat i\left(\frac{\partial b_3}{\partial y}-\frac{\partial b_2}{\partial z}\right) + \hat j\left(\frac{\partial b_1}{\partial z}-\frac{\partial b_3}{\partial x}\right) + \hat k\left(\frac{\partial b_2}{\partial x}-\frac{\partial b_1}{\partial y}\right) \right]. ]

Thus, [ \operatorname{Curl}(\vec a\pm\vec b)

\operatorname{Curl}\vec a\pm\operatorname{Curl}\vec b. ]

Hence, [ \boxed{\operatorname{Curl}(\vec a\pm\vec b)=\operatorname{Curl}\vec a\pm\operatorname{Curl}\vec b} ]

Both results follow from the linearity of the differential operator (\nabla).

Theorem (Sylvester’s law of inertia).
Let (V) be a finite-dimensional real vector space with a symmetric bilinear form (\langle,\rangle). Then there exists a unique integer (r\ge 0) such that for every orthogonal basis ({v_1,v_2,\dots,v_n}) of (V), exactly (r) of the numbers [ \langle v_1,v_1\rangle,\ \langle v_2,v_2\rangle,\ \dots,\ \langle v_n,v_n\rangle ] are positive.

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Proof

First note that an orthogonal basis always exists. We prove this by induction on (n=\dim V).

If (n=0), there is nothing to prove. Suppose (n>0). If the form is identically zero on (V), then every basis is orthogonal. Otherwise, there exist (u,w\in V) such that (\langle u,w\rangle\ne 0). Then [ \langle u+w,u+w\rangle=2\langle u,w\rangle,\qquad \langle u-w,u-w\rangle=-2\langle u,w\rangle. ] Hence at least one of (u+w) or (u-w) has nonzero scalar product with itself. So we can choose (v\in V) such that (\langle v,v\rangle\ne 0).

Let [ U=v^\perp={x\in V:\langle x,v\rangle=0}. ] Then (V=\operatorname{span}(v)\oplus U), because for any (x\in V), [ x-\frac{\langle x,v\rangle}{\langle v,v\rangle}v\in U. ] By the induction hypothesis, (U) has an orthogonal basis. Adding (v) to that basis gives an orthogonal basis of (V). Thus orthogonal bases exist.

Now let [ B={v_1,\dots,v_n},\qquad B'={w_1,\dots,w_n} ] be two orthogonal bases of (V). Let [ p=#{i:\langle v_i,v_i\rangle>0},\qquad p'=#{i:\langle w_i,w_i\rangle>0}. ] We prove (p=p').

Let [ A=\operatorname{span}{v_i:\langle v_i,v_i\rangle>0}. ] Then (\dim A=p). For any nonzero (x\in A), write [ x=\sum_{\langle v_i,v_i\rangle>0} a_i v_i. ] Since the basis is orthogonal, [ \langle x,x\rangle=\sum a_i^2\langle v_i,v_i\rangle>0. ] So the form is positive definite on (A).

Similarly, let [ C=\operatorname{span}{w_i:\langle w_i,w_i\rangle\le 0}. ] Then (\dim C=n-p'). For any (x\in C), [ \langle x,x\rangle=\sum_{\langle w_i,w_i\rangle\le 0} a_i^2\langle w_i,w_i\rangle\le 0. ] So the form is non-positive on (C).

If (x\in A\cap C), then from (x\in A) we get (\langle x,x\rangle>0) unless (x=0), while from (x\in C) we get (\langle x,x\rangle\le 0). Hence (x=0). Therefore [ A\cap C={0}. ] Thus [ \dim A+\dim C\le \dim V, ] so [ p+(n-p')\le n, ] which gives [ p\le p'. ] Interchanging the roles of (B) and (B'), we similarly get [ p'\le p. ] Hence [ p=p'. ]

Therefore the number of positive values (\langle v_i,v_i\rangle) is the same for every orthogonal basis. Denoting this common number by (r), we conclude that there exists a unique integer (r\ge 0) such that every orthogonal basis of (V) has exactly (r) vectors satisfying [ \langle v_i,v_i\rangle>0. ]

[ \boxed{\text{Hence proved.}} ]

Given
[ A^{-1} = B = \begin{bmatrix} -3 & \frac{13}{2} & \frac{5}{2} \[2mm] 3 & -5 & -2 \[2mm] \frac{1}{2} & \frac{3}{2} & -1 \end{bmatrix} ] We need (A = B^{-1}).

Write (B = \frac12 C), where
[ C = \begin{bmatrix} -6 & 13 & 5 \ 6 & -10 & -4 \ 1 & 3 & -2 \end{bmatrix} ] Then [ B^{-1} = \left(\frac12 C\right)^{-1} = 2,C^{-1}. ]

Step 1: Determinant of (C)
[ \det(C) = -6\begin{vmatrix}-10&-4\3&-2\end{vmatrix} -13\begin{vmatrix}6&-4\1&-2\end{vmatrix} +5\begin{vmatrix}6&-10\1&3\end{vmatrix} ] [ = -6(20+12) -13(-12+4) +5(18+10) ] [ = -6(32) -13(-8) +5(28) = -192 +104 +140 = 52. ]

Step 2: Cofactors of (C)
[ \begin{aligned} C_{11}&=32, & C_{12}&=8, & C_{13}&=28,\ C_{21}&=41, & C_{22}&=7, & C_{23}&=31,\ C_{31}&=-2, & C_{32}&=6, & C_{33}&=-18. \end{aligned} ] So the cofactor matrix is
[ \begin{bmatrix} 32 & 8 & 28 \ 41 & 7 & 31 \ -2 & 6 & -18 \end{bmatrix} ] and its adjugate (transpose) is
[ \operatorname{adj}(C) = \begin{bmatrix} 32 & 41 & -2 \ 8 & 7 & 6 \ 28 & 31 & -18 \end{bmatrix}. ]

Step 3: Inverse of (C)
[ C^{-1} = \frac{1}{\det(C)}\operatorname{adj}(C) = \frac{1}{52}\begin{bmatrix} 32 & 41 & -2 \ 8 & 7 & 6 \ 28 & 31 & -18 \end{bmatrix}. ]

Step 4: Inverse of (B)
[ A = B^{-1} = 2C^{-1} = \frac{1}{26}\begin{bmatrix} 32 & 41 & -2 \ 8 & 7 & 6 \ 28 & 31 & -18 \end{bmatrix}. ]

Thus
[ \boxed{ A = \begin{bmatrix} \dfrac{16}{13} & \dfrac{41}{26} & -\dfrac{1}{13} \[3mm] \dfrac{4}{13} & \dfrac{7}{26} & \dfrac{3}{13} \[3mm] \dfrac{14}{13} & \dfrac{31}{26} & -\dfrac{9}{13} \end{bmatrix} } ]

Verification:
[ B A = \begin{bmatrix} -3 & \frac{13}{2} & \frac{5}{2} \ 3 & -5 & -2 \ \frac{1}{2} & \frac{3}{2} & -1 \end{bmatrix} \begin{bmatrix} \frac{16}{13} & \frac{41}{26} & -\frac{1}{13} \ \frac{4}{13} & \frac{7}{26} & \frac{3}{13} \ \frac{14}{13} & \frac{31}{26} & -\frac{9}{13} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} ] so the result is correct.

Given

$$A=\begin{bmatrix} 3&-2&0&-1&-1\\ 0&2&2&1&-5\\ 1&-2&-3&-2&-1\\ 0&1&1&2&6 \end{bmatrix}$$

Reduce to echelon form by elementary row operations.

Swap (R_1) and (R_3):

$$A\sim \begin{bmatrix} 1&-2&-3&-2&-1\\ 0&2&2&1&-5\\ 3&-2&0&-1&-1\\ 0&1&1&2&6 \end{bmatrix}$$

Apply (R_3\to R_3-3R_1):

$$A\sim \begin{bmatrix} 1&-2&-3&-2&-1\\ 0&2&2&1&-5\\ 0&4&9&5&2\\ 0&1&1&2&6 \end{bmatrix}$$

Swap (R_2) and (R_4):

$$A\sim \begin{bmatrix} 1&-2&-3&-2&-1\\ 0&1&1&2&6\\ 0&4&9&5&2\\ 0&2&2&1&-5 \end{bmatrix}$$

Apply (R_3\to R_3-4R_2):

$$A\sim \begin{bmatrix} 1&-2&-3&-2&-1\\ 0&1&1&2&6\\ 0&0&5&-3&-22\\ 0&2&2&1&-5 \end{bmatrix}$$

Apply (R_4\to R_4-2R_2):

$$A\sim \begin{bmatrix} 1&-2&-3&-2&-1\\ 0&1&1&2&6\\ 0&0&5&-3&-22\\ 0&0&0&-3&-17 \end{bmatrix}$$

This is in echelon form. The number of non-zero rows is (4).

[ \boxed{\text{Rank of }A=4} ]

Given
[ A=(1,2,3),\qquad B=(-1,2,-3) ]

First find the dot product: [ A\cdot B=1(-1)+2(2)+3(-3)=-1+4-9=-6 ]

Magnitude of (A): [ |A|=\sqrt{1^2+2^2+3^2}=\sqrt{14} ]

Scalar projection of (B) onto (A)

[ \text{comp}_A B=\frac{A\cdot B}{|A|} ] [ =\frac{-6}{\sqrt{14}} =-\frac{6}{\sqrt{14}} =-\frac{3\sqrt{14}}{7} ]

Vector projection of (B) onto (A)

[ \text{proj}_A B=\left(\frac{A\cdot B}{|A|^2}\right)A ] [ =\left(\frac{-6}{14}\right)(1,2,3) ] [ =-\frac{3}{7}(1,2,3) ] [ =\left(-\frac{3}{7},-\frac{6}{7},-\frac{9}{7}\right) ]

Hence, [ \boxed{\text{Scalar projection}=-\frac{3\sqrt{14}}{7}} ] [ \boxed{\text{Vector projection}=\left(-\frac{3}{7},-\frac{6}{7},-\frac{9}{7}\right)} ]

The negative sign shows that (B) makes an obtuse angle with (A), so the vector projection points opposite to (A).

Solution:

Given equation: $x^4 - 2x^2 + 8x - 3 = 0$

It has no $x^3$ term. By Descartes’ method, assume it can be resolved into two quadratic factors of the form $(x^2 + ax + b)(x^2 - ax + d).$

So, $x^4 - 2x^2 + 8x - 3 = (x^2 + ax + b)(x^2 - ax + d).$

Expanding the right-hand side: $= x^4 + (b+d-a^2)x^2 + a(d-b)x + bd.$

Comparing coefficients with the given equation: $b+d-a^2 = -2 \quad (1)$ $a(d-b)=8 \quad (2)$ $bd=-3 \quad (3)$

Let $a^2 = y$. From (1), $b+d = y-2.$

From (2), $d-b = \frac{8}{a}.$

Now, $(d-b)^2 = (b+d)^2 - 4bd$ $\frac{64}{a^2} = (y-2)^2 + 12$ $64 = y\left[(y-2)^2 + 12\right]$ $64 = y(y^2 - 4y + 16)$ $y^3 - 4y^2 + 16y - 64 = 0$ $(y-4)(y^2+16)=0.$

Since $y=a^2$ is real and positive, we take $y=4 \implies a=2.$

Then, $d-b=4,\qquad b+d=2.$

Solving these, $d=3,\qquad b=-1.$

Hence the given equation factors as $(x^2+2x-1)(x^2-2x+3)=0.$

Now solve each quadratic factor.

First, $x^2+2x-1=0$ $x=\frac{-2\pm\sqrt{4+4}}{2}=\frac{-2\pm 2\sqrt2}{2}=-1\pm\sqrt2.$

Second, $x^2-2x+3=0$ $x=\frac{2\pm\sqrt{4-12}}{2}=\frac{2\pm 2i\sqrt2}{2}=1\pm i\sqrt2.$

Therefore, the roots are $\boxed{x=-1+\sqrt2,\quad -1-\sqrt2,\quad 1+i\sqrt2,\quad 1-i\sqrt2.}$

First alternative

Let
[ S=(\vec b\times \vec c)\cdot(\vec a\times \vec d)+(\vec c\times \vec a)\cdot(\vec b\times \vec d)+(\vec a\times \vec b)\cdot(\vec c\times \vec d). ]

Use the standard vector identity [ (\vec x\times \vec y)\cdot(\vec u\times \vec v) =(\vec x\cdot \vec u)(\vec y\cdot \vec v)-(\vec x\cdot \vec v)(\vec y\cdot \vec u). ]

Applying it to each term:

[ (\vec b\times \vec c)\cdot(\vec a\times \vec d) =(\vec b\cdot \vec a)(\vec c\cdot \vec d)-(\vec b\cdot \vec d)(\vec c\cdot \vec a), ]

[ (\vec c\times \vec a)\cdot(\vec b\times \vec d) =(\vec c\cdot \vec b)(\vec a\cdot \vec d)-(\vec c\cdot \vec d)(\vec a\cdot \vec b), ]

[ (\vec a\times \vec b)\cdot(\vec c\times \vec d) =(\vec a\cdot \vec c)(\vec b\cdot \vec d)-(\vec a\cdot \vec d)(\vec b\cdot \vec c). ]

Adding these three results,

[ \begin{aligned} S={}&(\vec a\cdot \vec b)(\vec c\cdot \vec d)-(\vec a\cdot \vec c)(\vec b\cdot \vec d)\ &+(\vec b\cdot \vec c)(\vec a\cdot \vec d)-(\vec a\cdot \vec b)(\vec c\cdot \vec d)\ &+(\vec a\cdot \vec c)(\vec b\cdot \vec d)-(\vec a\cdot \vec d)(\vec b\cdot \vec c). \end{aligned} ]

Every term cancels in pairs. Hence

[ \boxed{(\vec b\times \vec c)\cdot(\vec a\times \vec d)+(\vec c\times \vec a)\cdot(\vec b\times \vec d)+(\vec a\times \vec b)\cdot(\vec c\times \vec d)=0.} ]

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Deduction of the trigonometric identity

Take four unit vectors (\vec a,\vec b,\vec c,\vec d) in the same plane, making angles

[ \angle(\vec a)=0,\qquad \angle(\vec b)=\alpha+\beta,\qquad \angle(\vec c)=\beta,\qquad \angle(\vec d)=\alpha ]

with the positive (x)-axis. Let (\hat n) be the unit normal to the plane. For unit vectors (\vec u,\vec v) in the plane,

[ (\vec u\times \vec v)\cdot \hat n=\sin(\theta_v-\theta_u). ]

Also, since all cross products are perpendicular to the plane,

[ (\vec u\times \vec v)\cdot(\vec w\times \vec z) =\big[(\vec u\times \vec v)\cdot \hat n\big]\big[(\vec w\times \vec z)\cdot \hat n\big]. ]

Therefore,

[ (\vec b\times \vec c)\cdot(\vec a\times \vec d) =\sin(\beta-\alpha-\beta)\sin(\alpha-0) =\sin(-\alpha)\sin\alpha =-\sin^2\alpha, ]

[ (\vec c\times \vec a)\cdot(\vec b\times \vec d) =\sin(0-\beta)\sin(\alpha-\alpha-\beta) =(-\sin\beta)(-\sin\beta) =\sin^2\beta, ]

[ (\vec a\times \vec b)\cdot(\vec c\times \vec d) =\sin(\alpha+\beta)\sin(\alpha-\beta). ]

Using the proved identity,

[ -\sin^2\alpha+\sin^2\beta+\sin(\alpha+\beta)\sin(\alpha-\beta)=0. ]

Hence

[ \boxed{\sin(\alpha+\beta)\sin(\alpha-\beta)=\sin^2\alpha-\sin^2\beta.} ]

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OR alternative

The scalar product of three vectors is the scalar triple product. For three vectors (\vec a,\vec b,\vec c), it is defined as

[ [\vec a\ \vec b\ \vec c]=\vec a\cdot(\vec b\times \vec c). ]

It can also be written as the determinant

[ [\vec a\ \vec b\ \vec c]

\begin{vmatrix} a_1&a_2&a_3\ b_1&b_2&b_3\ c_1&c_2&c_3 \end{vmatrix}. ]

Geometrical meaning:
The absolute value of the scalar triple product,

[ |[\vec a\ \vec b\ \vec c]|, ]

is equal to the volume of the parallelepiped whose coterminous edges are represented by the vectors (\vec a,\vec b,\vec c). If the three vectors are coplanar, then the volume is zero, so

[ [\vec a\ \vec b\ \vec c]=0. ]

The sign of the scalar triple product indicates the orientation of the three vectors.

Let $\phi=(\vec r\times\vec a)\cdot(\vec r\times\vec b)$, where $\vec a,\vec b$ are constant vectors.

Step 1: simplify $\phi$. Using $(\vec p\times\vec q)\cdot(\vec s\times\vec t)=(\vec p\cdot\vec s)(\vec q\cdot\vec t)-(\vec p\cdot\vec t)(\vec q\cdot\vec s)$,

$\phi=(\vec r\cdot\vec r)(\vec a\cdot\vec b)-(\vec r\cdot\vec b)(\vec a\cdot\vec r)=r^2(\vec a\cdot\vec b)-(\vec r\cdot\vec a)(\vec r\cdot\vec b).$

Step 2: take the gradient. Since $\nabla r^2=2\vec r$, $\nabla(\vec r\cdot\vec a)=\vec a$ and $\nabla(\vec r\cdot\vec b)=\vec b$,

$\nabla\phi=2(\vec a\cdot\vec b)\vec r-\big[(\vec r\cdot\vec b)\vec a+(\vec r\cdot\vec a)\vec b\big].$

Step 3: expand the right-hand side. Using $\vec A\times(\vec B\times\vec C)=(\vec A\cdot\vec C)\vec B-(\vec A\cdot\vec B)\vec C$:

$(\vec b\times\vec r)\times\vec a=-\vec a\times(\vec b\times\vec r)=-\big[(\vec a\cdot\vec r)\vec b-(\vec a\cdot\vec b)\vec r\big]=(\vec a\cdot\vec b)\vec r-(\vec a\cdot\vec r)\vec b,$

$(\vec a\times\vec r)\times\vec b=-\vec b\times(\vec a\times\vec r)=-\big[(\vec b\cdot\vec r)\vec a-(\vec b\cdot\vec a)\vec r\big]=(\vec a\cdot\vec b)\vec r-(\vec b\cdot\vec r)\vec a.$

Adding,

$(\vec b\times\vec r)\times\vec a+(\vec a\times\vec r)\times\vec b=2(\vec a\cdot\vec b)\vec r-(\vec a\cdot\vec r)\vec b-(\vec b\cdot\vec r)\vec a=\nabla\phi.$

Hence $\text{grad}\,[(\vec r\times\vec a)\cdot(\vec r\times\vec b)]=(\vec b\times\vec r)\times\vec a+(\vec a\times\vec r)\times\vec b$. $\blacksquare$

Gauss's (Divergence) Theorem:

$\iint_S \vec{f}\cdot\hat{n}\,dS = \iiint_V (\nabla\cdot\vec{f})\,dV$

Given: $\vec{f} = (2x-z)\hat{i} + x^2y\,\hat{j} - xz^2\hat{k}$

Region: cube bounded by $x=0,1$; $y=0,1$; $z=0,1$.

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Step 1: Divergence of $\vec{f}$

$\nabla\cdot\vec{f} = \frac{\partial}{\partial x}(2x-z) + \frac{\partial}{\partial y}(x^2y) + \frac{\partial}{\partial z}(-xz^2)$

$= 2 + x^2 - 2xz$

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Step 2: Volume integral (RHS)

$\iiint_V (\nabla\cdot\vec{f})\,dV = \int_0^1\int_0^1\int_0^1 (2 + x^2 - 2xz)\,dx\,dy\,dz$

Integrate w.r.t. $x$: $= \int_0^1\int_0^1 \left[2x + \frac{x^3}{3} - x^2z\right]_0^1 dy\,dz = \int_0^1\int_0^1 \left(2 + \tfrac13 - z\right)dy\,dz$

$= \int_0^1\int_0^1 \left(\tfrac73 - z\right)dy\,dz = \int_0^1 \left(\tfrac73 - z\right)dz$

$= \left[\tfrac73 z - \tfrac{z^2}{2}\right]_0^1 = \tfrac73 - \tfrac12 = \frac{14-3}{6} = \frac{11}{6}$

$\boxed{\iiint_V \nabla\cdot\vec{f}\,dV = \frac{11}{6}}$

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Step 3: Surface integral (LHS)

The cube has 6 faces. Outward unit normals:

Face $x=1$: $\vec{f}\cdot\hat{i} = 2x-z = 2-z$ $\iint = \int_0^1\int_0^1 (2-z)\,dy\,dz = \int_0^1 (2-z)\,dz = \left[2z - \tfrac{z^2}{2}\right]_0^1 = \tfrac32$

Face $x=0$: $\vec{f}\cdot(-\hat{i}) = -(2x-z) = z$ (since $x=0$) $\iint = \int_0^1\int_0^1 z\,dy\,dz = \int_0^1 z\,dz = \tfrac12$

Face $y=1$: $\vec{f}\cdot\hat{j} = x^2y = x^2$ $\iint = \int_0^1\int_0^1 x^2\,dx\,dz = \int_0^1 x^2\,dx = \tfrac13$

Face $y=0$: $\vec{f}\cdot(-\hat{j}) = -x^2y = 0$ $\iint = 0$

Face $z=1$: $\vec{f}\cdot\hat{k} = -xz^2 = -x$ $\iint = \int_0^1\int_0^1 (-x)\,dx\,dy = -\int_0^1 x\,dx = -\tfrac12$

Face $z=0$: $\vec{f}\cdot(-\hat{k}) = xz^2 = 0$ $\iint = 0$

Total surface integral: $\iint_S \vec{f}\cdot\hat{n}\,dS = \frac32 + \frac12 + \frac13 + 0 - \frac12 + 0 = \frac32 + \frac13 = \frac{9+2}{6} = \frac{11}{6}$

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Step 4: Conclusion

$\iint_S \vec{f}\cdot\hat{n}\,dS = \frac{11}{6} = \iiint_V \nabla\cdot\vec{f}\,dV$

Hence Gauss's theorem is verified.

Let $V$ be a finite-dimensional real vector space with a symmetric bilinear form
$B(x,y)=\langle x,y\rangle .$ An orthogonal basis of $V$ is a basis $\{u_1,u_2,\dots,u_n\}$ such that $B(u_i,u_j)=\langle u_i,u_j\rangle=0\qquad (i\ne j).$

Statement to prove

There exists an integer $r\ge 0$ such that for every orthogonal basis $\{u_1,\dots,u_n\}$ of $V$, exactly $r$ of the numbers $\langle u_1,u_1\rangle,\ \langle u_2,u_2\rangle,\ \dots,\ \langle u_n,u_n\rangle$ are positive.

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Proof

Define the index of positivity of the form by $p(V)=\max\{\dim W:\ W\subseteq V,\ B|_W \text{ is positive definite}\}.$ That is, $p(V)$ is the largest dimension of a subspace on which the form is positive definite.

Let $\{u_1,\dots,u_n\}$ be any orthogonal basis of $V$. Put $\lambda_i=B(u_i,u_i)=\langle u_i,u_i\rangle .$ Let $I_+=\{i:\lambda_i>0\},\qquad I_-=\{i:\lambda_i\le 0\}.$ Let $P=\operatorname{span}\{u_i:i\in I_+\}.$ Then $\dim P$ is exactly the number of positive $\lambda_i$'s.

For any nonzero $v\in P$, write $v=\sum_{i\in I_+}a_i u_i.$ Since the basis is orthogonal, $B(v,v)=\sum_{i\in I_+}a_i^2\lambda_i>0.$ Hence $B|_P$ is positive definite. Therefore $\dim P\le p(V).$ So the number of positive diagonal entries in this orthogonal basis is at most $p(V)$.

Conversely, let $W\subseteq V$ be any subspace on which $B$ is positive definite. Let $N=\operatorname{span}\{u_i:i\in I_-\}.$ Every $w\in W$ can be written uniquely as $w=x+y,\qquad x\in P,\quad y\in N.$ Define a map $\pi:W\to P,\qquad \pi(w)=x.$ If $\pi(w)=0$, then $w=y\in N$. Since every $\lambda_i\le 0$ on $N$, $B(w,w)=B(y,y)=\sum_{i\in I_-}a_i^2\lambda_i\le 0.$ But $B|_W$ is positive definite, so $w=0$. Thus $\pi$ is injective. Hence $\dim W\le \dim P.$ Taking the supremum over all positive definite subspaces $W$, we get $p(V)\le \dim P.$ Therefore $\dim P=p(V).$ So every orthogonal basis has exactly $p(V)$ positive diagonal entries. Hence the required integer is $r=p(V).$ This proves that $r$ is independent of the choice of orthogonal basis.

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Definitions

Index of positivity

The index of positivity of the form is the maximum dimension of a subspace of $V$ on which the form is positive definite. Equivalently, it is the number of positive terms $\langle u_i,u_i\rangle>0$ in any orthogonal basis. It is usually denoted by $p$.

Index of nullity

The radical of the form is $V^\perp=\{v\in V:\langle v,w\rangle=0\ \text{for all }w\in V\}.$ The index of nullity is $q=\dim V^\perp.$ Equivalently, in any orthogonal basis, $q$ is the number of indices $i$ for which $\langle u_i,u_i\rangle=0.$ Also, $\operatorname{rank}(B)=n-q.$

Thus the form is completely described, up to congruence, by its signature $(p,s,q),$ where $p$ is the index of positivity, $s$ is the index of negativity, and $q$ is the index of nullity.