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The quadratic form can be written as

$\mathrm{q}\left(\stackrel{鈫�}{x}\right)=\stackrel{鈫�}{x}路A\stackrel{鈫�}{x}={\stackrel{鈫�}{x}}^{T}A\left(\stackrel{鈫�}{x}\right)$

Now,

$$\begin{gathered} L\left(\stackrel{鈫�}{x}\right)=q(\stackrel{鈫�}{x}+\stackrel{鈫�}{v})鈭�q\left(\stackrel{鈫�}{x}\right)鈭�q\left(\stackrel{鈫�}{v}\right) \\ =(\stackrel{鈫�}{x}+\stackrel{鈫�}{v}{)}^{\mathrm{鈯�}}A(\stackrel{鈫�}{x}+\stackrel{鈫�}{v})鈭�{\stackrel{鈫�}{x}}^{\mathrm{鈯�}}A\stackrel{鈫�}{x}鈭�{\stackrel{鈫�}{v}}^{\mathrm{鈯�}}A\stackrel{鈫�}{v} \\ =\left({\stackrel{鈫�}{x}}^{\mathrm{鈯�}}+{\stackrel{鈫�}{v}}^{\mathrm{鈯�}}\right)A(\stackrel{鈫�}{x}+\stackrel{鈫�}{v})鈭�{\stackrel{鈫�}{x}}^{\mathrm{鈯�}}A\stackrel{鈫�}{x}鈭�{\stackrel{鈫�}{v}}^{\mathrm{鈯�}}A\stackrel{鈫�}{v} \\ ={\stackrel{鈫�}{x}}^{\mathrm{鈯�}}A\stackrel{鈫�}{x}+{\stackrel{鈫�}{x}}^{\mathrm{鈯�}}A\stackrel{鈫�}{v}+{\stackrel{鈫�}{v}}^{\mathrm{鈯�}}A\stackrel{鈫�}{x}+{\stackrel{鈫�}{v}}^{\mathrm{鈯�}}A\stackrel{鈫�}{v}鈭�{\stackrel{鈫�}{x}}^{\mathrm{鈯�}}A\stackrel{鈫�}{x}鈭�{\stackrel{鈫�}{v}}^{\mathrm{鈯�}}A\stackrel{鈫�}{v} \\ ={\stackrel{鈫�}{x}}^{\mathrm{鈯�}}A\stackrel{鈫�}{v}+{\stackrel{鈫�}{v}}^{\mathrm{鈯�}}A\stackrel{鈫�}{x} \end{gathered}$$

Suppose we have column vectors of the same size,$\stackrel{鈫�}{a}$and $\stackrel{鈫�}{b}$, then ${\stackrel{鈫�}{a}}^T\stackrel{鈫�}{b}$is a $1脳1$matrix which we can think of as a scalar. Now, taking transposes, since ${\stackrel{鈫�}{a}}^T\stackrel{鈫�}{b}$is $1脳1$, it is its own transpose, so

${\stackrel{鈫�}{a}}^T\stackrel{鈫�}{b}=({\stackrel{鈫�}{a}}^T\stackrel{鈫�}{b}{)}^T={\stackrel{鈫�}{b}}^T({\stackrel{鈫�}{a}}^T{)}^T={\stackrel{鈫�}{b}}^T\stackrel{鈫�}{a}$

Here let$\stackrel{鈫�}{a}={\stackrel{鈫�}{x}}^T$and $\stackrel{鈫�}{b}=A\stackrel{鈫�}{v}$, so

$\begin{array}{r}L\left(\stackrel{鈫�}{x}\right)={\stackrel{鈫�}{x}}^{\mathrm{鈯�}}A\stackrel{鈫�}{v}+{\stackrel{鈫�}{v}}^{\mathrm{鈯�}}A\stackrel{鈫�}{x}\\ =(A\stackrel{鈫�}{v}{)}^{\mathrm{鈯�}}\stackrel{鈫�}{x}+{\stackrel{鈫�}{v}}^{\mathrm{鈯�}}A\stackrel{鈫�}{x}\\ ={\stackrel{鈫�}{v}}^{\mathrm{鈯�}}{A}^{\mathrm{鈯�}}\stackrel{鈫�}{x}+{\stackrel{鈫�}{v}}^{\mathrm{鈯�}}A\stackrel{鈫�}{x}\end{array}$

A is symmetric, so,$A=A^T$

$$\begin{gathered} ={\stackrel{鈫�}{v}}^{T}A\stackrel{鈫�}{x}+{\stackrel{鈫�}{v}}^{T}A\stackrel{鈫�}{x} \\ =\left(2{\stackrel{鈫�}{v}}^{T}A\right)\stackrel{鈫�}{x} \end{gathered}$$

From $L\left(\stackrel{鈫�}{x}\right)=\left(2{\stackrel{鈫�}{v}}^{T}A\right)\stackrel{鈫�}{x}$, it is clear that $\mathrm{L}(\stackrel{鈫�}{x})$ is linear with matrix $2{\stackrel{鈫�}{v}}^{T}A$.

b. We have

$\begin{array}{l}q\left(\stackrel{鈫�}{x}\right)=\stackrel{鈫�}{x}{A}^2\stackrel{鈫�}{x}={\stackrel{鈫�}{x}}^T{A}^2\stackrel{鈫�}{x}\\ =\stackrel{鈫�}{x}AA\stackrel{鈫�}{x}\end{array}$

from part. we mentioned that A is said to be a skew matrix if $A^T=-A$, so

$$\begin{gathered} ={\stackrel{鈫�}{x}}^T\left(-A^T\right)A\stackrel{鈫�}{x} \\ =\left(A\stackrel{鈫�}{x}\right).\left(A\stackrel{鈫�}{x}\right) \\ =-{||A{\stackrel{鈫�}{x}}^T||}^2 \\ 鈮�0 \end{gathered}$$

therefore, for all $x,q\left(\stackrel{鈫�}{x}\right)鈮�0$, which implies that$A^2$ is a negative definite (its eigenvalues are less than or equal to 0 ).

c. Let v be an eigenvector corresponding to the eigenvalue $位$,

$Av=位\mathrm{v}$

multiply both sides by$v^{-T}$

${\overline{v}}^{T}Av=位{\overline{v}}^T$

note: the bar" -" above v is for complex conjugation.

${\stackrel{炉}{v}}^{T}Av=位{||v||}^2$

for the left-hand side suppose we have column vectors of the same size, a and b, then $a^{T}b$is a $1脳1$matrix which we can think of as a scalar. Now, taking transposes, since$a^{T}b$is $1脳1$, it is its own transpose, so

${\mathrm{a}}^{\mathrm{T}}\mathrm{b}={\left({\mathrm{a}}^{\mathrm{T}}\mathrm{b}\right)}^{\mathrm{T}}={\mathrm{b}}^{\mathrm{T}}{\left({\mathrm{a}}^{\mathrm{T}}\right)}^{\mathrm{T}}={\mathrm{b}}^{\mathrm{T}}\mathrm{a}$

here let $a=\stackrel{炉}{v}$and $b=Av$, then for our left-hand side

$$\begin{gathered} {\left(Av\right)}^T\stackrel{炉}{v}=位{||v||}^2 \\ {v}^T{A}^T\stackrel{炉}{v}=位{||v||}^2 \end{gathered}$$

since A is skew-symmetric, we have$A^T=-A$. Substituting,

$-v^{T}A\overline{v}=位{||v||}^2$

taking the conjugate of$Av=位\mathrm{v}$ yields$A\stackrel{炉}{v}=\stackrel{炉}{位}\stackrel{炉}{v}$, notice we can substitute this into the left-hand side above

$$\begin{gathered} 鈭�v^{\mathrm{鈯�}}\stackrel{炉}{位}\stackrel{炉}{v}=位鈭�v{鈭�}^2 \\ 鈭�\stackrel{炉}{位}鈭�v{鈭�}^2=位鈭�v{鈭�}^2 \\ 鈭�\stackrel{炉}{位}=位 \end{gathered}$$

let,$位=a+ib$

$-a+ib=a+ib$which implies a=0 and $位=b$i. We also conclude that the zero matrix is the only skew-symmetric matrix that is diagonalizable over $\mathrm{鈩�}$.