91影视

First notice that$A^{-1}$is symmetric$\left({\left({\mathrm{A}}^{-}\right)}^{\mathrm{T}}={\left({\mathrm{A}}^{\mathrm{T}}\right)}^{-1}={\mathrm{A}}^1\right)$.

A matrix A has an inverse${\mathrm{A}}^{-1}$if and only if it does not have zero as an eigenvalue. Therefore ifis an eigenvalue of A and A exists, then$鈭�\stackrel{鈫�}{x}鈮�0$and$位鈮�0$.

$\begin{array}{l}A\stackrel{鈫�}{x}=位\stackrel{鈫�}{x}\\ A^{-1}\left(A\stackrel{鈫�}{x}\right)=A^{-1}\left(\stackrel{鈫�}{位x}\right)\end{array}$ $\mathrm{Multiply}\mathrm{both}\mathrm{sides}{\mathrm{byA}}^{-1}$

$\stackrel{鈫�}{x}=位A^{-1}\stackrel{鈫�}{x}$

$A^{-1}\stackrel{鈫�}{x}=\frac{1}{位}\stackrel{鈫�}{x}$

We can see that $\frac{1}{位}$is an eigenvalue of $A^{-1}$, so we can say that the eigenvalues of ${\mathrm{A}}^{-1}$are the reciprocals of the eigenvalues of A. Now, if the matrix A had:

- All eigenvalues positive, then so will${\mathrm{A}}^{-1}$, and both matrices are positive definite.

- All eigenvalues negative, then so will${\mathrm{A}}^{-1}$, and both matrices are negative definite (for proof look below).

- Positive and negative eigenvalues, then so will ${\mathrm{A}}^{-1}$, and both matrices are indefinite (for proof look below).

Note that if a matrix is invertible, it can鈥檛 be semidefinite, since we can't have an eigenvalue of 0.

Proof for matrices with all eigenvalues being negative: Let$\stackrel{鈫�}{x}$be an arbitrary nonzero vector in${\mathrm{鈩�}}^n$:

${\stackrel{鈫�}{\mathrm{x}}}^{\mathrm{T}}\mathrm{A}\stackrel{鈫�}{\mathrm{x}}={\stackrel{鈫�}{\mathrm{x}}}^{\mathrm{T}}{\mathrm{SDS}}^{\mathrm{T}}\stackrel{鈫�}{\mathrm{x}}$

Here$\mathrm{A}={\mathrm{SDS}}^{\mathrm{T}}$, where D is diagonal and contains eigenvalues of A, now let$\stackrel{鈫�}{\mathrm{v}}={\mathrm{S}}^{\mathrm{T}}\stackrel{鈫�}{\mathrm{x}}$, then this can be written as

$={\stackrel{鈫�}{\mathrm{v}}}^{\mathrm{T}}\mathrm{D}\stackrel{鈫�}{\mathrm{v}}$

let

$v=\left[\begin{array}{c}{v}_1\\ v_2\\ 鈰�\\ v_n\end{array}\right]$

so then we have

$=\left[\begin{array}{l}{v}_1鈥呪赌呪赌呪赌�v_2鈥呪赌呪赌呪赌�鈥�鈥呪赌呪赌呪赌�v_n\end{array}\right]\left[\begin{array}{cccc}{位}_1& 0& 0& 0\\ 0& {位}_2& 0& 0\\ 鈰�& 鈥�& 鈰�& 鈰�\\ 0& 0& 鈥�& {位}_n\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\\ 鈰�\\ v_n\end{array}\right]={位}_1{v}_1^2+{位}_2{v}_2^2+鈰�+{位}_n{v}_n^2$

Since we assumed that all eigenvalues of A are negative, the sum above is negative for all$\stackrel{鈫�}{v}鈮�\stackrel{鈫�}{0}$, hence${\stackrel{鈫�}{\mathrm{x}}}^{\mathrm{T}}\mathrm{A}\stackrel{鈫�}{\mathrm{x}}$is negative for any nonzero vector$\stackrel{鈫�}{x}$(we use that if$\stackrel{鈫�}{x}$is nonzero, then$\stackrel{鈫�}{v}=S^T\stackrel{鈫�}{x}$ is also nonzero). Hence, A is negative definite.

Using similar techniques we can prove that if A has positive and negative eigenvalues, then it is indefinite.