2 Linear algebra questions

**boromir** · January 6th 2012, 01:49 PM

1)Let A be an n x n symmetric matrix. Prove eigenvalues are real.

I can do it in the complex case by proving it is equal to it's conjugate but it doesn't work in the real case.

2) if A and B are similar, that is if AP=PB for invertible P, is there an easy way to find powers of A?

Thanks in advance.

**alexmahone** · January 6th 2012, 01:53 PM

Originally Posted by boromir

1)Let A be an n x n symmetric matrix. Prove eigenvalues are real.

I can do it in the complex case by proving it is equal to it's conjugate but it doesn't work in the real case.

2) if A and B are similar, that is if AP=PB for invertible P, is there an easy way to find powers of A?

Thanks in advance.

2) $AP=PB$

$A=PBP^{-1}$

$A^2=PBP^{-1}PBP^{-1}=PB^2P^{-1}$

By induction, we get $A^n=PB^nP^{-1}$

**Siron** · January 6th 2012, 10:47 PM

1) If $A$ is a real matrix then $A$ is an hermitian matrix if and only if $A$ is a symmetric matrix. You should know all the eigenvalues of an hermitian matrix are real (do you know why?). So if $A$ is symmetric and real then $A$ is an hermitian matrix thus the eigenvalues of $A$ are real.

**FernandoRevilla** · January 7th 2012, 12:11 AM

Originally Posted by boromir

1)Let A be an n x n symmetric matrix. Prove eigenvalues are real.

I suppose you mean $A=(a_{ij})\in M_{n\times n}(\mathbb{R})$ , otherwise the result is not true. Another way (without using the concept of hermitian matrix): let $\lambda$ be an eigenvalue of $A$ (a priori complex) and $0\neq X=(x_i)\in\mathbb{C}^n$ such that $AX=\lambda X$ . Then, $\bar{X}^tAX=\lambda \bar{X}^tX$ . But $\bar{X}X=\sum \bar{x_i}x_i=\sum |x_i|^2\in\mathbb{R}-\{0\}$ . On the other hand $\bar{X}^tAX=\sum a_{ij}x_i\bar{x_j}\in\mathbb{R}$ because its conjugate is the same: $\sum a_{ij}\bar{x_i}x_j\in\mathbb{R}$ ( $A$ is symmetric). So, $\lambda=\frac{\bar{X}^tAX}{\bar{X}^tX}\in\mathbb{R }$ .

**boromir** · January 7th 2012, 12:36 AM

Yes sorry, A is real matrix. So if A is symmetric, it is hermitian which means T(v)=Av is self adjoint w.r.t to standard inner product on ${C}^n$ . Hence,letting <.,.> denote the standard inner product and letting eigenvalue t have eigenvector v, we have,

t<v,v> =<tv,v>=<T(v),v>=<v,T(v)>=<v,tv>=t*<v,v>. v is non zero by definition so <v,v> is non zero by positive definite. Therefore t=t* and we are done.

Is this the correct reasoning?

**Siron** · January 7th 2012, 02:06 AM

Originally Posted by boromir

Yes sorry, A is real matrix. So if A is symmetric, it is hermitian which means T(v)=Av is self adjoint w.r.t to standard inner product on ${C}^n$ . Hence,letting <.,.> denote the standard inner product and letting eigenvalue t have eigenvector v, we have,

t<v,v> =<tv,v>=<T(v),v>=<v,T(v)>=<v,tv>=t*<v,v>. v is non zero by definition so <v,v> is non zero by positive definite. Therefore t=t* and we are done.

Is this the correct reasoning?

I think that is correct. I have a slightly different proof:
Let $A$ be an hermitian $(n,n)$ matrix and supposte $x\in \mathhb{C}^n$ is an eigenvector of $A$ and suppose $c$ is the corresponding eigenvalue. Then we have $A\cdot x=c\cdot x$ and thus $^H x\cdot A\cdot x=c\cdot ||x||^2$
$\Rightarrow \overline{c}\cdot ||x||^2=^H(^H x\cdot A\cdot x)=^H x \cdot A \cdot x=c\cdot ||x||^2$

Therefore $\overline{c}=c$ and thus $c$ is real.

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