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Thread: 2 Linear algebra questions

  1. #1
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    2 Linear algebra questions

    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.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: 2 Linear algebra questions

    Quote Originally Posted by boromir View Post
    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) $\displaystyle AP=PB$

    $\displaystyle A=PBP^{-1}$

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

    By induction, we get $\displaystyle A^n=PB^nP^{-1}$
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: 2 Linear algebra questions

    1) If $\displaystyle A$ is a real matrix then $\displaystyle A$ is an hermitian matrix if and only if $\displaystyle A$ is a symmetric matrix. You should know all the eigenvalues of an hermitian matrix are real (do you know why?). So if $\displaystyle A$ is symmetric and real then $\displaystyle A$ is an hermitian matrix thus the eigenvalues of $\displaystyle A$ are real.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: 2 Linear algebra questions

    Quote Originally Posted by boromir View Post
    1)Let A be an n x n symmetric matrix. Prove eigenvalues are real.
    I suppose you mean $\displaystyle A=(a_{ij})\in M_{n\times n}(\mathbb{R})$ , otherwise the result is not true. Another way (without using the concept of hermitian matrix): let $\displaystyle \lambda$ be an eigenvalue of $\displaystyle A$ (a priori complex) and $\displaystyle 0\neq X=(x_i)\in\mathbb{C}^n$ such that $\displaystyle AX=\lambda X$ . Then, $\displaystyle \bar{X}^tAX=\lambda \bar{X}^tX$ . But $\displaystyle \bar{X}X=\sum \bar{x_i}x_i=\sum |x_i|^2\in\mathbb{R}-\{0\}$ . On the other hand $\displaystyle \bar{X}^tAX=\sum a_{ij}x_i\bar{x_j}\in\mathbb{R}$ because its conjugate is the same: $\displaystyle \sum a_{ij}\bar{x_i}x_j\in\mathbb{R}$ ($\displaystyle A$ is symmetric). So, $\displaystyle \lambda=\frac{\bar{X}^tAX}{\bar{X}^tX}\in\mathbb{R }$ .
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  5. #5
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    Re: 2 Linear algebra questions

    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 $\displaystyle {C}^n$. 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?
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: 2 Linear algebra questions

    Quote Originally Posted by boromir View Post
    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 $\displaystyle {C}^n$. 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 $\displaystyle A$ be an hermitian $\displaystyle (n,n)$ matrix and supposte $\displaystyle x\in \mathhb{C}^n$ is an eigenvector of $\displaystyle A$ and suppose $\displaystyle c$ is the corresponding eigenvalue. Then we have $\displaystyle A\cdot x=c\cdot x$ and thus $\displaystyle ^H x\cdot A\cdot x=c\cdot ||x||^2$
    $\displaystyle \Rightarrow \overline{c}\cdot ||x||^2=^H(^H x\cdot A\cdot x)=^H x \cdot A \cdot x=c\cdot ||x||^2$

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