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Math Help - 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) AP=PB

    A=PBP^{-1}

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

    By induction, we get 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 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.
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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 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  } .
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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 {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 {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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