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Math Help - linear algebra proof

  1. #1
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    linear algebra proof

    I have no idea where to start to prove this.

    Let A=I_{n} -u^{T}u \in M_{n}(R) where I_{n} is the identity, u is a non-zero row vector in R^{n} and u^{T} denotes transpose of u. If A is symmetric, show that u^{T}u=I_{n}.

    I_{n}, u^{T}u both symmetric, then of course A is symmetrc. But how to show u^{T}u is equal to identity?
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  2. #2
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    Quote Originally Posted by deniselim17 View Post
    I have no idea where to start to prove this.

    Let A=I_{n} -u^{T}u \in M_{n}(R) where I_{n} is the identity, u is a non-zero row vector in R^{n} and u^{T} denotes transpose of u. If A is symmetric, show that u^{T}u=I_{n}.

    I_{n}, u^{T}u both symmetric, then of course A is symmetrc. But how to show u^{T}u is equal to identity?


    Without any further conditions this is false: take for example u=(12), then:

    A= I-u^tu=\begin{pmatrix}1&0\\0&1\end{pmatrix}-\begin{pmatrix}1&2\\2&4\end{pmatrix}=\begin{pmatri  x}0&\!\!\!-2\\\!\!\!-2&\!\!\!-3\end{pmatrix} , so A is symmetric but u^tu is not the unit matrix.

    As you say: sum/difference and product of symmetric matrices is symmetric, so the above difference will always be a symmetric matrix, no matter what u^tu is.


    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    Without any further conditions this is false

    Did you mean that u^{T}u=I_{n} is false?
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  4. #4
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    Quote Originally Posted by deniselim17 View Post
    Did you mean that u^{T}u=I_{n} is false?

    Yes.

    Tonio
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