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Thread: linear algebra proof

  1. #1
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    linear algebra proof

    I have no idea where to start to prove this.

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

    $\displaystyle I_{n}, u^{T}u$ both symmetric, then of course $\displaystyle A$ is symmetrc. But how to show $\displaystyle u^{T}u$ 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 $\displaystyle A=I_{n} -u^{T}u \in M_{n}(R)$ where $\displaystyle I_{n}$ is the identity, $\displaystyle u$ is a non-zero row vector in $\displaystyle R^{n}$ and $\displaystyle u^{T}$ denotes transpose of $\displaystyle u$. If $\displaystyle A$ is symmetric, show that $\displaystyle u^{T}u=I_{n}$.

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


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

    $\displaystyle A= I-u^tu=\begin{pmatrix}1&0\\0&1\end{pmatrix}-\begin{pmatrix}1&2\\2&4\end{pmatrix}=\begin{pmatri x}0&\!\!\!-2\\\!\!\!-2&\!\!\!-3\end{pmatrix}$ , so $\displaystyle A$ is symmetric but $\displaystyle u^tu$ 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 $\displaystyle u^tu$ 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 $\displaystyle u^{T}u=I_{n}$ is false?
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  4. #4
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    Quote Originally Posted by deniselim17 View Post
    Did you mean that $\displaystyle u^{T}u=I_{n}$ is false?

    Yes.

    Tonio
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