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Let K and L be symmetric n x n matrices. Prove that $x^TKx = x^TLx$ for all x $\in\mathbb{R}^n$ if and only if K = L.

It seems very obvious but how can I prove this. Will it be easier to prove it using contradiction?

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$x^TKx = x^TLx$ is the same as $x^T(K-L)x = 0$ for all $x$. This means all the eigenvalues of the symmetric matrix $K-L$ are zero. This happen iff $K-L$ is a zero matrix.

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  • $\begingroup$ Thank you very much , Fischer! $\endgroup$
    – diimension
    Oct 16 '12 at 0:28
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First consider $x = e_j$ where $e_j$ is a vector with $j^{th}$ component as $1$ and rest as zeros. This will help you conclude that the diagonal elements of $K$ and $L$ are the same.

Now consider $x = e_{ij}$ where $e_j$ is a vector with $i^{th}$ and $j^{th}$ component as $1$ and rest as $0$. This with the previous fact will help you conclude that the off-diagonal entries of $K$ and $L$ are also equal.

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  • $\begingroup$ Thank you very much Marvis! $\endgroup$
    – diimension
    Oct 16 '12 at 0:27

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