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It is given that a 2 by 2 matrix has eigenvalues of 0 and 1, and the corresponding eigenvectors are {{1},{2}} and {{2},{-1}}. How can I tell this is a symmetric matrix? I may think it is because the eigenvectors are orthogonal to each other. Is this right?

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up vote 1 down vote accepted

Well, let $V = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix}$ and $\Lambda = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$. Let $A$ be the matrix in question.

Then you are given $A V = V \Lambda$ from which you can see that $A=V \Lambda V^{-1}$. So you could work this out by just computation.

Alternatively, you could note that $W = \frac{1}{\sqrt{5}} V$ is orthogonal, that is $W^T W = I$. Hence $W^{-1} = W^T$, and in particular since $AW = W \Lambda$, we have $A = W \Lambda W^T$.

Then $A^T = (W \Lambda W^T)^T = W \Lambda W^T = A$.

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@Berci: Thanks for catching that! – copper.hat Nov 2 '12 at 15:12
But how can I tell this without any calculation? The Problem asks me to tell this is a symmetric matrix without any calculation. – Scorpio19891119 Nov 2 '12 at 15:14
Use my second suggestion – copper.hat Nov 2 '12 at 15:14
Do you mean the eigenspace is symmetric? Or something else? Would you please explain it explicitly? – Scorpio19891119 Nov 2 '12 at 15:18
My second suggestion was misleading. I have added more detail. – copper.hat Nov 2 '12 at 17:01

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