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?
How to tell this is a symmetric/self-adjoint matix?
0
$\begingroup$
linear-algebra
1 Answers
1
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 $A^T = (W \Lambda W^T)^T = W \Lambda W^T = A$.
-
0Hold on, I made a mistake, I will delete the answer, fix it and repost. – 2012-11-02