Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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

share|improve this answer
    
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.