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Consider you have two $n \times n$ matrices $A,B$ with the same eigenvalue $\pi$. Then $A-B$ has an eigenvalue of $0$.

The question is, is this correct or not?

I was looking for properties in my head and in the course text, but i didn't find anything useful. Cause we don't know if $\pi$ is die only eigenvalue. We also don't know if the eigen vectors corresponding to the eigenvalues are the same. So we don't know if the matrices are similar or not. The only thing I tried was:

Consider that they have the same eigen vector. Then you could write:

$Av-Bv$ = $(A-B)v$ and $\pi v-\pi v = (\pi-\pi) v$ and $\pi-\pi = 0$

So then you could proof this, but this is not the case unfortunately. I can also find no example that it's not true.

Thank you

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  • $\begingroup$ Ok, I registered now and now I can't comment on my own question anymore? Is there a more general solution to this problem. Because in the question they say the following: The questions are needed to be solved in general, and not with an example. (we need to test your insight) Now: A counter example, thats also general right ? $\endgroup$ – Thomas Jan 4 '15 at 10:29
  • $\begingroup$ you need to earn some reputation before you can leave comment. Please read the FAQ section. Also, please avoid posing non-answers as answers (which will promptly be deleted). $\endgroup$ – Ittay Weiss Jan 4 '15 at 10:37
  • $\begingroup$ But this is my OWN question, and i am asking a further explanation ? $\endgroup$ – Thomas Jan 4 '15 at 10:46
  • $\begingroup$ If you have to prove or disprove the statement "If $A$ and $B$ have the same eigenvalues, then $A-B$ has eigenvalue $0$", then one counterexample is sufficient to disprove the statement. If you want to prove that the statement is true, you'll need to show it for general matrices $A$ and $B$ and not just using one example. $\endgroup$ – Huy Jan 4 '15 at 11:07
  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. $\endgroup$ – Adam Hughes Jan 4 '15 at 11:23
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The matrices $$A = \begin{pmatrix} 1&0\\ 0&2 \end{pmatrix}, \, B = \begin{pmatrix} 2 & 0\\ 0 & 1 \end{pmatrix}$$ have the same eigenvalues, $1$ and $2$, yet $$A - B = \begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}$$ has eigenvalues $1$ and $-1$.

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  • $\begingroup$ So easy -.- Thank you very much :) $\endgroup$ – Thomas Dec 31 '14 at 11:37

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