# Question - eigenvector/eigenvalue

• Feb 19th 2010, 09:02 AM
krje1980
Question - eigenvector/eigenvalue
OK, first of all, I'm afraid I don't know how to write matrices on this forum, so I'll try to give the problem through a description.

I've been given a 3 X 3 Matrix A where the values in row 1 are 0, 2, 0. The values in row 2 are 1/2, 0, 0. And the values in row 3 are 0, 2/3, 0.

Further, I'm given a vector with entries (6,3,2). I'm supposed to discuss anything related to eigenvalues and eigenvectors with the given information.

I'm a bit confused though. The matrix A has 0 in positions a11, a22, and a33. Thus, I only get 0 as eigenvalue. However when I multiply matrix A with the given vector I end up with the same vector again (6, 3, 2). This indicates to me that (6,3,2) is an eigenvector with eigenvalue 1. But how can this be, when I, working with Matrix A on its own, only get eigenvalue 0? Is it just a freak occurrence that I get the same vector again? And it would be wrong to call it an eigenvector since 1 is not a possible eigenvalue for matrix A?

Any help in clarifying this would be greatly, greatly appreciated. This is the first time I've run into this situation in my eigenvalue/eigenvector calculations :) .
• Feb 19th 2010, 02:15 PM
tonio
Quote:

Originally Posted by krje1980
OK, first of all, I'm afraid I don't know how to write matrices on this forum, so I'll try to give the problem through a description.

I've been given a 3 X 3 Matrix A where the values in row 1 are 0, 2, 0. The values in row 2 are 1/2, 0, 0. And the values in row 3 are 0, 2/3, 0.

So we get that $A=\begin{pmatrix}0&2&0\\\frac{1}{2}&0&0\\0&\frac{2 }{3}&0\end{pmatrix}$ $\Longrightarrow p_A(x)=\det \begin{pmatrix}x&\!\!\!-2&0\\\!\!\!-\frac{1}{2}&x&0\\0&\!\!\!-\frac{2}{3}&x\end{pmatrix}$ $=x^3-x=x(x-1)(x+1) \Longrightarrow$ get from here the matrix's eigenvalues, and then check that the given vector below is one of the eigenvectors of the matrix corresponding to one of the eigenvalues above.

Further, I'm given a vector with entries (6,3,2). I'm supposed to discuss anything related to eigenvalues and eigenvectors with the given information.

I'm a bit confused though. The matrix A has 0 in positions a11, a22, and a33. Thus, I only get 0 as eigenvalue.

This is wrong: since the matrix is singular then zero is one of its eigenvalues, true; but there can be other eigenvalues, just as in this particular case.

Tonio

However when I multiply matrix A with the given vector I end up with the same vector again (6, 3, 2). This indicates to me that (6,3,2) is an eigenvector with eigenvalue 1. But how can this be, when I, working with Matrix A on its own, only get eigenvalue 0? Is it just a freak occurrence that I get the same vector again? And it would be wrong to call it an eigenvector since 1 is not a possible eigenvalue for matrix A?

Any help in clarifying this would be greatly, greatly appreciated. This is the first time I've run into this situation in my eigenvalue/eigenvector calculations :) .

.
• Feb 19th 2010, 10:03 PM
krje1980
Thanks a lot!

I always supposed that when you have zero in the diagonal, the only eigenvalue possble was 0. I really appreciate your help!