Eigenvalue/Eigenvector by Inspection

[BraedenP]

Homework Statement

Without calculation, find one eigenvalue and two linearly independent eigenvectors of A=\begin{bmatrix}<br /> 5 &amp; 5 &amp; 5\\ <br /> 5 &amp; 5 &amp; 5\\ <br /> 5 &amp; 5 &amp; 5<br /> \end{bmatrix}

Justify your answer.

Homework Equations

N/A

The Attempt at a Solution

This question would be incredibly easy if I could calculate the answers, but I'm not allowed to do that. How would I go about finding the eigenvalues/vectors simply by inspection?

I could use an eigenvalue of 0 to find one eigenvector, but given that the two eigenvectors I need have to be independent, I can't simply use another multiple of that eigenvector to be my second one.. I think I need to find the other eigenvalue, which I calculated to be 15, by inspection somehow.

 

[fzero]

Since all of the elements of A are the same, there's an obvious eigenvector with the same property. That gives you one nonzero eigenvalue. Also, since the columns of A are linearly dependent, you know at least one other eigenvalue. Since the 3 columns are actually equal, you know that this eigenvalue is repeated. Since the value of this repeated eigenvalue is very special, you can easily guess at the form of the other 2 eigenvectors.

 

[HallsofIvy]

We clearly have x+ y+ z= 0 for any eigenvector <x, y, z>. Taking x=0, z= -y and taking y= 0, z= -x. n That gives the two independent eigenvectors.