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What does it say about the eigenvectors of a matrix $A$ if the row-reduced form of the characteristic polynomal in coefficient matrix form has a row of 0's?

I know that it indicates something about the eigenvectors of $A$ but I can't remember what exactly...

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I believe you are talking about the companion matrix of the characteristic polynomial. The determinant of the companion matrix equals that of the original matrix $A$. If the row-reduced form of the companion matrix has an all-zero row, then it has determinant zero, so it tells you that zero is an eigenvalue of $A$. I can't imagine it tells you anything about the eigenvectors of $A$.

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But in this particular case I know that 0 isn't an eigenvalue. And that one of the eigenvalues is 1 with correlated eigenvector {0,0}. –  StickFigs May 3 '12 at 6:39
Sorry, that's nonsense. The zero vector is, by definition, never an eigenvector. You need to have another look at what you think you know about the problem. –  Gerry Myerson May 3 '12 at 7:20
I'm just going by what this says: wolframalpha.com/input/… which also suggests that 0 is not an eigenvalue. –  StickFigs May 3 '12 at 8:02
That's because the reduced row-echelon form of $\pmatrix{0&-1\cr1&2\cr}$ does not have a row of zeros. And wolfram is just flat out wrong, listing the zero vector as an eigenvector. If you allowed the zero vector to be an eigenvector, then every number would be an eigenvalue of every matrix. –  Gerry Myerson May 3 '12 at 12:54
@StickFigs, In your example case $\lambda=1$ is a double root of the characteristice equation, but the eigenspace is one-dimensional. Wolfram Alpha is simply programmed to tell you this fact by listing $(0,0)$ among the basis vectors. That is a tell-tale sign that something unusual happened, and it is the users responsibility to figure out what. As Gerry explained, zero is, by definition, never an eigenvector. –  Jyrki Lahtonen May 3 '12 at 13:32
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