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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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    @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.2012-05-03