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If a matrix $A$ has repeated eigenvalues, its eigenspace matrix would be dependent because there isn't enough eigenvector to tag along with the eigenvalue.

But if a matrix $A$ has repeated eigenvalues of $0$, that means the matrix $A$ is singular, then the dimension of the nullspace of matrix $A$ is the number of $0$ eigenvalues that matrix $A$ has. Although I read that the eigenvectors of the matrix A are dependent if there are repeated eigenvalues, I'm thinking why couldn't the eigenvectors be independent if the repeated eigenvalues were $0$?

Say in a matrix that has 2 eigenvalues of $0$, this implies that $dim(N(A))=2$, then wouldn't the eigenvector to each of this eigenvalue of $0$ be the 2 independent nullspace vectors? Then when putting all the eigenvectors together to form an eigenspace matrix, they will all be independent, wouldn't it?

What do I don't understand that lead me to this thought? Thanks for any help.

Update:

With the example suggested by Didier Piau, $A=\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}$ The $rank(A)=1$, $N(A)=\begin{bmatrix} 1\\ 0 \end{bmatrix}$, $eigenvals(A)=0, 0$ and true enough, the eigenvector can only be $\begin{bmatrix} 1\\ 0 \end{bmatrix}$ and therefore the eigenspace is dependent. There isn't enough eigenvectors to tag to the other eigenvalue of zero.

But say with another example of matrix $B$, $B=\begin{bmatrix} 1 & 2 & 3\\ 1 & 2 & 3\\ 2 & 4 & 6 \end{bmatrix}$ The $rank(B)=1$, the $N(A)=c_{1}\begin{bmatrix} -2\\ 1\\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} -3\\ 0\\ 1 \end{bmatrix}, c_{1}, c_{2} \in \mathbb{R}$, $eigenvals(B)=9, 0,0$ then the eigenvectors are... $B\begin{bmatrix} 0.5\\ 0.5\\ 1 \end{bmatrix}=9\begin{bmatrix} 0.5\\ 0.5\\ 1 \end{bmatrix}$ Then for the next two eigenvalues of $0$, which eigenvector do I choose to use as their eigenvector? Do I use one for each or only just one of the two or both of the two? $B\begin{bmatrix} 0.5\\ 0.5\\ 1 \end{bmatrix}=0\begin{bmatrix} -2\\ 1\\ 0 \end{bmatrix}$ and it could also be... $B\begin{bmatrix} 0.5\\ 0.5\\ 1 \end{bmatrix}=0\begin{bmatrix} -3\\ 0\\ 1 \end{bmatrix}$ Then for the eigenspace, should it be: $\begin{bmatrix} 0.5 & -2 & 3\\ 0.5 & 1 & 0\\ 1 & 0 & 1 \end{bmatrix}$ or $\begin{bmatrix} 0.5 & -2 & -2\\ 0.5 & 1 & 1\\ 1 & 0 & 0 \end{bmatrix}$ or $\begin{bmatrix} 0.5 & 3 & 3\\ 0.5 & 0 & 0\\ 1 & 1 & 1 \end{bmatrix}$?

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    Thanks a lot for your help! :)2011-08-16

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One can test one's understanding of this question on the matrix A=[0,1|0,0], since this example already exhibits many of the relevant features of the problem. For eample, one cannot find two linearly independent vectors v and w such that Av=Aw=0 (try to solve for v the equation Av=0) although 0 is a double root of the characteritic polynomial det(XI-A)=X^2.

This is not specific to the eigenvalue 0. Recall that for any eigenvalue one distinguishes its algebraic multiplicity n (in this case, n=2) from its geometric multiplicity m (in this case, m=1). See here for more details and a worked out example.

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    For B the eigenvalues are 9 (AM=1) and 0 (AM=2) (with AM=algebraic multiplicity). For the eigenvalue 9, GM=1 (GM=geometric multiplicity) because GM is always between 1 and AM hence when AM=1, AM=GM. For the eigenvalue 0, one could a priori have GM=1 or GM=2 but Bv=0 means that the coordinates of v are such that x+2y+3z=0 hence the eigenspace is a plane, in other words GM=2. Since AM=GM for every eigenvalue of B, B is diagonalizable (this is equivalent). (You should read carefully the WP page I linked to, it explains all this and much more.)2011-08-16
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you have 2 eigenvectors that represent the eigenspace for eigenvalue = 1 are linear independent and they should both be included in your eigenspace..they span the original space... note that if you have 2 repeated eigenvalues they may or may not span the original space, so your eigenspace could be rank 1 or 2 in this case.