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 $\begin{bmatrix} 0.5 & 3 & 3\\ 0.5 & 0 & 0\\ 1 & 1 & 1 \end{bmatrix}$?