If two matrices have the same eigenvalues and dimensions of eigenspaces, are they similar?

Question

My attempt at an informal proof for this would be: if they have the same eigenvalues and dimensions of eigenspaces, their generalised eigenspaces also have the same dimensions. This means they will have the same Jordan Canonical Form. Thus, both are similar to the same JCF, which means they are similar to one another.

Answer 1

The answer is no. As a counterexample, consider the matrices $$ A = \pmatrix{0&0&0&0\\0&0&1&0\\ 0&0&0&1\\0&0&0&0}, \quad B = \pmatrix{0 &0&1&0\\0&0&0&1\\0&0&0&0\\0&0&0&0}. $$ For both matrices, the only eigenvalue is $0$ and the associated eigenspace has dimension 2 with basis $\{(1,0,0,0),(0,1,0,0)\}$. To see that $A$ and $B$ are not similar, it suffices to note that $B^2 = 0$ but $A^2 \neq 0$.

Instead of the matrix $B$, we could also have used the Jordan normal form of $B$, which is $$ \pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&1\\0&0&0&0}. $$

More generally, the dimension of an the eigenspace associated with an eigenvalue $\lambda$ merely counts the total number of Jordan blocks associated with $\lambda$ that appear in a matrix's Jordan normal form. As the above example demonstrates, it is possible for matrices of the same size to have the same number of Jordan blocks associated with a given eigenvalue without being similar.

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Another counterexample:

$$ A = \pmatrix{1&0&0\\0&0&0\\0&1&0}, \quad B = \pmatrix{1&1&0\\0&1&0\\0&0&0}. $$