Possible Dimensions of Eigenspaces Given Characteristic Equation

Question

Let A be a $6$x$6$ matrix with characteristic equation $k^2(k-1)(k-2)^3 = 0$. What are the possible dimensions of the eigenspaces of A?

So here we have 3 eigenvalues: 0 with multiplicity 2, 1, and 2 with multiplicity 3. I know that but I don't know the relationship between the eigenvalues and the possible dimensions of the eigenspaces.

If I had to answer this question on a test, my intuition here is that since there are 3 distinct eigenvalues, then such matrix has 3 eigenvectors, so the eigenspace has 3 dimensions at best. But I'm not sure of that.

Answer 1

The dimensions are $2 ($the one associated with the $0$ eigenvalue$)$, $1 ($the one associated with the $1$ eigenvalue$),$ and $3($the one associated with the $2$ eigenvalue$).$ We get this directly from the powers above.

Eigenvalue $0$ can have either $1$ or $2$ dimensions, and eigenvalue $2$ can have $1,2$ or $3$ dimensions

Answer 2

There are many ways of realizing this characteristic equation into state-space, which would give different eigenspaces for each associated eigenvalue.

For instance, if we have \begin{equation} A=\begin{bmatrix} 0&0&0&0&0 &0 \\ 0&0&0&0&0 &0 \\ 0&0&1&0&0 &0 \\ 0&0&0&2&0 &0 \\ 0&0&0&0&2 &0\\0&0&0&0&0 &2 \end{bmatrix} \end{equation} then the eigenspace of the $0$ eigenvalue is $2$-dimensional, that of $1$ is $1$-dimensional and that of $2$ is $3$-dimensional.

However, if instead \begin{equation} A=\begin{bmatrix} 0&1&0&0&0 &0 \\ 0&0&0&0&0 &0 \\ 0&0&1&0&0 &0 \\ 0&0&0&2&1 &0 \\ 0&0&0&0&2 &1\\0&0&0&0&0 &2 \end{bmatrix} \end{equation} we still have the same characteristic equation but then the eigenspace of the $0$ eigenvalue is $1$-dimensional, that of $1$ is $1$-dimensional and that of $2$ is $1$-dimensional.

While distinct eigenvalues imply linear independent eigenvectors, the converse isn't true; for instance, the identity matrix.