Jordan form example clarification

Question

I am practicing Jordan forms and came across the following example: $$A=\left(\begin{array}{rrr} 1&1&1\\-1&-1&-1\\1&1&1\end{array}\right).$$ Jordan canonical form is $J=\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}$. My question is shouldn't $J=\begin{pmatrix}0&1&0\\0&0&0\\0&0&1\end{pmatrix}$ be the Jordan form?

I computed the eigenvalues which are 0 (multiplicity 2) and 1 (multiplicity 1). Now because of multiplicity 1 the block containing eigenvalue 1 is $1\times1$ matrix. With 0 we find the eigenvectors to be $E_1=\{\begin{pmatrix}-1\\1\\0\end{pmatrix}, \begin{pmatrix}-1\\0\\1\end{pmatrix}\}$. Since the dimension is 2 my Jordan block will have $2\times2$ block with diagonal entries 0 and thus we get the second form. Where am I making a mistake?

Answer 1

Just because you have a certain multiplicity of an eigenvalue does not mean you only have one Jordan block containing that eigenvalue! For example if you had an eigenvalue of multiplicity $4$ that could be $1$ block of $4$, it could be $1$ block of $3$ and $1$ of $1$, it could be blocks of size $2$ and $2$, it could be $2$, $1$, and $1$, or finally it could be $4$ blocks of size $1$!

Each block corresponds to one eigenvector in the basis, so the dimension of the eigenspace tells you the number of blocks that correspond to that eigenvector. In your case the eigenspace corresponding to $0$ is just the kernel of the matrix which has dimension $2$, so there are $2$ Jordan blocks corresponding to the eigenvalue $0$.

In your example knowing there are two blocks for $0$ and at least one block for the eigenvalue $1$ tells you that each of those blocks must be of size $1$. Note that in general knowing the number of blocks isn't enough to tell you the size of those blocks. For example multiplicity $4$ and $2$ blocks could be blocks of size $3$ and $1$ or blocks of size $2$ and $2$. So in general you'll need to look at the generalized eigenspaces to find out how big the blocks are.

Answer 2

The most straightforward way to see that your answer is incorrect is to calculate the rank of $A$, which is $1$, since the columns (equivalently, the rows) span a space of dimension $1$. Your answer has rank $2$.

Your argument is that a $2\times 2$ Jordan matrix with two linearly independent eigenvectors having eigenvalue zero must be $\begin{pmatrix}0&1\\0&0\end{pmatrix}$. In fact, you can check that this matrix has only one eigenvector up to scalar multiplication. The only Jordan matrix that gives you two is, in fact, $\begin{pmatrix}0&0\\0&0\end{pmatrix}$.

All of this can be stated very intuitively: if a $2\times 2$ matrix maps a space of dimension $2$ to zero, it has to be the zero matrix!