Jordan canonical forms and deficiency indices

I'm solving a homework question that asks me to do the following:

"List the five upper Jordan canonical forms for a $4\times 4$ matrix $A$ with a real eigenvalue $\lambda$ of multiplicity $4$ and give the corresponding deficiency indices in each case."

I can't seem to understand what they mean by "$5$ upper Jordan canonical forms"? Isn't the answer unique and straightforward with the canonical form:

$$J_= \left[ {\begin{array}{cccc} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 1 & 0 \\ 0 & 0 &\lambda & 1 \\ 0 & 0 & 0 & \lambda \end{array} } \right]$$

And what do they mean by deficiency indices? Are those the "$1$'s" that appear on top of each $\lambda$?

The other four Jordan canonical forms are $$\begin{bmatrix} \lambda & 0 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda \end{bmatrix} \quad \begin{bmatrix} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda \end{bmatrix} \quad \begin{bmatrix} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 1 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda \end{bmatrix} \quad \begin{bmatrix} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & \lambda \end{bmatrix}$$ The deficiency indices of a matrix $A\in\mathcal{M}_{n\times n}(\mathbb{C})$ are the numbers $$n_{\pm}(A)=\dim\ker(A^*\mp i\cdot I_n)$$ Since $\lambda$ is assumed to be real, computing these numbers should be straight forward.
• Do you know why for a $2x2$ matrix there are only 2 deficiency indices, for a $3x3$ matrix there are only 3 deficiency indices, yet for a $4x4$ there are 5? Perhaps the formula above holds the answer? – Arturo Feb 7 '14 at 7:31
• Why can't $J_= \left[ {\begin{array}{cccc} \lambda & 0 & 0 & 0 \\ 0 & \lambda & 1 & 0 \\ 0 & 0 &\lambda & 1 \\ 0 & 0 & 0 & \lambda \end{array} } \right]$ be a Jordan canonical form? – Arturo Feb 7 '14 at 8:24
• Oh ok, this makes sense, I can now see why $J_= \left[ {\begin{array}{cccc} \lambda & 0 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 &\lambda & 1 \\ 0 & 0 & 0 & \lambda \end{array} } \right]$ wouldn't produce an extra form since it is also a reordering of Jordan blocks in the second matrix! – Arturo Feb 7 '14 at 8:30