# Is this a 4x4 Jordan form matrix?

I was asked to list all $4 \times 4$ matrixes in Jordan form (over the complex numbers). I found $8$ but a solution online has only $7$ of my $8$. My "extra" one is:

$$\left( \begin{matrix} a & 0 & 0 & 0 \\ 1 & b & 0 & 0 \\ 0 & 0 & c & 0 \\ 0 & 0 & 1 & d \\ \end{matrix} \right)$$

where $a, b, c, d$ are complex numbers.

Is my matrix in Jordan form? Why/why not?

I think you have misunderstood something concerning the Jordan forms. Given a matrix $A \in {\rm GL}(V)$, for $V$ an $n$-dimensional complex vector space. Then the associated Jordan Normal form, say $J_{A}$, is a matrix in a block diagonal form $$\begin{bmatrix} J_1&&& \\ &J_2&&\\ & & \ddots & \\ &&&J_n \end{bmatrix},$$ where each one of the diagonal blocks has the following form $$J_i= \begin{bmatrix} \lambda_i&1&& \\ &\lambda_i&1&\\ & & \ddots & \\ &&&\lambda_i \end{bmatrix}.$$ Of course the values showed up are not random, but are the eigenvalues of $A$, and the dimension of each block corresponds to the multiplicity of the eigenvalue it represents. Now I have made all this introduction/reminder (which you might be able to find in many different sources, books, including the lecture notes of the module you participate) and let you think what you have written in your question and why is wrong/ambiguous. Also if you understand the definition, the answer is straightforward, so I recommend to elaborate it on your own. Cheers! (of course further questions are welcomed).
If $a=b$ and $c=d$, then the following matrix is in Jordan form. Otherwise, it is not!
$$\left( \begin{matrix} a & 0 & 0 & 0 \\ 1 & b & 0 & 0 \\ 0 & 0 & c & 0 \\ 0 & 0 & 1 & d \\ \end{matrix} \right)$$
where $a, b, c, d$ are complex numbers.
No, it needs $a=b$ and $c=d$ to be a Jordan matrix. Also, it's supposed to be upper triangular.