Why the first row of the product $MJM^{-1}$ is just the first row of $M$ multiplied by $JM^{-1}$? Let's define $A_{1,\cdot}$ as the first row of a matrix $A$. The Jordan form of $A$ is $J = M^{-1}AM$. Thus we can write $A = MJM^{-1}$.
Question is, what's $(MJM^{-1})_{1,\cdot}$? My textbook suggests (without providing ant proof, as it's quite obvious) that:
$$
A_{1,\cdot} = (MJM^{-1})_{1,\cdot} = M_{1,\cdot}JM^{-1}
$$
That is only the first row of the matrix $M$. Why? Sorry if this is obvious, but I can't get it...
 A: Let $\epsilon_j$ be the row vector with a $1$ in the $j$th slot and zeros everywhere else. Notice that, for any matrix $T$, $\epsilon_j T$ is the $j$th row of $T$. That means
$$\epsilon_j(MJM^{-1}) = (MJM^{-1})_{j,\cdot}$$
and
$$(\epsilon_j M)JM^{-1} = M_{j,\cdot}JM^{-1}$$
By associativity, the left-hand sides are equal!

Problems like this are often much easier to solve when you think about them abstractly—that's how I came up with the answer above.
To translate your matrix problem into an abstract linear algebra problem, you need three key facts:


*

*Let $V$ be a vector space over some field $\mathbb{k}$ (probably the field of complex numbers, in your case). The dual space $V^*$ is the space of linear maps from $V^*$ to $\mathbb{k}$. So, given a basis for $V$, you can write a dual vector (an element of $V^*$) as a $1 \times n$ matrix, where $n$ is the dimension of $V$. In other words, you can write a dual vector as a "row vector."

*If $f$ is a dual vector (a linear map from $V$ to $\mathbb{k}$), and $T$ is a linear map from $V$ to $V$, the composition $f \circ T$ is another dual vector.

*Let $e_1, \ldots, e_n$ be a basis for $V$. Let $\epsilon_j$ be the dual vector which sends $e_j$ to one and sends all the other basis vectors to zero. In terms of the basis $e_1, \ldots, e_n$, the matrix for $\epsilon_j \circ T$ is the $j$th row of the matrix for $T$.

A: Do you know how to multiply two matrices? Let $r_i$ denotes the $i$-th row of a matrix $R$ and $S_j$ be the $j$-th column of a matrix $S$. Then by definition of matrix multiplication, the $(i,j)$-th entry of $RS$ is given by $r_is_j$, (i.e. the dot product of them). So we have
$$
RS
=\begin{pmatrix}r_1s_1&r_1s_2&\cdots&r_1s_n\\
r_2s_1&r_2s_2&\cdots&r_2s_n\\
\vdots&&&\vdots\\
r_ns_1&r_ns_2&\cdots&r_ns_n\end{pmatrix}
=\begin{pmatrix}r_1S\\ r_2S\\ \vdots\\ r_nS\end{pmatrix}.
$$
That is, the first row of $RS$ is $r_1S$. Now put $R=M$ and $S=JM^{-1}$.
