How to prove that the matrix product is always zero? Lets say I have the $m \times n$ matrix A, and I create the matrix: $C = [A \ I]$, and the matrix $\begin{bmatrix}
-I \\ A
 \end{bmatrix}$.
Then we get that: $[A \ I]\begin{bmatrix}
-I \\ A
 \end{bmatrix}=-A+A=0$.
But lets say that I permute the columns in C, and I do the same permutations of the rows in $\begin{bmatrix}
-I \\ A
 \end{bmatrix}$. Then my book (in linear programming) uses that the product is still zero. Is there a way I can show or prove this?
 A: A more general thing is true: If you have two matrices $A_{m\times n}$ and $B_{n\times k}$ and you permute the columns in $A$ and then you apply the same permutation but on the rows of $B$, then the products $AB$ and $A'B'$ are equal.
The reason is because applying a permutation on the columns of a matrix $A_{m\times n}$ has the same effect as multiplying $A$ on the right by an invertible permutation matrix $P$ (we get $P$ by applying the same permutation we want for the columns of $A$ on the columns of the identity matrix $I$). On the other hand, if we want to permute the rows of a matrix $B$, we multiply on the left by the inverse matrix $P$ now.
Then $A'B'=APP^{-1}B=AB$.
Look at the wikipedia page on elementary matrices for details.
A: Let us write the columns of $C$ as $C_1,C_2,\ldots,C_n$ and let us denote the rows of the other matrix $R$ as $R_1,R_2,\ldots R_n$. 
What we will prove is that for any two matrices $C$ and $R$, for which the product $CR$ makes sense, the product $CR$ is same as the product of the newly formed matrices after operating with the permutation $\sigma$ as described above.
 So the new matrices are $\sigma C$ and $\sigma R$ the $i,j$ and $k,l$th element of the new matrices are $c_{i,\sigma^{-1}j}$ and $r_{\sigma^{-1}k,l}$ respectively. So the new product is $\sum c_{i,\sigma^{-1}k}r_{\sigma^{-1}k,l}$ which after a change of running variable is the same as the initial product. 
A: Let $X \in \Re^{m\times l}$ with $X_i$ being its column vectors,
$Y \in \Re^{l\times n}$ with $Y_i^T$ being its row vectors. Then there is a fact about the product of $X$ and $Y$:
$$
XY =\sum_{i=1}^{l}X_iY_i^T
$$
As the permutations are the same in your question, $XY$ will not change.
