Proof that $AI_n = A$ using $Ab_i$. I have this problem which I understand, but can't put into mathematical words.
Let $A$ denote a nxn matrix and let $b_i$ denote the nx1 column vector whose i-th entry is equal to 1 and whose other entries are equal to 0.
Prove that $Ab_i$ is equal to the i-th column of A.
Answer
The matrix A can be presented in the following form:
$A = \begin{pmatrix} a_{11}&a_{12}&\dots&a_{1i}&\dots&a_{1n}\\
a_{21}&a_{22}&\dots&a_{2i}&\dots&a_{2n} \\ \dots \end{pmatrix}$
Then, $Ab_i$ will yield the ith-column of A.

Now, let $I_n$ denote the nun matrix. Using the fact that $Ab_i$ is equal to the i-th column of A, prove that $AI_n = A$.
I want to say that $I_n = (b_1 + b_2 + \dots + b_n)$ but that is not true as far as matrix addition is concerned. I would need to replace $+$ by some "append" operator...
After, I would say
$$AI_n = A(b_1 + b_2 + \dots + b_n) = Ab_1 + Ab_2 + \dots + Ab_n = A$$ because each $Ab_i$ yields a unique column of A, thus "recreating $A$".
 A: Here we are defining the matrix as a row vector whose elements are column vectors.  So you define a couple of items;
Let $\mathbf{b}_i$ be the column vector $ \begin{bmatrix}
        b_1 \\
        \vdots \\
        b_i \\
        \vdots \\
        b_n \\
        \end{bmatrix}
$ whose $i$'th entry is $1$ and $0$ otherwise.  Thus, $\mathbf{b}_2=\begin{bmatrix}
        0 \\
        1 \\
        \vdots \\
        0 \\
        \end{bmatrix}$
Also, for our matrix $A$, we define $A_i=A\mathbf{b}_i$, which we know is the $i$'th row of our matrix $A$.  
Finally, define $I_n=(\mathbf{b}_1 ... \mathbf{b}_i ... \mathbf{b}_n)$.  This is a row vector whose elements are the column vectors.  There are no operations between these vectors like $+$ or $\cdot$.  Thus, you can see that our $I_n$ is just the identity matrix written a little differently.  Now using only properties of matrix multiplication,
$$AI_n=A(\mathbf{b}_1 ... \mathbf{b}_i ... \mathbf{b}_n)=(A\mathbf{b}_1...A\mathbf{b}_i...A\mathbf{b}_n)=(A_1...A_i...A_n)=A$$
Now, this entire argument is based on the fact that matrix multiplication on a vector is in fact a linear transformation.  I have not gone through the whole argument here for simplicity sake (the argument involves properties of Linear Tranformations, coordinate vectors, etc), but in reality think of what is happening here... Each row of $A$ is being multiplied by each individual column vector and producing the column of A, because of the make up of our defined $I_n$.  
If we define the rows of $A$ as $\mathbf{a}_i$, then matrix multiplication is just the dot product of rows of $A$ with columns of $I_n$, so the first element, $\mathbf{a}_1 \cdot \mathbf{b}_1=\sum_{k=1}^n{a_{1,k}b_k}=a_{1,1}b_1...+a_{1,n}b_n=a_{1,1}$, since $b_1=1$ and $b_j=0, j\neq{1}$.
