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I been staring at this for 2 hours trying to understand the last step, I can't figure out what they mean when they put the $\vec e_i$ to the left of the matrix.

Vector v can be represented in basis vectors, $ E = [\vec e_1 \vec e_2 \ldots \vec e_n]$ with coordinates $[v]_E = [v_1 v_2 \ldots v_n]$ :

$$\vec v = v_1 \vec e_1 + v_2 \vec e_2 + \cdots + v_n \vec e_n = \sum v_i \vec e_i = E [v]_E$$

$$A(\vec v) = A \left( \sum {v_i \vec e_i} \right) = \sum {v_i A(\vec e_i)} = [A(\vec e_1) A(\vec e_2) \ldots A(\vec e_n)] [v]_E =\; A \cdot [v]_E = [\vec e_1 \vec e_2 \ldots \vec e_n] \begin{bmatrix} a_{1,1} & a_{1,2} & \ldots & a_{1,n} \\ a_{2,1} & a_{2,2} & \ldots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \ldots & a_{n,n} \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n\end{bmatrix} $$

As far as I can tell this $ E = [\vec e_1 \vec e_2 \ldots \vec e_n]$ is just the matrix with basis vectors as columns. But that does not make sense to me. They finish everthing with:

The $a_{i,j}$ elements of matrix A are determined for a given basis E by applying A to every $\vec e_j = [0 0 \ldots (v_j=1) \ldots 0]^T$.

( what is $v_j$ doing inside this "thing"?) and then ending with:

And observing the response vector A $\vec e_j = a_{1,j} \vec e_1 + a_{2,j} \vec e_2 + \cdots + a_{n,j} \vec e_n = \sum a_{i,j} \vec e_i$. This equation defines the wanted elements, $a_{i,j}$, of j-th column of the matrix A.

For more info, https://en.wikipedia.org/wiki/Transformation_matrix#Finding_the_matrix_of_a_transformation

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  • $\begingroup$ $[v]_E$ should be $[v_1\ v_2\ \ldots\ v_n]^T$. That is, it should be a column matrix. $\endgroup$ – user137731 Oct 13 '15 at 15:20
  • $\begingroup$ @Bye_World I see that aswell..its actully e's on the left in the last equlity that startled me. $\endgroup$ – user1 Oct 13 '15 at 15:25
  • $\begingroup$ Let $ E = [\vec e_1 \vec e_2 \ldots \vec e_n]$ be a basis. Let $\vec v$ be a vector, then writing $$v= [\vec e_1 \vec e_2 \ldots \vec e_n] \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n\end{bmatrix} $$ is just a confusing way to say that, in basis $E$, $\vec v$ is expressed as the vector $[v_1 v_2 \ldots v_n]$. A much clear notation is to write $$[v]_E = \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n\end{bmatrix} $$ or $[v]_E = [v_1 v_2 \ldots v_n]$. $\endgroup$ – Ramiro Oct 13 '15 at 16:23
  • $\begingroup$ The expression $\vec e_j = [0 0 \ldots (v_j=1) \ldots 0]^T$ is again just a confusing notation to say that, in the basis $E$, $e_j$ is represented as a vector having all entries $0$ except in position $j$, where the entry is $1$. $\endgroup$ – Ramiro Oct 13 '15 at 16:25
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Am I right in assuming that this is the step you're having trouble with? $$A \cdot [v]_E = [\vec e_1 \vec e_2 \ldots \vec e_n] \begin{bmatrix} a_{1,1} & a_{1,2} & \ldots & a_{1,n} \\ a_{2,1} & a_{2,2} & \ldots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \ldots & a_{n,n} \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n\end{bmatrix}$$

Obviously $[v]_E = \begin{bmatrix}\vec e_1 & \vec e_2 & \ldots & \vec e_n\end{bmatrix}\begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n\end{bmatrix}$, right? Try multiplying it out if you don't understand this part.

That means that we just need to prove that $$\begin{bmatrix} a_{1,1} & a_{1,2} & \ldots & a_{1,n} \\ a_{2,1} & a_{2,2} & \ldots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \ldots & a_{n,n} \end{bmatrix}\left(\begin{bmatrix}\vec e_1 & \vec e_2 & \ldots & \vec e_n\end{bmatrix}\begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n\end{bmatrix}\right) = [\vec e_1 \vec e_2 \ldots \vec e_n] \begin{bmatrix} a_{1,1} & a_{1,2} & \ldots & a_{1,n} \\ a_{2,1} & a_{2,2} & \ldots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \ldots & a_{n,n} \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n\end{bmatrix}$$ I.e. that the block matrix of basis vectors commutes with the matrix $A$.


The line

The $a_{i,j}$ elements of matrix A are determined for a given basis E by applying A to every $\vec e_j = [0 0 \ldots (v_j=1) \ldots 0]^T$.

tells us that the vectors $e_i$ are also expressed with respect to the basis $E$ (I'm not at all sure why they labeled the $j$th component of $e_j$ as $v_j$ -- that seems to be a typo, but otherwise it's pretty clear what they meant). So because $e_1 = 1e_1 + 0e_2 + \cdots + 0e_n$, we see that $[e_1]_E = \begin{bmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ 0\end{bmatrix}$ and likewise $[e_2]_E = \begin{bmatrix} 0 \\ 1 \\ 0 \\ \vdots \\ 0\end{bmatrix}$, etc.

But then the matrix $\begin{bmatrix} e_1 & e_2 & \cdots & e_n\end{bmatrix}$ is just the identity matrix. And the identity matrix commutes with every other $n\times n$ matrix. So of course $A(Iv) = IAv$.

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  • $\begingroup$ I think its $E[v]_{E}$ after "Obviously" $\endgroup$ – user1 Feb 13 '16 at 6:23
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    $\begingroup$ Yeah, it looks like I made a mistake there. But either way a matrix whose columns are a basis represented with respect to themselves is always the identity matrix and thus you can stick $\pmatrix{e_1 & \cdots e_n}$ wherever you like. IMHO, Wikipedia uses awful notations here. $\endgroup$ – user137731 Feb 13 '16 at 14:51
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Interpret the $[\vec e_1 \vec e_2 \ldots \vec e_n]$ not as a matrix, but as a row of "arrows". Each vector $\vec e_i$ is just an arrow in space. Now when you multiply this "line" of "arrows" to a "column" of numbers you get: $x_1 \vec e_1 + x_2 \vec e_2 + \cdots + x_n \vec e_n$ - some big "arrow", which is the result vector.

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  • $\begingroup$ Im sorry I dont understand this. $\endgroup$ – user1 Sep 23 '15 at 4:01

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