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I want to see if my understanding of change of basis matrices is correct.

Say I have a matrix $T$ which is 2x3 in dimension and takes in a vector represented in the standard canonical basis $e$ in $\mathbb{R}^{3}$ and outputs a vector in $\mathbb{R}^{2}$ with the standard canonical basis $f$.

Now say I have a new set of basis vectors $e'$ in $\mathbb{R}^{3}$ such that $e'_1 = e_1$, $e'_2 = e_1 + e_2$, and $e'_3 = e_1 + e_2 + e_3$. Also a new basis in $\mathbb{R}^{2}$ called $f'$ such that $f'_1 = f_1$ and $f'_2 = f_1 + f_2$.

If I want the transformation matrix $T'$ which takes from $e'$ in $\mathbb{R}^{3}$ to $f'$ in $\mathbb{R}^{2}$ I need to construct the two matrices that take from $e'$ to $e$ and from $f$ to $f'$.

Let $E$ be the matrix that transforms the basis from $e'$ to $e$. To construct this matrix I must represent $e'$ basis vectors in $e$ as columns of $E$.

Let $F$ be the matrix that transforms the basis from $f$ to $f'$. To construct this matrix I must represent $f$ basis vectors in $f'$ as columns of $F$.

With that I get:

$E = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{bmatrix}$

$F = \begin{bmatrix} 1 & -1 \\ 0 & 1 \\ \end{bmatrix}$

My understanding is that $T' = FTE$, is this correct?

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    $\begingroup$ Exactly. Do you understand the thinking behind this formula? $\endgroup$ – Ofek Gillon Apr 29 '17 at 11:25
  • $\begingroup$ @OfekGillon Forgot to mention your username in my first comment. Yes, I see it as we have some vector $[v]_{e′}$ and to go to $e$ we apply $E[v]_{e′}$ to get $[v]_e$. Then by applying $T[v]_e$ we take it to $\mathbb{R}^{2}$ in $f$ where it becomes $[v]_f$. Now we have $[v]_f$ but want $[v]_{f′}$ so we apply $F[v]_f$ to get $[v]_{f′}$. Thus we have $[v]_{f′}$ = $FTE[v]_{e′}$. As for representing the basis $e′$ in $e$ as columns in $E$, it makes sense because $E_{ij}$ represents how much of vector $e_i$ is in $e'_j$. $\endgroup$ – unamehere1 Apr 29 '17 at 14:08
  • $\begingroup$ You understand the concept fully :) $\endgroup$ – Ofek Gillon Apr 29 '17 at 14:15
  • $\begingroup$ @OfekGillon Great, thanks for the help! $\endgroup$ – unamehere1 Apr 29 '17 at 14:17

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