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Problem: I have a question which has a linear map $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ defined by $T(x,y,z)=(0,x,y)$. With an ordered basis $F = \{(1,1,1)^T, (1,1,0)^T, (0,1,1)^T\}$ and $E$ the standard basis.

Part of the solution says that $_{F}T_{E}=\begin{pmatrix}0 & 0 & 0\\1 & 1 &0 \\1 & 1 & 1\end{pmatrix}$.

Attempt: However I thought that we know the matrix which takes the standard basis vectors to the their mapping under $T$ is $_{E}T_{E}=\begin{pmatrix}0 & 0 & 0\\1 & 0 &0 \\0 & 1 & 0\end{pmatrix}$. So the $F$ basis vectors must go to $_{E}T_{E}(1,1,1)^T= (0, 1, 1)^T,\hspace{2mm}_{E}T_{E}(1,1,0)^T= (0, 1, 1)^T,\hspace{2mm}_{E}T_{E}(0,1,1)^T= (0, 0, 1)^T$ which are the images represented in the $E$ basis, of the mappings under $T$. So the matrix which takes the $F$ basis vectors w.r.t. $F$ to the image of the $F$ basis vectors w.r.t. $E$ under the transformation (i.e. $_{E} T _{F}$) would be $_{E}T_{F}=\begin{pmatrix}0 & 0 & 0\\1 & 1 &0 \\1 & 1 & 1\end{pmatrix}$.

Why is what I've done the opposite of the solution?

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Consider the coordinate mappings:

$X_F:\mathbb{R}^3\to\mathbb{R}^3$ (sends a vector to its coordinates in the basis F) and $X_E:\mathbb{R}^3\to\mathbb{R}^3$ (sends a vector to its coordinates in the basis E).

We note that (since E is the elementary basis and our vector space is just $\mathbb{R}^3$ already) $X_E^{-1}$ is just the identity map $1_{\mathbb{R}^3}$. Write E and F respectively as follows: $\{E^1,E^2,E^3\}$. and $\{F^1,F^2,F^3\}$.

Note that multiplying a 3x3 matrix $M\in M_3(\mathbb{R})$ by $E^j$ will just give us the $j^{th}$ column of M. So we compute $_ET_F$ as follows:

$_ET_FE^j = X_F(T(X_E^{-1}(E^j))) = X_F(T(E^j))$.

Hence:

$_ET_FE^1 = X_F(T(E^1)) = X_F((0,1,0)^T) = (1,1,0)^T$ $_ET_FE^2 = X_F(T(E^2)) = X_F((0,0,1)^T) = (0,1,1)^T$ $_ET_FE^3 = X_F(T(E^3)) = X_F((0,0,0)^T) = X_F(0) = (0,0,0)^T$ so that

$_ET_F = \begin{pmatrix} 1 &0 &0\\ 1 &1 &0\\ 0 &1 &0\end{pmatrix}$.

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  • $\begingroup$ Thank you! My lecturer has never explained it like this and now I finally understand it. $\endgroup$
    – user857843
    Jan 13 '21 at 11:39

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