Linear transformations with respect to a basis

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?

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

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

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