[Math] Change of basis of a linear map defined by non-square matrix

change-of-basislinear algebralinear-transformationsmatrices

The given:

Let a linear map $L : V → U$ be given in the basis $(e_1, e_2, e_3)$ of $V$ and in the basis $(f_1, f_2)$ of $U$ by $\begin{pmatrix} 0 & 1 & 2 \\
3 & 4 & 5 \end{pmatrix}$
. Find the matrix of $L$ with respect to the bases
$(e_1, e_1 + e_2, e_1 + e_2 + e_3)$ and $(f_1, f_1 + f_2)$.

Now I know I am being stupid in some way, but I can't make this work. I want to say: we have new bases $(e_1,e_1+e_2,e_1+e_2+e_3)$ and $(f_1,f_1+f_2)$ which correspond to $C = \begin{pmatrix}
1 & 1 & 1 \\
0 & 1 & 1 \\
0 & 0 & 1
\end{pmatrix}$
and $D = \begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix}$
respectively. And Using our change of bases formula, one version of $L_1' = C^TAC$ and the other is $L_2' = D^TAD$. But here I run into a problem of dimension, the matrix multiplication does not work. I had thought perhaps to try $L'=DAC$, since this expression has workable dimension, but that's my only reason for trying it. The basis change examples I've seen before involve expressions with one matrix and either its transpose or inverse.

Apologies again for my stupidity here, and thanks very much in advance for any assistance.

Edit: Initial problem had typo on $D$, one-zero were swapped incorrectly.

Best Answer

I hate the change of basis formula. I think it confuses way too many people, and obscures the simple intuition going on behind the scenes.

Recall the definition of matrices for a linear map $T : V \to W$. If $B_1 = (v_1, \ldots, v_m)$ is a basis for $V$ and $B_2$ is a basis for $W$ (also ordered and finite), then we define $$[T]_{B_2 \leftarrow B_1} = \left([Tv_1]_{B_2} \mid [Tv_2]_{B_2} | \ldots | \, [Tv_n]_{B_2} \right),$$ where $[w]_{B_2}$ refers to the coordinate column vector of $w \in W$ with respect to the basis $B_2$. Essentially, it's the matrix you get by transforming the basis $B_1$, writing the resulting vectors in terms of $B_2$, and writing the resulting coordinate vectors as columns.

Such a matrix has the following lovely property (and is completely defined by this property):

$$[T]_{B_2 \leftarrow B_1} [v]_{B_1} = [Tv]_{B_2}.$$

This is what makes the matrix useful. When we compute with finite-dimensional vector spaces, we tend to store vectors in terms of their coordinate vector with respect to a basis. So, this matrix allows us to directly apply $T$ to such a coordinate vector to return a coordinate vector in terms of the basis on the codomain.

This also means that, if we also have $S : W \to U$, and $U$ has a (finite, ordered) basis $B_3$, then we have

$$[S]_{B_3 \leftarrow B_2}[T]_{B_2 \leftarrow B_1}[v]_{B_1} = [S]_{B_3 \leftarrow B_2}[Tv]_{B_2} = [STv]_{B_3},$$

and so

$$[ST]_{B_3 \leftarrow B_1} = [S]_{B_3 \leftarrow B_2}[T]_{B_2 \leftarrow B_1}.$$

Note also that, if $\mathrm{id} : V \to V$ is the identity operator, then

$$[\mathrm{id}]_{B_1 \leftarrow B_1}[v]_{B_1} = [v]_{B_1},$$

which implies $[\mathrm{id}]_{B_1 \leftarrow B_1}$ is the $n \times n$ identity matrix $I_n$. Moreover, if $T$ is invertible, then $\operatorname{dim} W = n$ and then

$$I_n = [\mathrm{id}]_{B_1 \leftarrow B_1} = [T^{-1}T]_{B_1 \leftarrow B_1} = [T^{-1}]_{B_1 \leftarrow B_2}[T]_{B_2 \leftarrow B_1}.$$

Similarly,

$$I_n = [\mathrm{id}]_{B_2 \leftarrow B_2} = [TT^{-1}]_{B_1 \leftarrow B_1} = [T]_{B_2 \leftarrow B_1}[T^{-1}]_{B_1 \leftarrow B_2}.$$

What this means is

$$[T]_{B_2 \leftarrow B_1}^{-1} = [T^{-1}]_{B_1 \leftarrow B_2}$$

From this, we can derive the change of basis formula. If we have a linear operator $T : V \to V$ and two bases $B_1$ and $B_2$ on $V$, then

\begin{align*} [T]_{B_2 \leftarrow B_2} &= [\mathrm{id} \circ T \circ \mathrm{id}]_{B_2 \leftarrow B_2} \\ &= [\mathrm{id}]_{B_2 \leftarrow B_1} [T]_{B_1 \leftarrow B_1} [\mathrm{id}]_{B_1 \leftarrow B_2} \\ &= [\mathrm{id}^{-1}]_{B_2 \leftarrow B_1} [T]_{B_1 \leftarrow B_1} [\mathrm{id}]_{B_1 \leftarrow B_2} \\ &= [\mathrm{id}]^{-1}_{B_1 \leftarrow B_2} [T]_{B_1 \leftarrow B_1} [\mathrm{id}]_{B_1 \leftarrow B_2} \\ \end{align*}

It's easy to see that, if $B_1$ is the standard basis for $V = \mathbb{F}^n$, then $[\mathrm{id}]_{B_1 \leftarrow B_2}$ is the result of putting the basis vectors in $B_2$ into columns of a matrix, and this particular case is the change of basis formula.

Now, this works for an operator on $\mathbb{F}^n$. You've got a linear map between two unspecified spaces, so this formula will not apply. But, we can definitely use the same tools. Let \begin{align*} B_1 &= (e_1, e_2, e_3) \\ B_1' &= (e_1, e_1 + e_2, e_1 + e_2 + e_3) \\ B_2 &= (f_1, f_2) \\ B_2' &= (f_1, f_1 + f_2). \end{align*} We want $[L]_{B_2' \leftarrow B_1'}$, and we know $[L]_{B_2 \leftarrow B_1}$. We compute

\begin{align*} [L]_{B_2' \leftarrow B_1'} &= [\mathrm{id} \circ L \circ \mathrm{id}]_{B_2' \leftarrow B_1'} \\ &= [\mathrm{id}]_{B_2' \leftarrow B_2} [L]_{B_2 \leftarrow B_1} [\mathrm{id}]_{B_1 \leftarrow B_1'}. \end{align*}

We know $[L]_{B_2 \leftarrow B_1}$, so we must compute the other two matrices. We have

$$[\mathrm{id}]_{B_1 \leftarrow B_1'} = \left([e_1]_{B_1} \mid [e_1 + e_2]_{B_1} | \, [e_1 + e_2 + e_3]_{B_1} \right) = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}.$$

Similarly,

$$[\mathrm{id}]_{B_2 \leftarrow B_2'} = \left([f_1]_{B_2} \mid \, [f_1 + f_2]_{B_2}\right) = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix},$$

and so

$$[\mathrm{id}]_{B_2' \leftarrow B_2} = [\mathrm{id}]^{-1}_{B_2 \leftarrow B_2'} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}.$$

Finally, this gives us,

$$[L]_{B_2' \leftarrow B_1'} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 & 2 \\ 3 & 4 & 5 \end{pmatrix} \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} -3 & -6 & -9 \\ 3 & 7 & 12 \end{pmatrix}.$$

Related Question