I'm reading some Linear algebra notes I found online, and am a bit confused about the following:
If $U,V$ are finite dimensional $\mathbb{C}$-spaces with bases $(\mathbf{u}_1,\dots,\mathbf{u}_m)$ and $(\mathbf{v}_1,\dots,\mathbf{v}_n)$ then there is a correspondence:
$$\mathrm{M}_{n\times m}(\mathbb{C})\longleftrightarrow \mathcal{L}(U,V);\;\;A\longleftrightarrow \alpha\;\text{where}\;\alpha(\mathbf{u}_i)=\sum_j A_{ji}\mathbf{v}_j$$
I'm confused about is the definition of $\alpha$. Why is $\mathbf{u}_i$ being sent to $\left(\mathbf{v}^{\text{T}}\cdot A^{\text{T}}\right)_i?$ (If I am reading that right that is). Why is $\alpha$ acting on $\mathbf{u}$ not the same as $A$ acting on $\mathbf{u}$ i.e. so $\left(A\cdot\mathbf{u}\right)_i$ instead?
Similarly it says that if $\alpha$ is represented by $A$ w.r.t. bases $\{\mathbf{u}_i\},\{\mathbf{v}_j\}$ and $\tilde A$ w.r.t. bases $\{ \mathbf{\tilde u}_i\},\{\mathbf{\tilde v}_j\}$ where $\mathbf{\tilde u}_i=\sum_k P_{ki}\mathbf{u}_k$ and $\mathbf{\tilde v}_j=\sum_k Q_{kj}\mathbf{v}_k$ then: $$\tilde A=Q^{-1}AP$$
Again it feels like things are the wrong way around. To me it looks like $P$ takes $\{\mathbf{u}_i\}\to\{\mathbf{\tilde u}_i\}$ and $Q$ takes $\{\mathbf{v}_i\}\to\{\mathbf{\tilde v}_i\}$ (both in a weird transpose/row vector way) but then the expression above suggests the opposite.
I realise this is really basic but I'm having a brain-freeze moment here.
$\endgroup$ 52 Answers
$\begingroup$Choose bases vectors $(\mathbf{u}_1,\dots,\mathbf{u}_m)$ and $(\mathbf{v}_1,\dots,\mathbf{v}_n)$, then the linear map will be completely determined by where it maps each $\mathbf{u}_i$ to. This can be written in matrix form: $$\alpha(\mathbf{u}_1 \mathbf{u}_2 \cdots \mathbf{u}_m)=(\mathbf{v}_1 \mathbf{v}_2 \cdots \mathbf{v}_n) \left(\begin{array}{cccc} A_{11} & A_{12} & A_{13} & \cdots & A_{1m} \\ A_{21} & A_{22} & A_{23} & \cdots & A_{2m} \\ \vdots & \vdots & \vdots &\cdots &\vdots \\ A_{n1} & A_{n2} & A_{n3} &\cdots & A_{nm}\end{array}\right)$$
Now general vectors can be written as $$\mathbf{u}=X_1\mathbf{u}_1+X_2\mathbf{u}_2+\cdots+X_m\mathbf{u}_m=(\mathbf{u}_1 \mathbf{u}_2\cdots\mathbf{u}_m)\left(\begin{array}{c} X_1\\X_2\\\vdots\\X_m\end{array}\right)$$ $$\mathbf{v}=Y_1\mathbf{v}_1+Y_2\mathbf{v}_2+\cdots+Y_n\mathbf{v}_n=(\mathbf{v}_1 \mathbf{v}_2\cdots\mathbf{v}_n)\left(\begin{array}{c} Y_1\\Y_2\\\vdots\\Y_n\end{array}\right)$$ After choosing a basis, the vectors can be identified with the column vectors in $\mathbb{F}^m$ and $\mathbb{F}^n$.
Now, by linearity $$\alpha(\mathbf{u})=\alpha\left[(\mathbf{u}_1 \mathbf{u}_2\cdots\mathbf{u}_m)\left(\begin{array}{c} X_1\\X_2\\\vdots\\X_m\end{array}\right)\right]$$ $$=(\mathbf{v}_1 \mathbf{v}_2 \cdots \mathbf{v}_n) \left(\begin{array}{cccc} A_{11} & A_{12} & A_{13} & \cdots & A_{1m} \\ A_{21} & A_{22} & A_{23} & \cdots & A_{2m} \\ \vdots & \vdots & \vdots &\cdots &\vdots \\ A_{n1} & A_{n_2} & A_{n3} &\cdots & A_{nm}\end{array}\right)\left(\begin{array}{c} X_1\\X_2\\\vdots\\X_m\end{array}\right)$$ $$=(\mathbf{v}_1 \mathbf{v}_2\cdots\mathbf{v}_n)\left(\begin{array}{c} Y_1\\Y_2\\\vdots\\Y_n\end{array}\right)$$ Hence $$\left(\begin{array}{c} Y_1\\Y_2\\\vdots\\Y_n\end{array}\right)=\left(\begin{array}{cccc} A_{11} & A_{12} & A_{13} & \cdots & A_{1m} \\ A_{21} & A_{22} & A_{23} & \cdots & A_{2m} \\ \vdots & \vdots & \vdots &\cdots &\vdots \\ A_{n1} & A_{n2} & A_{n3} &\cdots & A_{nm}\end{array}\right)\left(\begin{array}{c} X_1\\X_2\\\vdots\\X_m\end{array}\right)$$ This means $\alpha$ can be represented by a linear map from $\mathbb{F}^m$ to $\mathbb{F}^n$ by the above matrix multiplication.
Now, if we change basis to $$(\tilde{\mathbf{u}}_1 \tilde{\mathbf{u}}_2 \cdots \tilde{\mathbf{u}}_m)=(\mathbf{u}_1 \mathbf{u}_2 \cdots \mathbf{u}_m)\left(\begin{array}{cccc} P_{11} & P_{12} & P_{13} & \cdots & P_{1m} \\ P_{21} & P_{22} & P_{23} & \cdots & P_{2m} \\ \vdots & \vdots & \vdots &\cdots &\vdots \\ P_{m1} & P_{m2} & P_{m3} &\cdots & P_{mm}\end{array}\right)$$ $$(\tilde{\mathbf{v}}_1 \tilde{\mathbf{v}}_2 \cdots \tilde{\mathbf{v}}_n)=(\mathbf{v}_1 \mathbf{v}_2 \cdots \mathbf{v}_n)\left(\begin{array}{cccc} Q_{11} & Q_{12} & Q_{13} & \cdots & Q_{1n} \\ Q_{21} & Q_{22} & Q_{23} & \cdots & Q_{2n} \\ \vdots & \vdots & \vdots &\cdots &\vdots \\ Q_{n1} & Q_{n2} & Q_{n3} &\cdots & Q_{nn}\end{array}\right)$$ Then $$\mathbf{u}=(\mathbf{u}_1 \mathbf{u}_2\cdots\mathbf{u}_m)\left(\begin{array}{c} X_1\\X_2\\\vdots\\X_m\end{array}\right)=(\tilde{\mathbf{u}}_1 \tilde{\mathbf{u}}_2 \cdots \tilde{\mathbf{u}}_m)P^{-1}\left(\begin{array}{c} X_1\\X_2\\\vdots\\X_m\end{array}\right)=(\tilde{\mathbf{u}}_1 \tilde{\mathbf{u}}_2 \cdots \tilde{\mathbf{u}}_m)\left(\begin{array}{c} \tilde{X}_1\\\tilde{X}_2\\\vdots\\\tilde{X}_m\end{array}\right)$$ Hence $$\left(\begin{array}{c} \tilde{X}_1\\\tilde{X}_2\\\vdots\\\tilde{X}_m\end{array}\right)=P^{-1}\left(\begin{array}{c} X_1\\X_2\\\vdots\\X_m\end{array}\right)$$ Similarly, $$\left(\begin{array}{c} \tilde{Y}_1\\\tilde{Y}_2\\\vdots\\\tilde{Y}_n\end{array}\right)=Q^{-1}\left(\begin{array}{c} Y_1\\Y_2\\\vdots\\Y_n\end{array}\right)$$
In the new basis, $\alpha$ is determined by $$\alpha(\tilde{\mathbf{u}}_1 \tilde{\mathbf{u}}_2 \cdots \tilde{\mathbf{u}}_m)=(\tilde{\mathbf{v}}_1 \tilde{\mathbf{v}}_2 \cdots \tilde{\mathbf{v}}_n) \left(\begin{array}{cccc} \tilde{A}_{11} & \tilde{A}_{12} & \tilde{A}_{13} & \cdots & \tilde{A}_{1m} \\ \tilde{A}_{21} & \tilde{A}_{22} & \tilde{A}_{23} & \cdots & \tilde{A}_{2m} \\ \vdots & \vdots & \vdots &\cdots &\vdots \\ \tilde{A}_{n1} & \tilde{A}_{n2} & \tilde{A}_{n3} &\cdots & \tilde{A}_{nm}\end{array}\right)$$ Therefore we have $$\alpha\left((\mathbf{u}_1 \mathbf{u}_2 \cdots \mathbf{u}_m)P\right)=(\mathbf{v}_1 \mathbf{v}_2 \cdots \mathbf{v}_n)Q\tilde{A}$$ $$\alpha(\mathbf{u}_1 \mathbf{u}_2 \cdots \mathbf{u}_m)P=(\mathbf{v}_1 \mathbf{v}_2 \cdots \mathbf{v}_n)Q\tilde{A}$$ $$\alpha(\mathbf{u}_1 \mathbf{u}_2 \cdots \mathbf{u}_m)=(\mathbf{v}_1 \mathbf{v}_2 \cdots \mathbf{v}_n)Q\tilde{A}P^{-1}$$ Therefore, $$A=Q\tilde{A}P^{-1}$$ or $$\tilde{A}=Q^{-1}AP$$
$\endgroup$ $\begingroup$This system uses row vectors $u = (u_1, \dotsc, u_n)$. Then the definitions look a little bit different than with column vectors $u = (u_1, \dotsc, u_n)^T$, e.g. $$ x A = y $$ instead of $$ A x = y $$
$\endgroup$
