Proof of linear transform being a basis for set $L(U,V)$ meyer's book

Question

Hi, I have trouble understanding what the author means and how he extracts the information from the areas of the proof I highlighted. In the first rectangle he means $B_L=(B_{11}, B_{12}, ... B_{1m}..., B_{nm}) $ so it's a basis with size $m \cdot n$ ?
Then in the second rectangle, why $B_{ji}=0 $ for $j\neq k$ and $v_i$ for $j=k$ ?

Answer 1

• Yes, $B_L$ has $m*n$ elements.

  Imagine $U=\mathbb R^n$ and $V=\mathbb R^m$ and pick the standard basis', then the $m\times n$ matrices where exactly one entry is $1$ and all other are $0$ form a basis for $\mathcal L(U,V)$.

• We want to calculate $B_{ji}(u_k)$, where $u_k$ is the k'th element of $\mathcal B$. We calculate $[u_k]_{\mathcal B}$, which is the $n$-vector which has entry $1$ at position $k$ and is $0$ elsewhere, so $[u_k]_{\mathcal B}=\begin{pmatrix}\xi_1,\xi_2,\ldots,\xi_n\end{pmatrix}^T$ with $\xi_j=\delta_{jk}=\begin{cases}1& j=k\\0&j\neq k\end{cases}$. From the definition, $B_{ji}(u_k)=\xi_j v_i = \begin{cases}v_i &j=k\\0&j\neq k\end{cases}$.

As for the question why $[u_k]_{\mathcal B} = (0~0~\dots~\underbrace{1}_{\text{ k}}~0~\dots 0)^T$, we calculate $[u]_{\mathcal B}$ by decomposing $u$ with respect to $\mathcal B$: $$u=\sum_{i=1}^n \lambda_i u_i$$ Then $[u]_{\mathcal B}=(\lambda_1,\lambda_2,\dots,\lambda_n)^T$ by definition. Since $u_k$ is an element of $\mathcal B$, we obtain the result.

Let's add an example: We consider $U=\mathcal P_2(\mathbb R)$, the vector space of polynomial with real coefficients of degree $\leq 2$. $\mathcal B=\{\,1,1+X,1+X+X^2\,\}$ is a basis of $U$. If we take $p=1+2X^2$, a vector in $U$, we can calculate its coordinates with respect to $\mathcal B$: $$p=2(1+X+X^2) -2(X+1) + 1$$ therefore $[p]_{\mathcal B}=\begin{pmatrix}1&-2&2\end{pmatrix}^T$. The basis-element $1+X$ has coordinates $[1+X]_{\mathcal{B}}=\begin{pmatrix} 0&1&0\end{pmatrix}^T$.

Answer 2

$\mathscr L(U,V)$ should have dimension $mn$. So that's how many elements in the basis.

Next, $B_{i,j}(u)=\xi_jv_i$ where $\xi=(\xi_1,\dots,\xi_n)=[u]_{\mathcal B}$, by definition. That is (as the author states), "pick off the $j$th coordinate of $u$ and attach it to $v_i$".

Now note that for $u_k\in\mathcal B$, $[u_k]_{\mathcal B}=(0,\dots,0,1,0,\dots,0)$ , where the $1$ is in the $k$th coordinate.

Your statement follows.