(From Axler)Why does the basis satisfy the required conditions?

Question

Suppose $w1, ⋯⋯, wn$ is a basis of $W$ and $V$ is finite-dimensional. Suppose $T∈L(V,W)$. Prove that there exists a basis $v1 ⋯⋯, vm$ of $V$ such that all the entries in the first row of $M(T)$ (with respect to the bases $v1, ⋯⋯, vm$ and $w1, ⋯⋯, wn$) are 0 except for possibly a $1$ in the first row, first column.

Here is a proof.

http://linearalgebras.com/3C.html (Q5 in the page)

Could someone explain how to get the basis that I need? (i.e. the author seems to omit the process of obtaining the answer. He just gives the answer without explanations)

Thanks!

Answer 1

If $T$ is the zero map, the matrix will be the zero matrix. More generally, if the image of $T$ is in the span of $w_2,\ldots, w_n$, the first row will be all zero, no matter how we choose the basis of $V$. So we may assume that there exists some $v\in V$ such that $T(v)=a_1w_1+\ldots +a_nw_n$ with $a_1\ne 0$. Then with $v_1:=\frac1{a_1}v$ we will have $T(v_1)=w_1+\ldots$, i.e., a $1$ in the top left. Now extend $v_1$ to any basis $v_1,\tilde v_2.\tilde v_3,\ldots,\tilde v_n$. We have $T(\tilde v_k)=b_kw_1+\ldots$ and therefore if we let $v_k=\tilde v_k-b_kv_1$, we ensure that $T(v_k)=0\cdot w_1+\ldots$, as desired. Remains to see that this is a basis. But that is clear because we can obain the $\tilde v_k$ back.

Answer 2

The idea is this: If $T\in L(V,W)$. Let $B=\{u_1,...,u_m\}$ be a basis of $V$ and let $C=\{w_1,...,w_n\}$ the given basis of $W$. Matrix of $T$ with respect to $B$ and $C$ has $Tu_i$'s as columns. Applying elementary column operations on the matrix won't change the column space of $T$ and the corresponding operations on $u_i$'s won't affect their linear independence.

Let's proceed with the idea developed above. There exist scalars $c_{ij}$'s such that $Tu_i=\sum_{j=1}^n c_{ij}w_j$ for every $1\le i\le m$.

Suppose that $k$ is the smallest index in $\{1,2,...,m\}$ such that $c_{1k}\ne 0$.

If such a $k$ doesn't exist, then the required basis is $B$ and we are done.

So WLOG (Refer Note)let $k=1$. Let's let's turn all non zero $c_{1j}$ into $1$ by dividing by appropriate numbers.

Let's define: $u_1':=\frac {u_1}{c_{11}}, u_i':=\frac {u_i}{c_{1i}}$ for those $1\le i\le m$ for which $c_{1i}\ne 0$ and $u_i'=u_i$ for those $1\le i\le m$ for which $c_{1i}=0$. So now all non zero entries of first row of $T$ are converted to $1$.

To eliminate extra ones, let's again define $u_1'':=u_1', u_i'':=u_i'-u_1'$ for those $2\le i\le m$ such that $(1,i)$th entry of $T$ is non zero and $u_i'':= u_i'$ for those $2\le i\le m$ for which $(1,i)$th entry of $T$ is $0$.

Based on the idea developed in first para, it may be verified that $T(u_i'')'$s are linearly independent.

Note that for any $i\gt 1$, $Tu_i''= Tu_i'-Tu_1'=\frac {Tu_i}{c_{1i}}-\frac{Tu_1}{c_{11}}$ is in span $(w_2,...,w_n)$; and that $Tu_1''$ has coefficient of $w_1$ as $1$.

NOTE: If $k$ is not equal to $1$ then we may rearrange $u_i$'s to get $u_1'=u_k, u_i'=u_i$ for $m\ge i\ge 2$ and then instead of $B$, the proof will use $B':=\{u_1',u_2',...,u_m'\}$