1
$\begingroup$

Let $U$ and $V$ be finite dimensional vector spaces, and let $A\in \operatorname{Hom}(U,V)$. Show that if $A$ is injective, then there exists $B\in \operatorname{Hom}(U,V)$ such that $BA=1_U$.

Assume $A$ is injective. Let $\{y_1,\dots, y_n\}$ be a basis for $U$. Define $x_i=Ay_i$ for all $i$. Clearly, $x_i$ is unique. I do not know if $\{x_1,\dots, x_n\}$ spans $V$, but I guess that it is linearly independent, which will be shown. Let $\alpha_i\neq 0$ be some given scalars, then $0=\sum_{i=1}^n\alpha x_i=A(\sum_{i=1}^n \alpha y_i)$ by linearity of $A$. Then it implies that $\alpha_i=0$ for all $i$, which is a contradiction. Hence $\{x_1,\dots, x_n\}$ is linearly independent. Now we can define a linear map $B:V\to U$ by $Bx_i=y_i$ for all $i$. This clearly shows that $B$ is a left inverse of $A$, proving this problem. I am not sure if this answers the problem. If there is a way to fix this or another shorter way to prove, please enlighten me.

$\endgroup$
  • $\begingroup$ Instead of a basis for $U$, start with a basis for ${\cal R}A$ and extend to a basis of $V$. $\endgroup$ – copper.hat Oct 19 '16 at 20:16
0
$\begingroup$

The idea is correct, but the proof has some flaws.

The set $\{x_1,\dots,x_n\}$ is linearly independent. Indeed, if $$ 0=\sum_{i=1}^n \alpha_ix_i=\sum_{i=1}^n \alpha_iA(y_i)= A\biggl(\,\sum_{i=1}^n \alpha_iy_i\biggr) $$ from injectivity of $A$ we get $\sum_{i=1}^n\alpha_iy_i=0$. Therefore $\alpha_1=\dots=\alpha_n=0$ by linear independence of $\{y_1,\dots,y_n\}$.

It follows that $\{x_1,\dots,x_n\}$ can be extended to a basis $\{x_1,\dots,x_n,x_{n+1},\dots,x_m\}$ of $V$.

Defining $B$ by $$ B(x_i)=\begin{cases} y_i & \text{if $1\le i\le n$}\\[4px] 0 & \text{if $n<i\le m$} \end{cases} $$ provides the required left inverse, because, for $1\le i\le n$, $$ BA(y_i)=B(x_i)=y_i $$


It can happen that $m=\dim V=n=\dim U$, but this is not in the hypotheses. Certainly $m\ge n$, because we find a linearly independent set with $n$ elements in $V$.

The use of injectivity of $A$ and the possibility of extending a linearly independent set to a basis should be mentioned. We also use the fact that a linear map can be uniquely defined by its action on a basis.

If you have doubts whether mentioning $x_{n+1}$ is “legal”, split the proof into the cases $\dim V=\dim U$ and $\dim V>\dim U$. However the above argument is common: in the case $m=n$, it is understood that no vector is added.

$\endgroup$
0
$\begingroup$

In general, if $f:U \to V$ is injective, there is an inverse $g$ defined on the range by $g(f(x)) = x$. Injectivity shows that this is well defined.

For linear maps, the burden of definition is reduced somewhat since we only need to define the map on a linearly independent subset.

Let $y_1,...,y_r$ be a basis for ${\cal R} A$. Note that there are linearly independent $x_1,...,x_r$ such that $y_i = A x_i$. Define $B y_i = x_i$.

Note that $x_1,...,x_r$ form a basis for $U$. If not, this would contradict injectivity.

Now add $y_k$ such that $y_1,...,y_{\dim{V}}$ form a basis for $V$ and define $B y_i = 0$ for $i>r$.

Then $BA x = x$ for all $x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.