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.

  • $\begingroup$ 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

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.


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 $x$.


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