Inverse by left multiplication but not right?

Question

Suppose $T: V \rightarrow W$ and $U: W \rightarrow V$ are linear transformations. It is known that $ U = T^{-1}$ if $UT = I_V$ and $TU = I_W$.

Is it possible to then also have a transform $Z: W \rightarrow V$ such that $ZT = I_V$ but $TZ \neq I_W$ (and likewise, $TZ = I_W$ but $ZT \neq I_V$)?

It is not quite clear from the question: do you assume that $T^{-1}$ exists? — A.Γ., Oct 25 '18 at 17:18

score 2 · Accepted Answer · answered Oct 25 '18 at 17:14

It is indeed possible. For instance, take $W = \Bbb R^2, V = \Bbb R$, $$ Z(x,y) = x $$ If we define $T(x) = (x,0)$, then we indeed have $ZT = I_{V}$ but $TZ \neq I_W$. The reverse happens if we exchange the roles of $Z$ and $T$.

For a fixed $V,W$ with $\dim(V) = \dim(W) < \infty$, no such pair exists.

In the infinite dimensional setting, the left and right shift operators are an example of such a pair for which $V = W$.

score 2 · Answer 2 · answered Oct 25 '18 at 17:15

$\begin{bmatrix}1&0\end{bmatrix} \begin{bmatrix}1\\0\end{bmatrix}=1=I_1$ but $\begin{bmatrix}1\\0\end{bmatrix} \begin{bmatrix}1&0\end{bmatrix}= \begin{bmatrix}1&0\\0&0\end{bmatrix}\neq I_2$

It’s also worth pointing out that (for transformations over fields, anyway) that theorem only holds when $V$ and $W$ have equal dimension.

score 1 · Answer 3 · answered Oct 25 '18 at 17:23

If W, V are not the same dimension...e.g. suppose $V$ is $\mathbb R^3$ and $W$ is $\mathbb R^2$

Then there can exist $T, U$ such that $TU = I_W$ but since the rank of $T$ has rank less than $3$ it is not possible for $UT$ to be rank $3$ and be an the identity matrix of $\mathbb R^3$

Inverse by left multiplication but not right?

3 Answers3