Let $T:V \rightarrow W$ and $U:W \rightarrow Z$ be linear transformations. Give an example where $UT$ is one-to-one, but $U$ isn't. Give an example where $UT$ is onto, but $T$ isn't.

I know that in the first scenario, then $U$ must be one-to-one from a previous part of the question, and that $T$ must be onto in the second part of the problem.

I'm having trouble finding linear transformations for this because all the examples I can think of for the first part are a bijection.

I'm having similar problems for the second one, but where T is onto as well.

I think I can use the same set for all of these, like $V=W=Z$, but I can't think of any. I've done something similar, but with finite sets and normal functions. However, these sets must be infinite because they describe a vector space.

  • $\begingroup$ If $T$ is not injective, then there exists some non zero vector $\mathbf v$ such that $T\mathbf v =\mathbf 0$, which means that $UT\mathbf v= \mathbf 0$ for a non zero vector $\mathbf v$. So $UT$ cannot be injective. $\endgroup$ – lEm Sep 23 '16 at 3:36
  • $\begingroup$ Sorry, I mistyped that. $U$ is the one which is not injective there. Edited. $\endgroup$ – AndroidFish Sep 23 '16 at 3:46
  • $\begingroup$ You can't use the same space for all three. For the first one, you must have $T$ 1-1 but not onto, and then $U$ be 1-1 on the image of $T$ (but not on all of $W$. $\endgroup$ – Gerry Myerson Sep 23 '16 at 7:11

$UT$ is injective means that $UT\mathbf v=\mathbf 0$ iff $\mathbf v =\mathbf 0$. Then it means that $U\mathbf w=\mathbf 0$ is only possible if $\mathbf w$ does not belong to $\mathrm{im}(T)-\{\mathbf 0\}$. Also notice that $T$ must be injective as I commented before. So all you need to find is just an injectivr $T$ and a non-injective map $U$ such that $\mathrm{ker}(U) \cap \mathrm{im}(T)=\{\mathbf 0\}$.

I also consider the case where $V,W,Z$ are finite dimensional. In which case, $UT$ is injective if $\dim V \leq \dim Z$. For $T$ to be injective, I also need $\dim W \geq \dim V$.

A simple example (with respect to standard basis) could be: $T:\mathbb R^2 \to \mathbb R^4$ given by

$$\begin{pmatrix} 1&0 \\ 0&1 \\0&0 \\ 0&0 \end{pmatrix}$$

And take $U: \mathbb R^4 \to \mathbb R^3$

$$\begin{pmatrix} 1&1&0&0\\ -1&1&0&0 \\ 1&1&0&0 \end{pmatrix}$$

Check that $UT: \mathbb R^2 \to \mathbb R^3$ has the matrix: $$\begin{pmatrix} 1&1 \\ 1&-1 \\ 1&1\end{pmatrix}$$

Now use a similar argument for the second part, $UT$ is surjective but $T$ is not. $\forall \mathbf z \in Z$, there exists $\mathbf v \in V$ such that $UT \mathbf v =\mathbf z$. Take $\mathbf w \in \mathrm{im}(T) \neq W$, then $U \mathbf w= \mathbf z$ implies that $U$ must be surjective. $\dim (\mathrm{ker}(U)=\dim W-\dim (\mathrm {im}(U))$ implies that $U$ is also non injective.

It is logical to take $\dim W > \dim Z$ for $U$ to not be injective.

Let $U: \mathbb R^4 \to \mathbb R^2$ given by:

$$\begin{pmatrix} 1&0&0&0 \\ 0&1&0&0 \end{pmatrix}$$

Take $T: \mathbb R^3 \to \mathbb R^4$

$$\begin{pmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \\ 0&0&0 \end{pmatrix}$$

Verify that: $UT: \mathbb R^3 \to \mathbb R^2 $ is surjective.

$$\begin{pmatrix} 1 &0&0 \\ 0&1&0 \end{pmatrix}$$


Take $V=W=Z=\mathbb R^{\mathbb N}$ the vector space of real sequences.

$U : (x_1, x_2, \dots) \mapsto (x_2, x_3, \dots)$ is the left shift operator and $T : (x_1, x_2, \dots) \mapsto (0, x_1, x_2, \dots)$ the right shift operator.

$U$ is not one-to-one and $T$ isn't onto. However $UT$ is the identity operator, so is one-to-one and onto.

For some more information, you can have a look here: MathCounterexamples.

  • $\begingroup$ No need to go to infinite-dimensional vector spaces, though. $\endgroup$ – Gerry Myerson Sep 23 '16 at 7:11

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