If $\varphi|_W$ and $\bar{\phi}$ are nonsingular prove that $\varphi$ is nonsingular.

Question

Here is the question I am trying to answer the second and the third part of it:

If $W$ is a subspace of the vector space $V$ stable under the linear transformation $\varphi$(i.e., $\varphi(W) \subseteq W$), show that $\varphi$ induces linear transformations $\varphi |_{W}$ on $W$ and $\bar{\varphi}$ on the quotient vector space $V/W.$ If $\varphi|_W$ and $\bar{\phi}$ are nonsingular prove that $\varphi$ is nonsingular. Prove the converse holds if $V$ has finite dimension and give a counterexample with $V$ infinite dimensional.

I know that a linear transformation $f$ is nonsingular iff ker f = 0, but still I do not know how to prove the second and the third part of the above question. Any hint will be greatly appreciated.

Edit: Here is my trial depending on the given comments below:

Since we know that a linear transformation $f$ is nonsingular iff $\ker f = 0.$ and since we are given that ker $\bar{\varphi} = 0$ and ker $\varphi|_W =0$ and since my definition for $\bar{\varphi}(x)$ is $ \varphi (x) + W$ then $x \in ker \bar{\varphi}$ implies that $\bar{\varphi}(x) = W$ which means that $\varphi(x) \in W.$ but then how can I use the assumption that ker $\varphi|_W =0$? I know that from this assumption, I can conclude that $\varphi(w) = 0$ iff $w=0$ but then how can I complete the proof till the end? Could someone help me please?

Answer 1

Some hints, assuming in your problem "nonsingular" means injective (or equivalently having zero kernel):

1. If $\varphi|_W$ and $\overline{\varphi}$ are both nonsingular, suppose that $\varphi(x)=0$. What can you say about $\overline{\varphi}(\overline{x})=\overline{\varphi(x)}$, where here $\overline{z}=z+W\in V/W$. What does this tell you about $\overline{x}$? Why does that tell you that $x\in W$? Finally, why does that tell you that $x=0$?

2. If $V$ is finite-dimensional and $\varphi$ is nonsingular, why is it immediate that $\varphi|_W$ is also nonsingular? Why does this tell you that $\varphi(W)=W$? Now suppose $\overline{\varphi}(\overline{x})=0$ where $x\in V$. Why does this tell you that $\varphi(x)\in W$? Why does this tell you that $x\in W$? Finally, why does that tell you that $\overline{x}=0$?

3. Consider the space $V$ of infinite sequences $(x_1,x_2,\ldots)$ and $\varphi:V\to V$ given by $\varphi(x_1,x_2,\ldots)=(0,x_1,x_2,\ldots)$.

Answer 2

Addition for the 3 above: Suppose $V=\mathbb R^\infty$, and $\varphi((x_1,x_2,\cdots))=(0,x_1,x_2,\cdots)$. Then $Ker(\varphi)=0$, that is $\varphi$ is nonsingular. Suppose $W=\{(0,x_2,x_3,x_4,\cdots)|x_i\in\mathbb R,i\ge2\}$, then $V/W=\{(x_1,0,\cdots)+W|x_1\in\mathbb R\}$. Then $\bar\varphi(V/W)=0$, $Ker(\bar\varphi)=V/W$, that is $\bar\varphi$ is not nonsingular.