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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?

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    $\begingroup$ Here is a hint for the second part. Suppose that $x \in \ker\phi$. What can you say about $\overline{\phi}(x+W)$, and can you use that to show $x \in W$? $\endgroup$ Dec 3, 2023 at 22:07
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    $\begingroup$ Related fact: if $V$ is finite-dimensional, $\det(\varphi)=\det(\varphi|_W)\det(\overline{\varphi})$. $\endgroup$
    – blargoner
    Dec 3, 2023 at 22:19
  • $\begingroup$ @blargoner why is that correct? How this will help me in the proof? $\endgroup$
    – Intuition
    Dec 4, 2023 at 0:27
  • $\begingroup$ @ChrisEagle I have edited my question, do you have a justification for the question I asked in the edit? $\endgroup$
    – Intuition
    Dec 4, 2023 at 0:43
  • $\begingroup$ @blargoner I have an edit for my question that contains a question, do you have an answer for the question in the edit? $\endgroup$
    – Intuition
    Dec 4, 2023 at 0:45

1 Answer 1

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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 $\varphi(x_1,x_2,\ldots)=(0,x_1,x_2,\ldots)$.

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  • $\begingroup$ Nonsingular ought to mean invertible, which is not equivalent to injective in the infinite dimensional case. $\endgroup$
    – Allen Bell
    Dec 5, 2023 at 6:01
  • $\begingroup$ @AllenBell Take it up with the dead linear algebra book authors (like Hoffman and Kunze, etc.) who define "nonsingular" to mean having zero kernel or injective, and who separately define invertibility. Since the OP said he knew the equivalence with zero kernel, I assumed such a definition was probably used in his book. You're free to use the word in a different way in your own answer, as Humpty Dumpty suggests. $\endgroup$
    – blargoner
    Dec 5, 2023 at 13:24
  • $\begingroup$ Can you please elaborate more on 3? $\endgroup$
    – Intuition
    Dec 13, 2023 at 11:01

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