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So I'm trying to prove or disprove the following:

Given a finite dimensional linear space $V$, two linear transformation $T,S$ from $V$ to $V$. Can $T$ be as such that $\ker TS =\{0\}$, but $\ker T \neq \{0\}$.

My thoughts are as follows:

I have to find a $T$ that is not injective but a $TS$ that is injective. I also know that you can represent any linear transformation with a matrix. So maybe I need to show that it is possible to multiply two matrices and get the identity matrix but $T$ is not as such. Or maybe two matrices that their product's determinant is not zero but $T$'s determinant is zero. I'm not sure if all my thoughts are correct or how to go about proving this.

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  • $\begingroup$ Intuitively, since linear transformations must map the zero vector to itself, composing a linear transformation with another can never make the kernel smaller. $\endgroup$
    – amd
    Jan 13, 2018 at 1:26

3 Answers 3

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If $\ker(TS)=\{0\}$, then $TS$ is invertible and therefore $\det(TS)\neq0$. But $\det(TS)=\det(T)\det(S)$, and therefore $\det(T)\neq0$. This is impossible, since $\ker(T)\neq\{0\}$.

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Your thoughts about determinants are precisely the correct way of thinking about this.

We want to:

$T$ is not injective but $T\circ S$ is injective

In terms of matrices, this is equivalent to:

$T$ is singular but $T\circ S$ is non-singular

Here, "Singular" just means it has $0$ as an eigenvalue, or there's some $v\neq 0$ such that $T(v) = 0v = 0$. It's well-known that failing to be injective is equivalent to being singular.

Since we've turned this into a question about eigenvalues, looking at $\det T$ now makes sense, as $\det T = \prod_i \lambda_i$ is the product of the eigenvalues. So, transforming the question into this language, we get:

$\det T = 0$ but $\det (TS)\neq 0$

You may know that $\det (AB) = \det(A)\det(B)$ (Note that it's important here that $A,B$ are both $V\to V$. If this isn't the case we must be more careful). I claim that this is enough to get your result. I'd encourage you think about it yourself, and below I'll put the answer for completeness.

We want $T$ such that $\det T = 0$, but $\det (TS)\neq 0$. As we have $\det(TS) = \det(T)\det(S)$, we clearly have that both conditions can't be satisfied (as if we have $\det(T) = 0$, then $\det(TS) = \det(T)\det(S) = 0\det(S) = 0$).

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Since $TS$ is a map from $V$ to itself, injectivity means bijectivity, so $T$ and $S$ are bijective.

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