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$V$ is a finite-dimensional vector space. $S,T$ are linear maps from $V$ to itself (i.e. operators).

I have to show that :

$ST$ is invertible $\Rightarrow$ Both $S,T$ are invertible.

[An operator $P$ on a vector space $V$ is said to be invertible iff there exists an operator $Q$ on $V$ such that $PQ(v)=QP(v)=v$ for all $v \in V$.]

Any help would be greatly appreciated.

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    Do you know that an operator is invertible if and only if it is injective and surjective? – 2017-02-13

2 Answers 2

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Hint If $L$ is the inverse of $ST$ then $$LST=I_V \, \mbox{ and } \, STL =I_V$$

Can you find an operator $L_0$ such that $L_0T=I_V$? Since $V$ is finite dimensional, this also implies that $TL_0=I_V$.

For $S$ try to find $L_1$ such that $SL_1 =I_V$.

Edit

Lemma If $T_1 ,T_2 : V \to V$ are linear, $V$ is finite dimensional and $T_1 T_2=I_V$ then $T_2T_1=I_V$.

Proof Let $B= \{ v_1, v_2,..,v_n \}$ be a basis in $V$ and let $$w_j=T_2 v_j$$

Then, since $T_1(w_1)=v_1,.., T_1(w_n)=v_n$ is a basis in $V$, it is trivial to show that $w_1,..,w_n$ are linearly independent, hence a basis in $V$.

Now, $$T_2 T_1 (w_j)=w_j$$ which shows that $T_2T_1 =I_v$ on a basis of $V$, hence on $V$.

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    I don't get the step $L_0T=I_V \implies TL_0=I_V$ from the fact that $V$ is finite-dimensional. How do I show this? – 2017-02-13
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    @Dragon Is the added lemma clear? – 2017-02-13
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    Yes. Now it's clear to me. Thanks! – 2017-02-13
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This can be proven in different ways. Although, depending on where you are in your course and what you are or are not allowed to use, some may be more appropriate than others.

Here's another one (although it's kind of "cheating", but that's what I like about it). Pick a basis for this finite-dimensional vector space, and then each operator can be expressed as a square matrix with respect to this basis. By the Invertible Matrix Theorem, an operator is invertible iff the determinant of its matrix is nonzero. And then use the property of the determinant of a product.

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    Thanks. That was a wonderful cheat so +1. But I'm looking for a proof based on linear maps rather than matrix and determinants. – 2017-02-13