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Understanding a statement about composite linear maps

Mathematics Asked on November 29, 2021

I know the definition of a bijective linear map $L: V rightarrow W$ (one that is both injective and surjective). However, my book states that it is equivalent to assert that its inverse $K: W rightarrow V$ satisfies $K circ L = Id_V$ and $L circ K = Id_W$.

Why? This is given in my book completely out of the blue, so a hint/explanation in how to prove it would be much appreciated.

EDIT

Let’s show that $K circ L = text{Id}_{V}$ implies that $L$ is injective.

Let $v, u in V$ be arbitrary and suppose that $L(v) = L(u)$. Applying $K$ to both sides, we see that

$$ (K circ L)(u) = K(Lu) = K(Lv) = (K circ L)(v). $$

Now since $K circ L = text{Id}_{V}$, we see that

$$ u = (K circ L)(u) = (K circ L)(v) = v $$

so $u = v$ and $L$ is injective.

QED.

Let’s show that $L circ K = text{Id}_{W}$ implies that $L$ is surjective.

Let $w in W$ be arbitrary. Now since $L circ K = text{Id}_{W}$, we see that $L(K(w)) = w$. Now notice that $K(w) in V$. Setting $v = K(w)$, we conclude that there exists a $v$ in $V$ such that $L(v) = w$. Since $w$ was arbitrary, we can conclude that $L$ is surjective.

QED.

One Answer

We can show that a linear map $L: V rightarrow W$ is bijective if and only if it is invertible. Invertible meaning that there exists some linear map $K: W rightarrow V$ such that

$$ K circ L = text{id}_{V} $$ and

$$ L circ K = text{id}_{W} .$$

$(Rightarrow)$ Suppose $L$ is a bijective linear map. It follows that $L$ has an inverse linear map (do you know why?). Furthermore, the inverse is unique so we denote it by $L^{-1}$. Then

$$ L^{-1} circ L = text{id}_{V} $$ and

$$ L circ L^{-1} = text{id}_{W} .$$

$(Leftarrow)$ Now suppose that $L$ is an invertible linear map, so there exists some linear map $K: W rightarrow V$ such that

$$(1) hspace{0.5cm} K circ L = text{id}_{V} $$ and

$$(2) hspace{0.5cm} L circ K = text{id}_{W} .$$

You can check that $(1)$ above implies that $L$ is injective and $(2)$ implies that $L$ is surjective, so $L$ must be a bijection.

Answered by Kevin Aquino on November 29, 2021

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