Verify my proof: For an invertible $A$ , $V\in F^n$ is an invariant subspace $V$ of $A$ $\iff$ $V$ is an invariant subspace $V$ of $Adj(A)$

Question

$$ v_0,v_1\in V $$ $$ V \text{ is an }A\text{-invariant subspace} $$ $$ Av_0=v_1 $$ $$\iff$$ $$ A^{-1}Av_0=v_0=A^{-1}v_1 $$

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$\text{Using }A^{-1}=\frac{1}{\operatorname {det}(A)}\operatorname{Adj}(A)\text{ :}$

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$$\iff$$ $$\frac{1}{\operatorname {det}(A)}\operatorname{Adj}(A)v_1=v_0$$ $$\iff$$ $$\operatorname{Adj}(A)v_1={\operatorname {det}(A)}v_0$$ $${\operatorname {det}(A)}v_0\in V$$

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Does this seal the proof that $V$ is an $\operatorname {Adj}(A)$ invariant subspace?

Answer 1

As already mentioned, I think your proof is correct though slightly confusing. I'd propose the following:

We're given that for all $\;v\in V\;$ there exists $\;u\in V\;$ such that $\;Av=u\;$ , and from here we get that $\;A^{-1}u=v\implies V\;$ is also $\;A^{-1}\,-$ invariant. Observe that since that since $\;A\;$ is an isomorphism, for any $\;u\in V\;$ there is always one single $\;v\in V\;$ such that $\;Av=u\;$, and this is so because we're given $\;AV\subset V\implies AV=V\;$ as we're in finite dimension and $\;A\;$ is bijective.

After that what you did is just fine, I believe.