Statements about the correlation between invertibility of an operator and the corresponding matrix

Question

Let $A\in M_n$. Prove the following:

$(a)\;T\in L\left(M_{n\times 1}\right),\;T(X)=AX$ is invertible $\iff\; A$ is invertible.

$(b)\;S\in L\left(M_n\right),\;S(X)=AX$ is invertible $\iff\; A$ is invertible.

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My attempt:

$(a)$

Since $T\in L\left(M_{n\times 1}\right)$, $X$ is a column matrix, i.e. $$X=\begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}$$ If $A$ is regular, then $\operatorname{rank}{A}=n\;\implies\;$ all the columns of $A$ are linearly independent.

Let $c_j$ denote columns of $A$.

$$AX\in M_{n\times 1}\implies AX=\displaystyle\sum_{j=1}^n\left(x_j\sum_{i=1}^n a_{ij}\right)=\sum_{j=1}^nx_jc_j$$ $$\sum_{j=1}^nx_jc_j=0\implies x_j=0\;\forall j\in\{1,\ldots,n\}\implies T(X)=0\iff X=0_{n\times 1}$$ $$\implies \dim Ker (T)=0\implies T\;\text{is a monomorphism}\iff T\;\text{is an isomorphism}$$ because $M_{n\times 1}\cong M_{n\times 1}$.

Opposite direction:

Let $\{B_1,\ldots,B_n\}$ be some basis for $M_{n\times 1}$.

Let $X\in M_{n\times 1}$ be arbitrary and $\alpha_j\in\mathbb F$. Then

$$X=\sum_{j=1}^n\alpha_j B_j$$ $$T(X)=T\left(\sum_{j=1}^n\alpha_j B_j\right)=A\left(\sum_{j=1}^n\alpha_j B_j\right)$$ $\dim Ker(T)=0 \iff$ the corresponding homogeneous system $A\left(\displaystyle\sum_{j=1}^n\alpha_j B_j\right)=0$ has a unique trivial solution $\iff\;\det A\ne 0\;\land\;\alpha_i=0\forall\;j\in\{1,\ldots,n\}$. Since the basis is preserved, $T$ must be an isomorphism .

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I thought I could use Binet-Cauchy for $(b)$ as the main tool so as to prove the statement by contradiction in both directions at once: $$\det(AX)=0\iff \det A\cdot\det X=0$$ $$\det A=0\implies\;S(X)=0\;\text{for an arbitrary}\; X\in M_n\implies\;\dim Ker(S)\geqslant 1$$

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May I ask for advice on how to improve my proof for $(a)$ and how to start the proof for $(b)$ if Binet-Cauchy isn't the right choice? Thank you in advance!

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Update - notation explanation:

$T$ and $S$ are linear operators;

$T\in L\left(M_{n\times 1}\right)\equiv T: M_{n\times 1}\to M_{n\times 1}$

Analogously,

$S\in L\left(M_n\right)\equiv S: M_n\to M_n$

$M_n\equiv M_{n\times n}$

$\cong$ means isomorphic, so both $M_{n\times 1}$ and $M_n$ are isomorphic to themselves because the relation '$\cong$' is reflexive.

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Extra update:

We haven't formally gone neither through the spectrum of a linear operator nor the characteristic polynomial of the matrix. My apology for not mentioning it.

Answer 1

Here is a proof plan. Let $A\in M_n(K)$.

a) $T:u\in K^n\mapsto Au$.

$T$ (or $A$) invertible iff $\{Au=0\implies u=0\}$.

b) $S:X\in M_{n,p}(K)\mapsto AX\in M_{n,p}(K)$.

$S$ invertible iff $\{AX=0\implies X=0\}$ iff $\{A(C_1,\cdots,C_p)=(AC_1,\cdots,AC_p)=0\implies C_1=\cdots=C_p=0\}$

iff $A$ invertible (use $p$ times a)).

Answer 2

Now I add a note:

$S\in L\left(M_n\right)$, and any linear operator $\Lambda\in L(V)$ in general is $\color{red}{\text{invertible}}\iff\color{red}{\text{regular}}\iff\color{red}{\text{epimorphism}}\iff\color{red}{\operatorname{rank}A=n}\iff A$ is invertible