If $\Lambda_A\in L(M_n),\;\Lambda_A(T)=AT$, show $\sigma\left(\Lambda_A\right)=\sigma (A)$

Question

Let $A\in M_n$ and let $\Lambda_A\in L\left(M_n\right)$ $$\Lambda_A(T)=AT.$$ Show $\sigma\left(\Lambda_A\right)=\sigma(A).$

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Definition 1:

Let $V$ be a vector space over field $\mathbb F$ and $A\in L(V$). Scalar $\lambda_0\in\mathbb F$ is an eigenvalue of $A$ if $\exists x\in V,x\ne 0$ s.t. $Ax=\lambda_0 x$. Set $\sigma(A)$ of all eigenvalues of $A$ is called spectrum.

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Definition 2:

Let $A\in L(V,W)$. $$Ker A= A^{-1}\left(A\{0\}\right)=\{x\in V: Ax=0\}\leqslant V$$

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Definition 3:

Let $E_A(\lambda_0)=\{x\in V:Ax=\lambda_0 x\}$. $E_A(\lambda_0)$ is the eigenspace of an eigenvalue $\lambda_0$. Note: $$E_A(\lambda_0)=Ker(A-\lambda_0I).$$ $\lambda_0$ is an eigenvalue of $A$ $\iff$ $A$ is singular.

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Definition 4:

Let $A\in M_n(\mathbb F)$. Polynomial $k_A(\lambda)\in\mathbb F[\lambda]$, $k_A=\det(A-\lambda_0I)$ is the characteristic polynomial of $A$.

Note: $\lambda_0$ is an eigenvalue of $A$ $\iff$ $k_A(\lambda_0)=0$,i.e., $\lambda_0$ is a root of the polynomial $k_A$.

$\deg(k_A)\leqslant n$

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My attempt:

We have to prove: $$(\forall\lambda\in\mathbb F)\lambda\in\sigma\left(\Lambda_A\right)\implies\lambda\in\sigma(A)$$ $$\&$$ $$(\forall\lambda\in\mathbb F)\lambda\in\sigma(A)\implies\lambda\in\sigma\left(\Lambda_A\right)$$ i.e. $$\sigma\left(\Lambda_A\right)\subseteq\sigma(A)\;\&\;\sigma(A)\subseteq\sigma\left(\Lambda_A\right)$$ For $\Lambda_A(T)$ $$AT=\lambda_0T\iff AT=\lambda_0 IT\iff(A-\lambda_0I)T=0$$

No matter how obvious this seems, I haven't come up to any justified step to prove both the matrix $A$ and the linear operator have to be singular. Everything seemed circular. I know: $$T\ne 0\nRightarrow \det(A-\lambda_0I)=0$$ because $T$ might be a zero-divisor and hence $\det T=0$. Also, $\color{red}{\text{determinant isn't a linear form}}$, so I can't figure out how to end the proof. What would be the next step? Thank you in advance!

Answer 1

In general, given a matrix $B\in M_{n}$ whose columns are $(v_{1},...,v_{n})$, we have that the product $AB$ is the matrix whose columns are $(Av_{1},...,Av_{n})$.

$\Rightarrow$ Assume $\lambda\in\sigma(A)$. Let $v$ be a non-zero eigenvector of $A$ corresponding to $\lambda$. Consider the matrix $B$ whose columns are $(v,\overset{\rightarrow}{0},\overset{\rightarrow}{0},...,\overset{\rightarrow}{0})$. Then, $\text{Λ}_{A}(B)=AB$ is the matrix whose columns are $(Av,\overset{\rightarrow}{0},\overset{\rightarrow}{0},...,\overset{\rightarrow}{0})$, i.e., it is $\lambda B$.

$\Leftarrow$ Assume $\lambda\in\sigma(\text{Λ}_{A})$. Let $B$ be a non-zero eigenvector of $\text{Λ}_{A}$ corresponding to $\lambda$. Let $v$ be a non-zero column of $B$. Then, $Av=\lambda v$, so that $\lambda\in\sigma(A)$.

Answer 2

Note that $\Lambda_A$ admits $n^2$ eigenvalues (taking into account the multiplicities). There are two definitions of $spectrum(A)$ and, in my opinion, it's a very bad idea.

DEF 1. We consider $spectrum(A)$ as a SET, that is, we don't take multiplicities into account. Here $spectrum(diag(1,1,2,3))=\{1,2,3\}$.

DEF 2. We consider $spectrum(A)$ as a LIST, that is, we take multiplicities into account. Here $spectrum(diag(1,1,2,3))=\{1,1,2,3\}$. It is sometimes called "multispectrum". Of course, this second definition gives (much) more information about $A$.

There is a well-known result that shows the interest of Def 2. If $f$ is an holomorphic function, then $spectrum(f(A))=f(spectrum(A))$.

Clearly, this last result is also true for DEF 1. But what do you do with it? You cannot calculate neither $det(f(A))$ nor $trace(f(A))$!!

In the considered exercise, we are facing the same blockage: how to calculate $\det(\Lambda_A),trace(\Lambda_A)$?

Here the result (for DEF 2) is $\sigma(\Lambda_A)=n$ copies of $\sigma(A)$.

For example, if $A=diag(1,1,2,3)$, then $\sigma(\Lambda_A)=\{1,1,1,1,1,1,1,1,2,2,2,2,3,3,3,3\}$ (a list of $16$ elements).

When you are a scholar, you write $\Lambda_A=A\otimes I$. For a generalization, see

https://en.wikipedia.org/wiki/Kronecker_product