Characteristic polynomial of a mapping from matrices space to matrices space

Question

Let $T$ be the linear map from $M_n \to M_n$ given by TX=AX, while A is as well a matrix $n \times n$

(a) Write out the characteristic polynomials for $T$

(b) Show that if A is diagonalzible then T is also diagonalizble

I know that the characteristic polynomial of B, some $n \times n$ matrix is the expansion of $$\det(B - I \lambda ).$$

The problem is with finding the matrix for $T$.

On one hand it seems so natural to say that A itself is the matrix that represents T by the natural base.

On the other hand, if I do take the natural base of matrixes $n \times n$ (Where every vector in the base has a 1 somewhere and the rest of the matrix is zeros), the dimension of the base is $n^2$ so my matrix $T_A(x)$ is the size of $n^2$?

I'm not sure if I explained myself correctly, but I don't even know where to start.. if A is simply the matrix that represent T, the answer to (b) is quite obvious, it's the declaration for T to be digonizable. (or at least a sentence that we already proved in lesson) Moreover, what can I write as the characteristic polynomial? just $det(A-\lambda I)$

I would really appreciate any help understanding these things. Thanks.

Answer 1

Let $E_{ij}$ the matrix which all its entries are $0$ except the entry located at the $i^{th}$ row and $j^{th}$ column is equal $1$ then we know that $$\left(E_{ij}\right)_{1\le i,j\le n}=(E_{11},\ldots,E_{n1},E_{12},\ldots,E_{n2},\ldots,E_{1n},\ldots,E_{nn})$$ is a basis of $\mathcal M_n(\Bbb R)$ and we see that the matrix of $T$ relative to this basis is the diagonal block matrix: $$ M=\begin{pmatrix} \large A & && & \\ & \ddots & &{\huge 0}&& \\ & & \ddots & & \\ & {\huge 0} & & \ddots & \\ & & & &A \\ \end{pmatrix} $$ so $$\chi_T(x)=\det(M-xI_{n^2})=\prod_{k=1}^n\det(A-xI_n)=\left(\chi_A(x)\right)^n$$ Now we have $$T^2X=T(TX)=T(AX)=A^2X$$ hence we see easily that (we denote $\pi_A$ the minimal polynomial of $A$) $$\pi_A(T)=0$$ so $\pi_T|\pi_A$ hence if $A$ is diagonalizable then $\pi_A$ has a simple roots so $T$ is also diagonalizable and $$\pi_T=\pi_A$$

Answer 2

For b):

Consider a basis $(x_i)$ of eigenvectors of $A$: $$ Ax_i = a_ix_i $$

Then consider the matrices $$ X_{i, j} = (0, \ldots, e_{i}\ldots, 0) \\ TX_{i,j} = a_iX_{i, j} $$ (with $e_i$ in the $j^{th}$ position).

and then check that $(X_{i, j})$ is a basis (rather easy).

You conclude that the eigenvalues are (a_i), each with multiplicity multiplied by $n$.

For a):

Use the density of the diagonalizable matrices and the continuity of the application giving the characteristic polynomial.

You conclude that the following relationship between characteristic polynomials remains in general: $$ \Pi_T(X) = \left(\Pi_A(X)\right)^n $$