# Find the spectral decomposition of $R_D$, where $R_D$ is right multiplication by a diagonal matrix

Question: Let $M_n(C)$ denote the vector space of complex $n\times n$ matrices. For $B$ in $M_n(C)$, denote by $\overline{B}^t$ the conjugate transpose of $B$, and let $\text{tr}(B)$ denote the trace. The following defines an inner product on $M_n(C):$ $$\langle A,B\rangle = \text{tr}(A\overline{B^t}).$$

a) Let $R_D$ denote the linear operator on $M_n(C)$ given by right multiplication by $D$, thus $R_D(A) = AD$. Now, let $D$ be a diagonal element of $M_n(C)$. Show that $R_D$ is a normal operator.

b) Still with diagonal $D$, express $R_D$ as a linear combination of orthogonal projections.

My attempt:

a) First, we prove that $R_D^*(A) = AD^*$, right multiplication by the conjugate transpose of $D$. To show this, notice that $$\langle R_D(A),B\rangle = \text{tr}(AD\overline{B^t}),$$ and we also have $$\langle A,BD^*\rangle = \text{tr}(A\overline{(BD^*)^t})$$ $$=\text{tr}(AD\overline{B^t}),$$ so $\langle R_D(A),B\rangle = \langle A,BD^*\rangle$, which means that $R_D^* = BD^*$. Now, $$R_DR_D^*(A) = AD^*D = ADD^* = R_D^*R_D(A),$$ so $R_D$ is a normal operator, as desired.

b) (Edited) So, I've found that the matrix representation of $R_D$ using the standard ordered basis of $M_n(C)$ is:

$$\left[ {\begin{array}{*{20}{c}} D&0& \ldots &0\\ 0&D& \ldots &0\\ \vdots & \vdots & \ddots & \vdots \\ 0&0& \ldots &D \end{array}} \right],$$

which makes the eigenvalues $d_1,d_2,\dots,d_n$, where $D$ has the form:

$$\left[ {\begin{array}{*{20}{c}} {{d_1}}&0& \ldots &0\\ 0&{{d_2}}& \ldots &0\\ \vdots & \vdots & \ddots & \vdots \\ 0&0& \ldots &{{d_n}} \end{array}} \right].$$

Now, this makes $E_1$, the eigenspace corresponding to the eigenvalue $d_1$, the space of vectors spanned by the set:

$$\left\{ {\left[ {\begin{array}{*{20}{c}} 1&0& \ldots &0\\ 0&0& \ldots &0\\ \vdots & \vdots & \ddots & \vdots \\ 0&0& \ldots &0 \end{array}} \right],\left[ {\begin{array}{*{20}{c}} 0&0& \ldots &0\\ 1&0& \ldots &0\\ \vdots & \vdots & \ddots & \vdots \\ 0&0& \ldots &0 \end{array}} \right], \ldots ,\left[ {\begin{array}{*{20}{c}} 0&0& \ldots &0\\ 0&0& \ldots &0\\ \vdots & \vdots & \ddots & \vdots \\ 1&0& \ldots &0 \end{array}} \right]} \right\},$$

and $E_2$ is the space of vectors spanned by the set:

$$\left\{ {\left[ {\begin{array}{*{20}{c}} 0&1& \ldots &0\\ 0&0& \ldots &0\\ \vdots & \vdots & \ddots & \vdots \\ 0&0& \ldots &0 \end{array}} \right],\left[ {\begin{array}{*{20}{c}} 0&0& \ldots &0\\ 0&1& \ldots &0\\ \vdots & \vdots & \ddots & \vdots \\ 0&0& \ldots &0 \end{array}} \right], \ldots ,\left[ {\begin{array}{*{20}{c}} 0&0& \ldots &0\\ 0&0& \ldots &0\\ \vdots & \vdots & \ddots & \vdots \\ 0&1& \ldots &0 \end{array}} \right]} \right\},$$

and etcetera. We can find, then that $\left.T\right|_{E_1}$ has the matrix representation: (using the basis for $E_1$ that was found above)

$$\left[ {\begin{array}{*{20}{c}} {{d_1}}&0& \ldots &0\\ 0&{{d_1}}& \ldots &0\\ \vdots & \vdots & \ddots & \vdots \\ 0&0& \ldots &{{d_1}} \end{array}} \right],$$

and that the matrix representation of $\left.T\right|_{E_2}$ is

$$\left[ {\begin{array}{*{20}{c}} {{d_2}}&0& \ldots &0\\ 0&{{d_2}}& \ldots &0\\ \vdots & \vdots & \ddots & \vdots \\ 0&0& \ldots &{{d_2}} \end{array}} \right],$$

and so on, so by the spectral theorem, we can represent $T$ as:

$$T = d_1^2I + d_2^2I + \dots d_n^2I.$$

But, this can't be right, because that representation of $T$ is $n\times n$, and the one I found originally was $n^2 \times n^2$. Am I doing something completely wrong? Have I just mixed up my dimensions somewhere?

## 1 Answer

Ah, ok, I seem to have answered my own question. The mistake I was making was thinking that the spectral theorem used restrictions of operators on eigenspaces, when in fact it uses orthogonal projections. So, using the definitions of $E_1,E_2,\dots, E_n$ from the question, we let $T_1,T_2,\dots,T_n$ be the orthogonal projections of $V$ onto $E_1,E_2,\dots,E_n$, respectively. Then, we can write

$$T = d_1T_1 + d_2T_2 + \dots + d_nT_n,$$

as desired.

Now, the above is only valid if the entries of $D$ are all distinct. If they are not, so if $d_i = d_j$ for any $i \neq j$, we let $f_1,f_2,\dots,f_k$ be the distinct diagonal entries of $D$, and we will have eigenspaces: $F_1, F_2,\dots,F_k$, where each eigenspace corresponds to an eigenvalue in $\{f_1,f_2,\dots,f_k\}$ , and the $F_\ell$, $1 \le \ell k$, are defined as: $$F_\ell = \bigcup_{d_i = f_\ell} E_i.$$

In this case we let $R_\ell$ be the orthogonal projection of $V$ on $F_\ell$, and then we can write, by the spectral theorem, $$T = f_1R_1 + f_2R_2 + \dots + f_kR_k,$$ as desired.

I am unsure as to how to find the actual matrix forms of these $T_i$, so if someone has an explicit form of these or a hint on how to find them, that would be much appreciated.