In $\mathbb{C^{3x3}}$, with the inner product of matrices defined as $$\langle A,B\rangle = \mbox{tr}(A^*B)$$ find the orthogonal complement of the subspace of diagonal matrices.

Then, considering the following matrices $\in$ $\mathbb{C^{3x3}}$

$$A=\begin{pmatrix} a_{11}&0&0\\0&a_{22}&0\\0&0&a_{33}\end{pmatrix}$$


$$B=\begin{pmatrix} b_{11}&0&0\\0&b_{22}&0\\0&0&b_{33}\end{pmatrix}$$

I concluded by the statement and the definition of orthogonal complement:

$$\langle A,B\rangle=Tr\begin{pmatrix} \overline{a_{11}}b_{11}&0&0\\0&\overline{a_{22}} b_{22}&0\\0&0&\overline{a_{22}}b_{33}\end{pmatrix}=0$$

After, I get the trace of the $3$-by-$3$ square matrix of the inner product $\langle A,B\rangle$

$$\overline{a_{11}} b_{11}+ \overline{a_{22}}b_{22}+ \overline{a_{33}}b_{33}=0$$

I'm stuck from here, how can I do to get the orthogonal complement?


Consider the matrix$$A=\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}.$$Then $A$ belongs to the orthogonal complement $D^\bot$ of the space $D$ of diagonal matrices, since$$\left\langle A,\begin{pmatrix}a&0&0\\0&b&0\\0&0&c\end{pmatrix}\right\rangle=\operatorname{tr}\begin{pmatrix}0&b&0\\0&0&0\\0&0&0\end{pmatrix}=0.$$For the same reason, every matrix with one and only one entry whose value is $1$, which is outside the main diagonal, and such that all other entries are equal to $0$ belongs to $D^\bot$. The space $V$ spanned by all these matrices is the space of the matrices such that all entries of the main diagonal are equal to $0$. Since the $\dim V=6$, $\dim D=3$ and the whole space has dimension $9$, the space that you're after is $V$. In other words,$$D^\bot=\left\{\begin{pmatrix}0&a&b\\c&0&d\\e&f&0\end{pmatrix}\,\middle|\,a,b,c,d,e,f\in\mathbb{C}\right\}$$



$$\mathcal D_3 (\mathbb C) := \{ \mbox{diag} (\mathrm c) \mid \mathrm c \in \mathbb C^3 \}$$

be the set of $3 \times 3$ diagonal matrices over $\mathbb C$. Its orthogonal complement is the set of all matrices $\mathrm X \in \mathbb C^{3 \times 3}$ such that $\langle \mbox{diag} (\mathrm c), \mathrm X \rangle = 0$ for all $\mathrm c \in \mathbb C^3$, i.e.,

$$\begin{bmatrix} | \\ \mathrm c \\ | \end{bmatrix}^* \begin{bmatrix} x_{11}\\ x_{22}\\ x_{33}\end{bmatrix} = 0$$

for all $\mathrm c \in \mathbb C^3$. Hence, we conclude that $\color{blue}{x_{11} = x_{22} = x_{33} = 0}$. The orthogonal complement is, thus, the set of $3 \times 3$ matrices over $\mathbb C$ with zeros on the main diagonal

$$\mathcal D_3^\perp (\mathbb C) := \{ \mathrm X \in \mathbb C^{3 \times 3} \mid x_{11} = x_{22} = x_{33} = 0 \}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.