Linear Algebra – Orthogonal Complement of Symmetric Matrices

functional-analysislinear algebra

How do I can prove that the orthogonal complement of space of symmetric matrices is the space of skew-symmetric matrices?

With the inner product $\langle A,B\rangle = \mbox{tr}(A^TB)$. Thanks in advance.

Best Answer

Notice that $\langle A, B \rangle = \sum_{i,j} [A]_{ij} [B]_{ij}$. Let $E_{ij} = e_i e_j^T$, ie, the zero matrix except for a one in the $(i,j)$ position.

Suppose $\langle A, S \rangle = 0$ for all symmetric matrices, then it is true for $S=E_{ij}+E_{ji}$. This gives $\langle A, E_{ij} \rangle + \langle A, E_{ji} \rangle = 0$, which gives $[A]_{ij} + [A]_{ji} = 0$, from which it follows that $A = -A^T$.

Now suppose that $A$ is skew symmetric, and $S$ is symmetric. Then we can write $S = U^T + \Lambda + U$, where $\Lambda$ is diagonal, and $U$ is strictly upper triangular. Then $\langle A, \Lambda \rangle = 0$ because $[A]_{ii} = 0$. Since $\operatorname{tr} M = \operatorname{tr} M^T$ we also have $\langle A, U^T \rangle = \langle A^T, U \rangle$, and skew symmetry gives $\langle A^T, U \rangle = - \langle A, U \rangle$. Hence $\langle A, S \rangle = \langle A, U^T \rangle + \langle A, \Lambda \rangle - \langle A, U^T \rangle = 0$.

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