# Orthogonal complement

Let $V=M_n(\mathbb{R})$ and define an inner product on $V$ such that $$\langle A,B \rangle \colon= tr(A^tB)$$ Let $W$ be a subspace of $V$ consisting of all symmetric matrices. Show that its orthogonal complement $W^{\perp}$ contains all antisymmetric matrices.

In the solution to this problem they simply show that if $A$ is antisymmetric then $tr(A^tB)=0.$

Could someone explain why this is the solution?

• Do you know how $W^\perp$ is defined? They showed that if $A$ is antisymmetric then $A\in W^\perp$. – Ranc May 7 '16 at 13:21

If $S$ is a symmetric matrix and $A$ is an antisymmetric matrix.

$$\displaystyle \operatorname{tr}(SA^t) = \sum_{i,j}S_{ij}A_{ji} = -\sum_{i,j}S_{ji}A_{ij} = -\operatorname{tr}(S^tA)$$

Since $\operatorname{tr}(SA^t) = \operatorname{tr}(S^tA)$, we conclude that

$\langle S,A \rangle = 0$ for all $S \in W$. So $A \in W^\perp$.

• Where do you get trace of $S^tA$ equals product of trace of $S$ and trace of $A$? Trace of a product isn't always the product of the traces. – Gerry Myerson May 7 '16 at 13:39
• @Steven Why is $tr(S^tA)=tr(S)\cdot tr(A)?$ – Si.0788 May 7 '16 at 13:39
• @GerryMyerson : I fixed it. – steven gregory May 7 '16 at 14:01
• @Si.0788 - I fixed it. – steven gregory May 7 '16 at 14:01
• @StevenGregory Why is $tr(SA^t)=tr(S^tA)?$ – Si.0788 May 21 '16 at 20:19