4
$\begingroup$

Problem: Prove that $\left\{ A \in \mathbb{R}^{n \times n} \mid A \text{ is symmetric}\right\}^{\bot} = \left\{ A \in \mathbb{R}^{n \times n} \mid A \ \text{is skew-symmetric}\right\}$ with $\langle A, B \rangle = Tr(A^T B)$.

Attempt at proof: Let $A$ be symmetric, and $B$ skew-symmetric. I want to prove that $\langle A, B \rangle = 0$. So this is what I had so far: \begin{align*} \langle A, B \rangle &= Tr(A^T B) \\ &= Tr(AB) \\ &= \sum_{i=1}^n (AB)_{ii} \\ &= \sum_{i=1}^n \sum_{k=1}^n (a_{ik} b_{ki}) \end{align*} Now I need to use somewhere the fact that $b_{ii} = 0$, i.e. the diagonal elements of a skew-symmetric matrix are zero. But I don't know how to split up the summations? Help would be appreciated!

$\endgroup$
1

1 Answer 1

6
$\begingroup$

You're on the right track. You don't need to break it down into components, assuming you already know the following: $$ \text{Tr}(AB) = \text{Tr}(BA)~,~~\text{Tr}(A^TB) = \text{Tr}(B^TA)~,~~\text{hence}~ \langle A, B\rangle = \langle B, A\rangle $$ I assume you've already proved these things previously, by breaking down into components. So now we can proceed like at the start of your attempt: $$ \langle A,B\rangle = \text{Tr}(A^TB) = \text{Tr}(AB) = \text{Tr}(BA) = \text{Tr}(-B^TA) = \langle -B, A\rangle = - \langle A, B\rangle $$

To show that the spaces of symmetric and anti-symmetric matrices are actually orthogonal complements of each other, we also need to show that any matrix has a unique decomposition as a sum of a symmetric matrix and an anti-symmetric one. But this is easy: $$ A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T) $$ The first term is symmetric; the second is anti-symmetric.

$\endgroup$
5
  • 1
    $\begingroup$ I see thanks. And from this it follows that $\langle A, B \rangle$ must be zero? $\endgroup$
    – Kamil
    Jul 27, 2015 at 21:18
  • 1
    $\begingroup$ It shows that it is its own negative. :-) $\endgroup$
    – Rhys
    Jul 27, 2015 at 21:20
  • $\begingroup$ But in this way you only show that the two spaces - of symmetric matrices and of skew-symmetric are orthogonal. But you do not show that the one is the orthogonal complement of the other, i.e the direct sum of the two spaces gives the entire space of matrices $\in \mathbb R^n$ $\endgroup$
    – Svetoslav
    Jul 27, 2015 at 21:22
  • $\begingroup$ Example: in $\mathbb R^3$ every two perpendicular lines are orthogonal, but they do not make up the whole space, i.e their direct sum is not $\mathbb R^3$, or in other words "they do not span $\mathbb R^3$" $\endgroup$
    – Svetoslav
    Jul 27, 2015 at 21:24
  • $\begingroup$ Yes, but the question (despite the title), asked just for this. Let me add your suggestion for completeness though. $\endgroup$
    – Rhys
    Jul 27, 2015 at 21:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .