# Symmetric matrix

• January 28th 2010, 12:51 PM
Swlabr
Symmetric matrix
Let $M=(g_{i, j})$ by a symmetric matrix. Then is it true in general that $u^t M v = v^t M u$ for $u$ and $v$ column vectors?

If not, when is it true?

(I'm trying to show that $=u^t M v$ forms an inner product, but as far as I can see $ \neq $...)
• January 28th 2010, 02:09 PM
clic-clac
It is $u^t \times M\times v,$ isn't it? This is a scalar (a 1-1 matrix).

Therefore it is equal to its transpose, and $u^tMv=(u^tMv)^t=v^tM^tu=v^tMu$ whenever $M$ is symmetric.
• January 28th 2010, 02:19 PM
Swlabr
Quote:

Originally Posted by clic-clac
It is $u^t \times M\times v,$ isn't it? This is a scalar (a 1-1 matrix).

Therefore it is equal to its transpose, and $u^tMv=(u^tMv)^t=v^tM^tu=v^tMu$ whenever $M$ is symmetric.

Thanks - I was thinking it gave a column vector for some reason.

I do have one more question though: does this always give us a positive value if $u=v$?
• January 28th 2010, 07:07 PM
NonCommAlg
Quote:

Originally Posted by Swlabr

does this always give us a positive value if $u=v$?

no. it depends on $u$ and $M.$ for example if $u=e_1 = [1 \ 0 \ 0 \cdots \ 0]^T$ and $g_{11} < 0,$ then $u^TMu=g_{11} < 0.$
• January 28th 2010, 11:52 PM
Swlabr
Quote:

Originally Posted by NonCommAlg
no. it depends on $u$ and $M.$ for example if $u=e_1 = [1 \ 0 \ 0 \cdots \ 0]^T$ and $g_{11} < 0,$ then $u^TMu=g_{11} < 0.$

Hmm. Well, I've been asked to prove that the assignment,

$<\cdot, \cdot>: T_pM \times T_pM \rightarrow \mathbb{R}$, $(u, v) \mapsto u^t (g_{i, j}) v^t$ is an inner product, where $g_{i, j} = e_i . e_j$ and the $e_i$ are the basis vectors for $T_pM$, the tangent space of a manifold at a point $p$.

So I need to prove that it is positive definite, but I can't see why this holds. I suppose the point to use would be that elements on the diagonal are all (strictly) positive, but I don't see how to use this.
• January 29th 2010, 01:33 AM
Opalg
Quote:

Originally Posted by Swlabr
I've been asked to prove that the assignment,

$<\cdot, \cdot>: T_pM \times T_pM \rightarrow \mathbb{R}$, $(u, v) \mapsto u^t (g_{i, j}) v^t$ is an inner product, where $g_{i, j} = e_i . e_j$ and the $e_i$ are the basis vectors for $T_pM$, the tangent space of a manifold at a point $p$.

So I need to prove that it is positive definite

If the matrix M has entries $e_i.e_j$, and the vector u has components $u_1,\ldots,u_n$, then $u^{\textsc t}Mu = \textstyle\sum_i\sum_ju_i(e_i.e_j)u_j = \Bigl(\sum_iu_ie_i\Bigr).\Bigl(\sum_ju_je_j\Bigr) \geqslant0$. So M is positive semi-definite.