Skew symmetric matrix question

[TTob]

Let A in n x n real matrix.
For every x in R^n we have <Ax,x>=0 where < , > is scalar product.
prove that A^t=-A (A is skew symmetric matrix)

 

[tiny-tim]

Let A in n x n real matrix.
For every x in R^n we have <Ax,x>=0 where < , > is scalar product.
prove that A^t=-A (A is skew symmetric matrix)

Hi TTob!

Hint: just write out <Ax,x>=0 in terms of A_ij etc …

then jiggle it around a bit!

What do you get?

 

[TTob]

Thank you.

note $$x=(x_1,...,x_n)$$ and $$A=(a_{ij})$$.
then $$(Ax)_i=$$ $$\sum_{\substack{0\leq j\leq n}} a_{ij}x_j$$
then
$$<Ax,x>=$$ $$\sum_{\substack{0\leq j\leq n \\ 0\leq i\leq n}} a_{ij}x_i x_j=0$$

when you put $$x=e_i$$ you get $$a_{ii}=0$$ for all i.
when you put $$x=e_i+e_j$$ you get $$a_{ii}+a_{ij}+a_{ji}+a_{jj}=0$$ for all i,j .

so $$a_{ji}=-a_{ij}$$ for all i,j and then $$A^t=-A$$.

 

[tiny-tim]

Hi TTob!

Yes … or if you want to avoid using components …

$$0\ =\ <A(x+y),x+y>\ -\ <Ax,x>\ -\ <Ay,y>\ \ =\ \ <Ax,y>\ +\ <Ay,x>\ \ =\ \ <(A + A^T)x,y>$$

 

[HallsofIvy]

Or use the fact that <Ax, y>= <x, A^Ty>. (That is a more general definition of "adjoint")

From that, <Ax, x>= <x, A^Tx>. If A is skew-symmetric, <x, A^{T x>= -<Ax, x> so <Ax, x>= -<Ax, x> and <Ax, x>= 0.}

 

[tiny-tim]

… <Ax, x>= <x, A^Tx>. If A is skew-symmetric, <x, A^T x>= -<Ax, x> so <Ax, x>= -<Ax, x> and <Ax, x>= 0.

Hi HallsofIvy!

erm …

I think you've proved the transpose of the original theorem! ​