Skew Symmetric Matrix property

I want to show: If a matrix $A$ is skew symmetric, that is if $A^t=-A$ then $x^tAx=0$ for all vectors $x$.

Please give hints and thought processes for the proof. I am quite stuck on this question. Please answer the following question:

1. In general, when you are proving an algebraic statement, how do you get intuition? A statement like this means nothing to me other than manipulating a bunch of symbols. This sounds like a "strategy" setup for failure...
• Someone could argue that maths itself is manipulation of symbols :). In this case however, I can provide a quick application. You could think at $A$ as at the skew part of a stress tensor, $S$ i.e. $A = 1/2(S^\intercal - S)$. In this case a statement like that can tell you something about which is the effective part of such a tensor as of producing surface tension. – JosephK Jul 18 '13 at 13:56

$$x^TAx=d\to(x^TAx)^T=d^T=d\to x^TAx=-d\Rightarrow\begin{cases} x^TAx=d \\ x^TAx=-d \\ \end{cases}\color{red}{\Rightarrow}2(x^TAx)=0\to x^TAx=0$$

• Is this literally then just that no number can be positive and negative and so $d=0$? – CodeKingPlusPlus Jul 18 '13 at 11:47

Hint

If $c$ is a real then $c^t=c$ so let $c=x^tAx$....

Remark There is no magic method in mathematics just try to use the hypothesis and think and think and think.

Try to use the property through which the transpose matrix is defined, i.e. $$\bf {A a \cdot b = a \cdot A^\intercal b },$$ where $\bf a$ and $\bf b$ are vectors, $\bf A$ is a matrix and by the dot $\cdot$ I mean the inner product defined by $\textbf {a$\cdot$b }= \sum a_i b_i$.

Now, you can write $x^\intercal A x$ in a different way - it is just notation, but I believe that it helps in this case - as $\bf A x \cdot x$ and apply the above property.