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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...
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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

3 Answers 3

up vote 2 down vote accepted

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

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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.

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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.

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