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I was reading this proof from our text book. I didnt get one step.

Q. If A is a skew symmetric matrix and X is a column matrix, then show X'AX is a null matrix.

Proof

Since A is skew symmetric, A'=-A Let A be square matrix of order n, X of nx1 and X' of 1xn, then X'AX is a matrix of order 1x1.

Let X'AX = B which will be of order 1x1 and hence symmetric i.e. B' = B.

Now,

(X'AX)' = B'

$\therefore $ X'A'X'' = B' ...I didnt get this step, I think it should be X''A'X' = B'

But A' = -A and X'' = X and B' = B

$\therefore$ X'(-A)X'' = B'

$\therefore$ -(X'AX) = B

-B = B

$\therefore$ 2B = 0

$\therefore$ B = 0

Mahesha999
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    The step you didn't get uses the fact that $(AB)^T=B^TA^T$, not $A^TB^T$. –  Nov 08 '14 at 09:18

1 Answers1

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The term $X'AX$ is also $$ \sum_{k,l} x_k A_{k,l} x_l $$

which is, when reversing the variables of summation, $$ \sum_{k,l} x_l A_{l,k} x_k = X'A'X $$

your mistake is that $(CD)^T $ is $D^TC^T$, not $C^TD^T$.

mookid
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