Linear Algebra question.

[Steven G]

I need help with parts c and d.
Here is what I know (I think). A is skew-symmetric if A^T = -A. This was actually given in the problem and I understand it.
A is an orthogonal matrix if A*A^T = I (So A_T = A^-1). I understand that.

I played games with things like x^T(Ax) = (x^TA)x = (A^Tx)^Tx = (-Ax)^Tx= -(Ax)^T)x =-(x^TA^T)x but I get no where. I want to get (x^TA)x = -(x^TA)x but no luck.

 

[Romsek]

\(\displaystyle (Ax)\cdot x = (x^T A^T) \cdot x = -(x^T A) \cdot x = -x^T\cdot (Ax) = -Ax \cdot x\)

\(\displaystyle \text{The only number for which $c = -c$ is $c=0$}\)

 

[Steven G]

\(\displaystyle (Ax)\cdot x = (x^T A^T) \cdot x = -(x^T A) \cdot x = -x^T\cdot (Ax) = -Ax \cdot x\)

\(\displaystyle \text{The only number for which $c = -c$ is $c=0$}\)

I do have a concern with your work. Why is \(\displaystyle (Ax)\cdot x = (x^T A^T) \cdot x\) true?
\(\displaystyle (x^T A^T) = (Ax)^T\) and does that really equal \(\displaystyle (Ax)\). If yes, then why?

 

[Steven G]

I do have a concern with your work. Why is \(\displaystyle (Ax)\cdot x = (x^T A^T) \cdot x\) true?
\(\displaystyle (x^T A^T) = (Ax)^T\) and does that really equal \(\displaystyle (Ax)\). If yes, then why?


Edit: Well of course \(\displaystyle (Ax)^T\) does not equal \(\displaystyle (Ax)\)but why is ok what you did?

 

[Romsek]

You can only dot vectors with each other one way. It doesn't matter if you call one of them a transpose.

 

[Steven G]

You can only dot vectors with each other one way. It doesn't matter if you call one of them a transpose.

So for example you can dot any vector v with v^T?
Thanks for your time. I appreciate it.

 

[Romsek]

\(\displaystyle
\text{Generally if you're given two vectors, $u,~v$ and asked to find $u\cdot v$ it's understood that}\\
\text{1) $u$ and $v$ are column vectors; and }\\
\text{1) $u\cdot v = u^T v$}
\)

let's see if I can be a bit more rigorous in my argument

\(\displaystyle
(Ax)x = (Ax)\cdot x = (Ax)^T x = x^T A^T x = -x^T A x = -x^T (Ax) = -x\cdot (Ax) = -(Ax)\cdot x = -(Ax)x
\)

 

[Harry_the_cat]

Did you find in (a) that all skew-symmetric matrices are of the form

A =\(\displaystyle \begin{pmatrix}0&a\\-a&0\\\end{pmatrix}\) ?

(Note: That's negative a in 2nd row, 1st column. Can't get the neg sign right. Any help appreciated.)

So letting vector x = \(\displaystyle \begin{pmatrix}m\\n\\\end{pmatrix}\)

Ax =\(\displaystyle \begin{pmatrix}an\\-am\\\end{pmatrix} \)

Then it's easy to show the dot product is 0 and hence vectors are orthogonal.