# Thread: Matrix operator proof - expression clarification

1. ## Matrix operator proof - expression clarification

Hi everyone,

The question I am trying to solve is as follows:

"ax is the matrix operator that, when operating on any other vector, produces the effect of cross product of that vector with the vector a. That is axb = a X b for any vector b. Show that ax^2 = aaT - |a|^2I"

where:
"^2" means squared
"ax" is a subscript x
"a X b" means 'a' cross product 'b'
"aT" is 'a-transpose'

I understand |a|^2 to equal a1^2 + a2^2 + a3^3 + ... (i.e. first element of a squared, plus second element of a squared, plus ... etc). Is this correct?

What does ax^2 mean? Is it simply a X a (cross product)?

Can someone please provide a little insight as to what the above expression (both left and right sides) actually means.

Thank you.

2. The operator $\displaystyle \mathbf u\mapsto\mathbf a\times\mathbf u$, where $\displaystyle \mathbf a=\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$, has the matrix $\displaystyle \mathbf A=\begin{pmatrix}0&-a_3&a_2\\a_3&0&-a_1\\-a_2&a_1&0\end{pmatrix}$.

Thus $\displaystyle \mathbf A^2=\begin{pmatrix}0&-a_3&a_2\\a_3&0&-a_1\\-a_2&a_1&0\end{pmatrix}\begin{pmatrix}0&-a_3&a_2\\a_3&0&-a_1\\-a_2&a_1&0\end{pmatrix}=\begin{pmatrix}-a_2^2-a_3^2&a_1a_2&a_1a_3\\a_1a_2&-a_1^2-a_3^2&a_2a_3\\a_1a_3&a_2a_3&-a_1^2-a_2^2\end{pmatrix}$.

Thus $\displaystyle \mathbf A^2=\begin{pmatrix}a_1a_1&a_2a_1&a_3a_1\\a_1a_2&a_ 2a_2&a_3a_2\\a_1a_3&a_2a_3&a_3a_3\end{pmatrix}-(a_1^2+a_2^2+a_3^2)\begin{pmatrix}1&0&0\\0&1&0\\0& 0&1\end{pmatrix}=\mathbf a\mathbf a^{\mathbf T}-|\mathbf a|^2\mathbf I$.