I have some questions regarding vector Transpose.
Let $u \in \Bbb R^n$, such that $||u||=1$ and let $A=uu^T$.
So I understand that $A$ is a $\Bbb n ~{\Bbb x}~\Bbb n$ matrix.
I want to find $\Bbb {nul}A$.
So for $u=\begin{bmatrix} a_1\\ \vdots \\ a_n \end{bmatrix}$ we have $uu^T=\begin{bmatrix} a_1a_1 &\cdots & a_1a_n\\ \vdots & \ddots & \vdots \\ a_na_1 &\cdots & a_na_n \end{bmatrix}$
Then $\Bbb {nul}A= \Bbb {nul}\begin{bmatrix} a_1a_1 &\cdots & a_1a_n\\ \vdots & \ddots & \vdots \\ a_na_1 &\cdots & a_na_n \end{bmatrix}=\Bbb {nul} \begin{bmatrix} a_1 &\cdots & a_n\\ \vdots & \ddots & \vdots \\ a_1 &\cdots & a_n \end{bmatrix}=\Bbb {nul} \begin{bmatrix} a_1 &\cdots & a_n\\ \vdots & \ddots & \vdots \\ 0 &\cdots & 0 \end{bmatrix} ~~ \Rightarrow$
$\Rightarrow ~~a_1=-a_2-a_3-...-a_n$
$\Rightarrow \Bbb{nul}A =\Bbb{span}\begin{Bmatrix} -1 & -1 & \cdots & -1\\ 1 & 0& \cdots & 0\\ 0 & 1& \cdots & 0\\ \vdots & \vdots & \cdots & \vdots \\ 0 & 0 & \cdots & 1 \end{Bmatrix}$
Which should be $= \Bbb{span}\left \{ u \right \}^{\perp}$
Is that correct?
Thanks!