My understanding of how to prove things with null spaces is beyond limited.
I need help proving the following.
Given a unit vector
$$
\mathbf{x} \in \begin{bmatrix} x_{1} \\ \vdots \\ x_{n}\end{bmatrix}
$$
where,
$$
\mathbf{Y} \in null(\mathbf{x}^{\top})
$$
such that,
$$
\mathbf{Y}^{\top}\mathbf{x} = \begin{bmatrix} 0 \\ \vdots \\ 0\end{bmatrix}
$$
show that
$$
\mathbf{Y}\mathbf{Y}^{\top} + \mathbf{x}\mathbf{x}^{\top} = \mathbf{I}
$$
In my case, $\mathbf{x}$ is a 3×1 column vector and $\mathbf{Y}$ is a 3×2 matrix.
Thanks
Best Answer
I fear that this is not true under the current hypotheses. Consider the real $3 \times 2$ matrix $$A = \begin{pmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{pmatrix}.$$ We have that $v = \langle 1, -2, 1 \rangle^t$ satisfies $v^t A = \langle 0, 0 \rangle$ and $A^t v = \langle 0, 0 \rangle^t,$ hence $w = \frac 1 {||v||} v = \frac 1 {\sqrt 6} v$ is a unit vector satisfying $w^t A = \langle 0, 0 \rangle$ and $A^t w = \langle 0, 0 \rangle^t;$ however, it is not the case that $$AA^t + ww^t = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$ Explicitly, we have that $$AA^t = \begin{pmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} = \begin{pmatrix} 17 & 22 & 27 \\ 22 & 29 & 26 \\ 27 & 36 & 45\end{pmatrix} \text{ and } ww^t = \begin{pmatrix} \phantom{-} 1/6 & -1/3 & \phantom{-} 1/6 \\ -1/3 & \phantom{-} 2/3 & -1/3 \\ \phantom{-} 1/6 & -1/3 & \phantom{-} 1/6 \end{pmatrix}.$$