Suppose I have the following rank-one matrix $$uv^T=\begin{bmatrix} u_1\\u_2\\u_3 \end{bmatrix} \begin{bmatrix} v_1 & v_2 & v_3 \end{bmatrix}$$
Suppose I have the following lifting operator $\mathcal{A}: \mathbb{R}^{3\times 3}\rightarrow \mathcal{S}^{4\times 4}$ $$\mathcal{A}(Y)=\left[\begin{array}{cccc} Y_{11}-Y_{22}-Y_{33} & Y_{12}+Y_{21} & Y_{13}+Y_{31} & Y_{23}-Y_{32} \\ Y_{12}+Y_{21} & -Y_{11}+Y_{22}-Y_{33} & Y_{23}+Y_{32} & -Y_{13}+Y_{31} \\ Y_{13}+Y_{31} & Y_{23}+Y_{32} & -Y_{11}-Y_{22}+Y_{33} & Y_{12}-Y_{21} \\ Y_{23}-Y_{32} & -Y_{13}+Y_{31} & Y_{12}-Y_{21} & Y_{11}+Y_{22}+Y_{33} \end{array}\right]$$
Can we say something about the rank and column space of $\mathcal{A}(uv^T)$? For example, the rank of $\mathcal{A}(uv^T)$ is one.
I believe so far it is obvious that
- it is symmetric
- trace$(\mathcal{A}(Y))=0$
Thanks!
$\def\R{{\mathbb R}}\def\m#1{ \left[\begin{array}{c}#1\end{array}\right] }$Let $\epsilon\in\R^{3\times 3\times 3}$ denote the Levi-Civita symbol and $I\in\R^{3\times 3}$ denote the identity matrix. Use these to define the following scalar, vector, matrix functions of the $Y$ matrix $$\eqalign{ \alpha &= {\rm trace}(Y) \;=\; I:Y \\ a &= \epsilon:Y \\ A &= Y + Y^T - \alpha I \;=\; A^T\\ }$$ where $(:)$ denotes the double-contraction product.
The following constants will also be useful $$\eqalign{ z &= \m{0\\0\\0} \qquad E &= \m{I\\z^T} \qquad e &= \m{0\\0\\0\\1} \\ }$$ Then your lifting operator can be expressed in a form which lends itself to algebraic manipulation $$\eqalign{ {\cal A}(Y) &= \m{A&a\\a^T&\alpha} \;\doteq\; EAE^T + Eae^T + ea^TE + \alpha ee^T \\ }$$ Setting $Y=uv^T$ yields $$\eqalign{ \alpha &= u^Tv \qquad a &= u\times v \qquad A &= uv^T + vu^T - (u^Tv)I \\ }$$ where $(\times)$ denotes the vector cross product.