I need help with the following excercise about rank-$1$ matrices and their geometric interpretation. I think I managed to show the analytic parts but I struggle with the geometric interpretation parts. I've included below what I have done so far, but it might not be correct at all.
In the following, let $A$ be $m \times n$ with $\mbox{rank}(A)=1$.
a) Show there are non-zero vectors $\mathbf{u} \in \Bbb R^m$ and $\mathbf{v} \in \Bbb R^n$ such that $A=\mathbf{u}\mathbf{v}^T$. Describe the geometry of the four fundamental subspaces in terms of $\mathbf{u}$ and $\mathbf{v}$.
Since $\text{rank}(A)=1$ all columns of $A$ a scalar multiples of one specific column say $\mathbf{a}_i$ (for a fixed $i$). Using the column properties of matrix multiplication we can write
$$A = \begin{bmatrix} \vert & & \vert \\ c_1\mathbf{a}_i & \ldots & c_n\mathbf{a}_i \\ \vert & & \vert \end{bmatrix} = \begin{bmatrix} \vert \\ \mathbf{a}_i \\ \vert\end{bmatrix} \begin{bmatrix} c_1 & \ldots & c_n \end{bmatrix} = \mathbf{u} \mathbf{v}^T$$
where we define $\mathbf{u} = \mathbf{a}_i$ and $\mathbf{v} = \begin{bmatrix} c_1 \ldots c_n \end{bmatrix}^T$.
Alternatively also all rows of $A$ are scalar multiples of one specific row say $\mathbf{A}_j$ (for a fixed $j$) and using the row properties of the matrix multiplication we find again
$$A = \begin{bmatrix} \text{—} & d_1\mathbf{A}_j & \text{—} \\ & \vdots & \\ \text{—} & d_m\mathbf{A}_j & \text{—} \end{bmatrix} = \begin{bmatrix} d_1 \\ \vdots \\ d_m \end{bmatrix} \begin{bmatrix} \text{—} & \mathbf{A}_j & \text{—} \end{bmatrix} = \mathbf{u} \mathbf{v}^T$$
where we define $\mathbf{u} = \begin{bmatrix} d_1 \ldots d_m \end{bmatrix}^T$ and $\mathbf{v} = \mathbf{A}_j$.
Regarding the geometry I am not sure. I have
- The column space $C(A)$ is a line in $\Bbb R^m$ (since it's one dimensional) which is spanned by this one column vector $\mathbf{a}_i$.
- The row space $R(A)$ is a line in $\Bbb R^n$ (since it's one dimensional) which is spanned by this one row vector $\mathbf{A}_j$.
- The nullspace $N(A)$ is the orthogonal complement of the row space in $\Bbb R^n$ and therefore $n-1$ dimensional.
- The left nullspace $N(A^T)$ is the orthogonal complement of the column space in $\Bbb R^m$ and therefore $m-1$ dimensional.
b) Let $T: \Bbb R^n \to \Bbb R^m$ with matrix $A$ and $S: \Bbb R^m \to \Bbb R^n$ with matrix $A^T$. For each $\mathbf{x} \in R(A)$ is $(S \circ T)(\mathbf{x})=\mathbf{x}$ if and only if $\left\lVert \mathbf{u} \right\rVert \left\lVert \mathbf{v} \right\rVert = 1$. Interpret $T$ geometrically.
First of all $(S \circ T)(\mathbf{x}) = A^TA\mathbf{x}$ where by part a)
$$A^TA = (\mathbf{u}\mathbf{v}^T)^T \mathbf{u}\mathbf{v}^T = \mathbf{v} \mathbf{u}^T \mathbf{u}\mathbf{v}^T = \mathbf{v} (\mathbf{u} \cdot \mathbf{u}) \mathbf{v}^T = \mathbf{v} \left\lVert \mathbf{u} \right\rVert^2 \mathbf{v}^T = \left\lVert \mathbf{u} \right\rVert^2 \mathbf{v}\mathbf{v}^T$$
Let $\mathbf{x} \in R(A)$, then $\mathbf{x} = c\mathbf{A}_j$ for some scalar $c$. Let's choose the second interpretation of a) so that $\mathbf{v} = \mathbf{A}_j$. Then it follows
$$(S \circ T)(\mathbf{x}) = A^TA\mathbf{x} = \left\lVert \mathbf{u} \right\rVert^2 \mathbf{A}_j\mathbf{A}_j^T\mathbf{x} = \left\lVert \mathbf{u} \right\rVert^2 \mathbf{A}_j \left(\mathbf{A}_j \cdot c\mathbf{A}_j\right) = \left\lVert \mathbf{u} \right\rVert^2 \left\lVert \mathbf{A}_j \right\rVert^2 c \mathbf{A}_j = \left\lVert \mathbf{u} \right\rVert^2 \left\lVert \mathbf{v} \right\rVert^2 \mathbf{x}$$
And so $(S \circ T)(\mathbf{x}) = \mathbf{x}$ if and only if $\left\lVert \mathbf{u} \right\rVert^2 \left\lVert \mathbf{v} \right\rVert^2=1$ which means $\left\lVert \mathbf{u} \right\rVert \left\lVert \mathbf{v} \right\rVert=1$.
Again regarding the geometric interpretation of $T$ I don't really know. We have $T(\mathbf{x})=A\mathbf{x}=\mathbf{u}\mathbf{v}^T\mathbf{x}=\mathbf{u}(\mathbf{v} \cdot \mathbf{x})=(\mathbf{v} \cdot \mathbf{x})\mathbf{u}=t_x \mathbf{u}$, where $t_x := \mathbf{v} \cdot \mathbf{x}$ is a scalar dependent on $\mathbf{x}$. And so $T$ is sending every $\mathbf{x}$ on a scalar multiple of $\mathbf{u}$, so $\text{image}(T) = \text{span}(\mathbf{u})$.