If $A\in\mathbb{R}^{n\times n}$ is a matrix with rank $1\leq k < n$ there exists two matrices $X\in\mathbb{R}^{n\times k}$ and $Y\in\mathbb{R}^{n\times k}$ both of which have full column rank $k$ such that $$A = XY^T$$ This is called full rank factorization of $A$. The reverse is also true, i.e., if there exists two matrices $X\in\mathbb{R}^{n\times k}$ and $Y\in\mathbb{R}^{n\times k}$ both of which have full column rank $k$ such that $A = XY^T$, then rank of $A$ is $k$.
i.) Show that the full rank factorization of $A$ is not unique.
ii.) Find a basis for $\mathcal{R}(A)$, the range of $A$.
iii.) Characterize a vector in the null space $\mathcal{N}(A)$.
Attempted solution i.) Since the rank of $A$ is $k$ and $A\in\mathbb{R}^{n\times n}$ with rank $1\leq k < n$ then clearly rank of $A$ cannot be $n$ hence $A$ is not invertible which implies that the full rank factorization of $A$ is not unique.
Attempted solution ii.) We need to find the span of the columns of $A$. Consider a nonsingular matrix $M\in\mathbb{R}^{k\times k}$ then we have $$A = XY^T = XMM^{-1}Y^T = (XM)(YM^{-T})^{T} = \tilde{X}\tilde{Y}^T$$
For $x\in\mathbb{R}^n$ consider $$Ax = XY^T x = Xv$$ Since $Y$ us rank $k$ so is $Y^T$ and therefore for any $d\in\mathbb{R}^k$ there exists at least one $x\in\mathbb{R}^n$ such that $d = Y^Tx$ then as a result $\mathcal{R}(A) = \mathcal{R}(X)$. Not sure if this is correct.
Attempted solution iii.) For $x\in\mathcal{R}(A)$ then $Ax = 0$ and $XY^T x = 0$. Since $X$ is full rank i.e., linearly independent column, this implies $Y^T x = 0$ so $x\perp \mathcal{R}(Y)$ and $\mathcal{R}^\perp(Y) = \mathcal{N}(A)$. I am not sure if this is right though.
Best Answer
This is my view point:
i.) For example, we can write $A=XY^T=\left(2X\right)\left(\frac{1}{2}Y^T\right)$. Since $X$ and $Y$ have both rank $k\ge1$, they must not be zero matrices. Thus $X\ne 2X$ and $Y\ne\frac{1}{2}Y$, which implies we have the two different factorizations of $A$.
ii.) Let $\{{\bf a}_1,{\bf a}_2,\ldots,{\bf a}_n\}\subseteq\mathbb{R}^n$ and $\{{\bf x}_1,{\bf x}_2,\ldots,{\bf x}_k\}\subseteq\mathbb{R}^n$ be the sets of column vectors of $A$ and $X$, respectively, and let $$Y^T= \begin{pmatrix} y_{11}&y_{12}&\cdots&y_{1n}\\ y_{21}&y_{22}&\cdots&y_{2n}\\ \vdots&\vdots&&\vdots\\ y_{k1}&y_{k2}&\cdots&y_{kn}\\ \end{pmatrix}.$$ It is clear that the $j$th column of $A$ is $${\bf a}_j=\sum_{i=1}^ky_{ij}{\bf x}_i,\quad \forall j=1,2,\ldots,n.$$ That is, each column of $A$ is a linear combination of columns of $X$, and hence $\mathcal{R}(A)=\mathcal{R}(X)$.
iii.) Your answer is very nice!