Find Matrix $A$ of rank 1, if $Ax=y$ and $x$ and $y$ are defined as:

Question

$x = \begin{bmatrix}5\\ 3\\ 9 \end{bmatrix}$

$y = \begin{bmatrix}2\\ 0\\ 1 \end{bmatrix}$

x and y are column matrices.

I have tried to solve then using inversion and direct multiplication, but I got indeterminant form like 0/0.

Answer 1

Because $A$ has rank 1, it must be of the form $A = v w^T$ for some vectors $v$ and $w$. Since $y = Ax = vw^T x = (w^T x) \cdot v$, $v$ must be a multiple of $y$. If we take $v = y$, then this forces $w^T x = 1$, and we can take $w = \frac{1}{x^T x} x$ so that this is satisfied. In summary, $A = \frac{1}{x^Tx} y x^T$ does the trick.

Answer 2

Take the transpose of each side and then solve $$x^T\begin{bmatrix}a\\b\\c\end{bmatrix}=y^T,$$ which is the usual matrix form for 3 equations in the unknowns $a,b,c$.

Answer 3

If $x_1\neq 0,$ $$A=\begin{bmatrix}{y_1\over x_1} &0&0\\{y_2\over x_1} &0&0\\{y_3\over x_1} &0&0 \end{bmatrix} $$ Whenever $x_k$ is nonzero, $A$ can be defined so that only the k-th column is nonzero. (Please disregard my earlier posting, which I tried to delete.)