Matrix decomposition into column vector times its transpose

Question

I would like to decompose the matrix, \begin{equation*} \mathbf{C}=\sigma_w^2 \begin{bmatrix} f_2^2&f_1f_2&0\\[10pt] f_1f_2&f_1^2+f_2^2&f_1f_2\\[10pt] 0&f_1f_2&f_1^2 \end{bmatrix} \end{equation*} into, $$\mathbf{C}=\mathbf{f}.\mathbf{f}^T\sigma_w^2$$

Is there a technique to solve for $\mathbf{f}$?

Answer 1

Since the determinant of

$\mathbf{C}=\sigma_w^2 \begin{bmatrix} f_2^2&f_1f_2&0\\[10pt] f_1f_2&f_1^2+f_2^2&f_1f_2\\[10pt] 0&f_1f_2&f_1^2 \end{bmatrix} $

is null, the determinant of $\mathbf{f}$ must also be null.

Therefore,

$\mathbf{f}=\begin{bmatrix} f_2&0&0\\[10pt] f_1&f_2&0\\[10pt] 0&f_1&0 \end{bmatrix}$.

Answer 2

Let $f$ be the column vector $\begin{bmatrix}a&b&c\end{bmatrix}^T$. If $C$ is the outer product $ff^T\!$ of $f$ with itself (as opposed to the inner product $f^T\!f$, which results in a scalar), then

$$ C = ff^T = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \cdot \begin{bmatrix} a & b & c \end{bmatrix} = \begin{bmatrix} a^2 & ab & ac \\ ab & b^2 & bc \\ ac & bc & c^2 \end{bmatrix} = \begin{bmatrix} f_2^2 & f_1f_2 & 0 \\ f_1f_2 & f_1^2+f_2^2 & f_1f_2 \\ 0 & f_1f_2 & f_1^2 \end{bmatrix}. $$

To find the values of $a$, $b$ and $c$ al you need to do is solve the following system of equations.

$$ \left\{\;\begin{array}{ccc} a^2 & = & f_2^2 \\ b^2 & = & f_1^2 + f_2^2 \\ c^2 & = & f_1^2 \\ ab & = & f_1f_2 \\ ac & = & 0 \\ bc & = & f_1f_2 \end{array} \right. $$

You might be interested in checking out the wikipedia article for the outer product.

Answer 3

(Since OP did not mention what is $\sigma_w^2$, assuming it to be a scalar) This cannot be done unless both $f_1,f_2=0$. Note that the matrix $C/\sigma_w^2$ (assuming $\sigma_w^2$ to be a scalar) has rank more than or equal to 2 (the first and the last rows are independent unless $f_1=f_2=0$). But any such decomposition should have rank less than or equal to 1. So it is not possible.