Show that the $3 \times 3$ matrix $A$ can be writtin as $TT'$ for some $3 \times 2$ matrix $T$

Question

$A=\begin{bmatrix} 5 & 4 & 0 \\ 4 & 5 & 3 \\ 0 & 3 & 5 \end{bmatrix}$. There is a theorem in my book that is an $m \times m$ nonnegative definite matrix of rank $r$ can be written as $TT'$ for some $m \times r$ matrix $T$. The theorem doesn't describe how to find $T$ though. The matrix above has rank $2$ and is nonnegative definite so I know $T$ exists but don't know where to start to find it. Any hints?

Answer 1

You can diagonalize $A$ by an orthogonal matrix $O$, that is $$A = O D O^{-1}=ODO'$$ where $O^{-1}=O'$ and $D$ diagonal with first $k$ diagonal entries $>0$ and the rest $0$. $$A = (OE)(E'O')=(OE)(OE)'$$ Now the matrix $OE$ has the last $n-k$ columns $0$. Remove them to get $T$. Check that $TT'=(OE)(OE)'=A$.