Find the singular value decompostion of the summation of the product of vectors

Question

the normal SVD is given by:

Let $A$ be an n x m matrix with rank($A$)=r. Then $A = PDQ'$ but now i saw a new form of writing this is $A = PDQ'= \sum_{i=0}^r d_i p_i q_i'$ where $P$ is n x r orthonormal and $Q$ m x r orthonormal.

Now the problem is that i have to find the singular value decomposition of matrix $A$ given $A=aa'-bb'$, where $a=(2,2,2)'$ and $b=(-1,2,-1)'$

Is there any quick way using vector $a$ and $b$ to find the singular value decomposition?

The only way i can think of is writing out the matrix but then i can't figure out the eigenvalues so i'm stuck here. Does someone knows what to do here?

Answer 1

As $a$ and $b$ are already orthogonal, all you need to do is to scale them to orthonormal vectors: $$u:= \frac{1}{\sqrt{12}}a\;\;\text{and}\;\;v:=\frac{1}{\sqrt{6}}b$$ Then you have $A=u (12) u'+ (-v) 6 v'$ and you can choose $$P=\begin{pmatrix} & \\ u & -v \\ & \end{pmatrix}\;\;\text{and}\;\;Q=\begin{pmatrix} & \\ u & v \\ & \end{pmatrix}$$ and $$D=\begin{pmatrix} 12 & 0 \\ 0 & 6 \end{pmatrix}$$