1.1.10 Calculating the SVD

As we will be deal­ing with square ma­tri­ces, we choose the ex­am­ple for a matrix $M\in \mathbb{C}$, to be

We start by com­put­ing $M^{\dagger}$

then cal­cu­late the matrix prod­uct $M^{\dagger}M$

To de­ter­mine our sin­gu­lar value matrix $\Sigma$, we find the eigen­val­ues of $M^{\dagger}M$, take their square roots, and arrange them in de­scend­ing order

and de­ter­mine the in­verse of $\Sigma$, equals

To de­ter­mine the $V$ matrix, we solve for each eigen­val­ue of $M^{\dagger}M$, in order to find the values of the columns of $V$, en­sur­ing these are nor­mal­ized. For the first eigen­val­ue $4.8769,$we have

A sim­i­lar cal­cu­la­tion for the second eigen­val­ue $13.1231$, re­sults in

where the $V$ matrix is then

Taking the SVD de­com­po­si­tion

we solve the SVD for $U$, to find

which for the case ex­am­ple, we find $U$ equals

and the SVD of the matrix $M$, is

To finish, we look at how the SVD de­com­po­si­tion of $M$ acts on the vector $v = (i,1)^{T}$ $\in \mathbb{C}$. Start­ing with the prod­uct

we note that the two-norm of the start­ing and re­sul­tant vec­tors is $\sqrt{2}$, mean­ing in polar co­or­di­nates, the start­ing vector $(i,1)^{T}$ has the same radial dis­tance (mod­u­lus) from the origin as the re­sult­ing vector $V^{\dagger}v \approx (0.8703,\ 1.1147)^{T}$. The trans­for­ma­tion per­formed by the matrix $V^{\dagger}$on $v$, is a ro­ta­tion that maps both com­po­nents of the vector into the real field. Scal­ing $V^{\dagger}v$ by $\Sigma$, we find

noting that our first vector el­e­ment $v_{1}(0.8703)\ has\ been\ scaled\ by\ \sigma_{1}(2.2084)$, equal­ing $1.9220$, and our second vector el­e­ment $v_{2}(1.1147)\ has\ been\ scaled\ by\ \sigma_{2}(3.6226)$, to now equal to $4.0381$. The last step is to mul­ti­ply the above result by the matrix $U$, find­ing

where the $U$ ro­ta­tion gives us the real and imag­i­nary vector el­e­ments once again.