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This should be easy, but...

I would like to express the singular value decomposition of a 2 x 2 complex matrix $A$ as function of its coefficients $A_{ij}$. In "closed form", no intermediate values, straight up.

What I mean is that if I express the SVD in terms of eigenvalues, say, then I have to write those eigenvalues in terms of the $A_{ij}$'s and substitute. In other words, I don't want to write the solution to SVD in terms of the solution of something else.

Possibly useful background:

In the Wikipedia page on SVD, if you write $A$ as a linear combination of the Pauli matrices (and identity), the singular values are expressed in terms of the corresponding coefficients. But then I don't know how you find those coefficients, so I can't tell if they themselves are in "closed form".

http://en.wikipedia.org/wiki/Singular_value_decomposition

Thanks.

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  • $\begingroup$ Have you tried yourself? $\endgroup$ – sebas Mar 4 '14 at 23:29
  • $\begingroup$ Is this a structure tensor in the form of $[a, b; b, c]$ ? $\endgroup$ – Tolga Birdal Mar 5 '14 at 17:33
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The Pauli matrices and the identity matrix form an orthogonal basis of the space of $2\times 2$ matrices, so finding the expansion coefficients amounts to just a projection onto this basis (i.e., you need to form the inner product of your matrix onto each element of this basis).

But, for $2\times 2$ matrices, the answer is simple enough to write things down right away. The eigenvalues are those values $\lambda$ for which $$ \det (A-\lambda I) = \det \begin{pmatrix} a_{11}-\lambda & a_{12} \\ a_{21} & a_{22}-\lambda \end{pmatrix} = 0. $$ This is a quadratic equation in $\lambda$: $$ (a_{11}-\lambda)(a_{22}-\lambda)-a_{12}a_{21} = 0 $$ for which you can write down the roots immediately as a function of the elements of $A$.

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