Singular value decomposition of adjoint

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Homework Statement


Let T be a linear transformation over V, a finite-dimensional complex inner product space (with inner product < , >). Suppose T has the singular value decomposition
[tex] Tv = \sum_j^n s_j \left\langle v,e_j \right\rangle f_j [/tex]
where s_1, ... s_n are the singular values of T and (e_1, ..., e_n) and (f_1, ..., f_n) are orthonormal bases of V.

I'm supposed to show that (* denotes adjoint)
[tex] T^*v = \sum_j^n s_j \left\langle v, f_j\right\rangle e_j. [/tex]



The Attempt at a Solution


(I'll suppress the upper limit on the sums. Overline denotes complex conjugation.)
[tex] T^*v = \sum_j \left\langle T^*v, e_j \right\rangle e_j [/tex]
Require <Te_i, v> = <e_i, T*v> for i = 1, ..., n.
Calculating these, I get:
[tex]
\left\langle Te_i,v \right\rangle = \left\langle \sum_j s_j \left\langle e_i, e_j \right\rangle f_j, v \right\rangle = \sum_j s_j \left\langle e_i, e_j \right\rangle \left\langle f_j, v \right\rangle = \sum_j s_j \delta_{ij} \left\langle f_j, v \right\rangle = s_i \left\langle f_i, v \right\rangle
[/tex]
And:
[tex]
\left\langle e_i, T^*v \right\rangle = \left\langle e_i, \sum_j \left\langle T^*v, e_j \right\rangle e_j \right\rangle = \sum_j \overline{\left\langle T^*v, e_j \right\rangle} \left\langle e_i, e_j \right\rangle = \sum_j \overline{\left\langle T^*v, e_j \right\rangle} \delta_{ij} = \overline{\left\langle T^*v, e_i \right\rangle}. [/tex]

Hence for i = 1, ..., n, I obtain:
[tex]
\left\langle T^*v, e_i \right\rangle = \overline{s_i}\left\langle v, f_i \right\rangle
[/tex]

This *almost* gives the desired expression, except for I'm off by the complex conjugate of the s_i's. Where did I go wrong?
 
Last edited:

Answers and Replies

  • #2
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I figured it out while sleeping. The singular values of T are the eigenvalues of [itex] \sqrt{T^*T} [/itex] which is a self-adjoint operator, and therefore has real eigenvalues. So [itex] s_i = \overline{s_i} [/itex] for each i. The desired relation follows.
 

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