2

Let $A=(a_{ij})$ be an infinite matrix. Consider $|A|=(A^*A)^{1/2}$ and $A'=(|a_{ij}|)$.

Is there any relation between $|A|$ and $A'$?

niki
  • 3,372
  • 1
  • 15
  • 35
  • 4
    Do you really mean infinite (or maybe indefinite)? While in some context infinite matrices may be considered, it does not look like this is one. – Marc van Leeuwen Jan 26 '17 at 08:06
  • 1
    You call determinant the absolute 'value of a matrix'? – StubbornAtom Jan 26 '17 at 08:09
  • @MarcvanLeeuwen : I mean infinite matrice – niki Jan 26 '17 at 08:13
  • If I am right, the Eigenvalues of $|A|$ are the absolute values of those of $A$, and have nothing to do with those of $A'$. –  Jan 26 '17 at 08:21
  • @niki I found a connection in the case of Hilbert-Schmidt operators and added it in the answer, in this case the sum $\sum_j |Ae_j|^2=\sum_{ij}|(Ae_i,e_j)|^2=\sum_{ij}|a_{ij}^2$ does not depend on the choice of the basis – Peter Melech Jan 26 '17 at 09:13
  • 1
    @niki Then you should state more context. What every you mean by $A^$, I am pretty sure $A^ A$ is not defined for all infinite matrices (take for instance one with all entries equal to $1$). – Marc van Leeuwen Jan 26 '17 at 09:59
  • @Marc van Leeuwen I think what is ment is the adjoint of the operator represented by the matrix, that always exists by Riesz´s representation theorem – Peter Melech Jan 26 '17 at 14:04
  • @PeterMelech The problem is not the adjoint, it is the multiplication of infinite matrices. – Marc van Leeuwen Jan 26 '17 at 14:15
  • @Marc van Leeuwen I´m sorry. Of course You´re right,it only makes sense if the matrix can be interpreted as the representation of a bounded linear map! – Peter Melech Jan 26 '17 at 15:03

1 Answers1

3

The notation $|A|$ for $(A^*A)^{\frac{1}{2}}$ is due to an analogy of the polar decomposition of the matrix $A=U|A|$ where $U$ is a partial isometry to the polar decomposition of a complex number $z=e^{i\arg(z)}|z|$. There is no obvious connection to $A^{'}$.

The notation should not be interpreted as an absolute value. One does not have in general

$|A+B|\leq |A|+|B|$ in the sense that the difference is a positive matrix. If $A$ represents a Hilbert-Schmidt operator there is the following connection between $|A|$ and $A^{'}$:

$\sum_{1\leq ij<\infty}|a_{ij}|^2=\sum_{j=1}^{\infty}\sigma_j^2(A)$

where $\sigma_j(A)$ are the singular numbers of $A$, i.e. the eigenvalues of $|A|$.

Peter Melech
  • 4,353