Math Problems Vault

Problem 134

If $A$ is a bounded linear transformation from a Hilbert space $\mathbf{H}$ to a Hilbert space $\mathbf{K}$ , then there exists a partial isometry $U$ (from $\mathbf{H}$ to $\mathbf{K}$ ) and there exists a positive operator $P$ (on $\mathbf{H}$ ) such that $A = UP$ . The transformations $U$ and $P$ can be found so that $\ker U = \ker P$ , and this additional condition uniquely determines them.

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Remark.

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The representation of $A$ as the product of the unique $U$ and $P$ satisfying the stated conditions is called the polar decomposition of $A$ , or, more accurately, the right-handed polar decomposition of $A$ . The corresponding left-handed theory ( $A = PU$ ) follows by a systematic exploitation of adjoints.

Problem 135

Every bounded linear transformation is the product of a maximal partial isometry and a positive operator.

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Problem 136

What are the extreme points of the closed unit ball in the space of operators on a Hilbert space?

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Remark.

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The closed unit ball in the space of operators is convex. For every interesting convex set, it is of interest to determine the extreme points.

Problem 137

An operator with polar decomposition $UP$ is quasinormal if and only if $UP = PU$ .

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Remark.

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An operator $A$ is called quasinormal if $A$ commutes with $A^{*}A$ . Every normal operator is quasinormal, but the converse is false: if $A$ is an isometry, then $A^{*}A = 1$ and therefore $A$ commutes with $A^{*}A$ , but if $A$ is not unitary, then $A$ is not normal (a concrete example is the unilateral shift).

Problem 138

Is it true for every operator $A$ that

\left| \left\langle Af, g \right\rangle \right|^{2} \leq \left\langle \sqrt{A^{*}A}f, f \right\rangle \cdot \left\langle \sqrt{A^{*}A}g, g \right\rangle ?

What if $\sqrt{AA^{*}}$ is used in place of $\sqrt{A^{*}A}$ ? What if they are both used, one in each factor?

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Remark.

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If $A$ is a positive operator, then $\left\langle Af, g \right\rangle$ defines an inner product (not necessarily strictly positive); it follows that $A$ satisfies the Schwarz-like inequality

\left| \left\langle Af, g \right\rangle \right|^{2} \leq \left\langle Af, f \right\rangle \cdot \left\langle Ag, g \right\rangle .

(To feel comfortable about this relation it helps to recall that the positiveness of $A$ implies $(\sqrt{A})^{2} = A$ .) If $A$ is not positive, then the inner-product expression $\left\langle Af, g \right\rangle$ still defines something (sesquilinear, yes, positive and symmetric, probably not), but since the equation $(\sqrt{A})^{2} = A$ is not available, it is not clear what, if any, Schwarz-like inequality prevails. A possible guess is to replace $A$ on the right (majorant) side by $\sqrt{A^{*}A}$ . (Reason: if $z$ is a complex number, then $\sqrt{z^{*}z} = \left|z\right|$ .) Objection: why $\sqrt{A^{*}A}$ ? Why not $\sqrt{AA^{*}}$ ?

Problem 139

Which weighted shifts are quasinormal?

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Problem 140

The set of all operators that have either a left or a right inverse is dense, but the set of all operators that have both a left and a right inverse (i.e., the set of all invertible operators) is not. What about the set of left-invertible operators?

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Problem 141

The set of all invertible operators is connected.

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