Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 23 September 2014

Lecture 19: Polar decomposition

Let $f : V \to V$ be a linear transformation. Let $⟨, ⟩ : V \times V \to ℂ$ be a positive definite Hermitian form. Show that the following are equivalent.

(a)	$f$ is self adjoint and all eigenvalues are positive.
(b)	There exists $g : V \to V$ such that $g$ is self adjoint and $f = g^{2} .$
(c)	There exists $h : V \to V$ such that $f = h h^{*} .$
(d)	$f$ is self adjoint and $⟨ f (v), v ⟩ \geq 0$ for all $v \in V .$

Proof.

To show: (a) $\Rightarrow$ (b) $\Rightarrow$ (c) $\Rightarrow$ (d) $\Rightarrow$ (a).

(a) $\Rightarrow$ (b) Assume $f$ is self adjoint and all eigenvalues are positive.
To show: There exists $g : V \to V$ such that $g$ is self adjoint and $f = g^{2} .$
Since $f$ is self adjoint, $f$ is normal.
By the spectral theorem, there exists an orthonormal basis $B = {b_{1}, b_{2}, \dots, b_{k}}$ such that $B_{f} = (\begin{matrix} d_{1} & 0 \\ d_{2} \\ ⋱ \\ 0 & d_{k} \end{matrix}) .$ Since all eigenvalues of $f$ are positive, $d_{1}, d_{2}, \dots, d_{k} \in ℝ_{\geq 0} .$
Let $B_{g} = (\begin{matrix} \sqrt{d_{1}} & 0 \\ ⋱ \\ 0 & \sqrt{d_{k}} \end{matrix})$ be the matrix of $g : V \to V .$

To show:

(1)	$g$ is self adjoint.
(2)	$f = g^{2} .$

(1)	To show: $g = g^{} .$ $B_{g^{}} = {\overline{B_{g}}}^{t} = (\begin{matrix} \overline{\sqrt{d_{1}}} & 0 \\ ⋱ \\ 0 & \overline{\sqrt{d_{k}}} \end{matrix}) = (\begin{matrix} \sqrt{d_{1}} & 0 \\ ⋱ \\ 0 & \sqrt{d_{k}} \end{matrix}) = B_{g},$ since $\sqrt{d_{1}}, \dots, \sqrt{d_{k}} \in ℝ .$ So $g^{*} = g .$
(2)	To show: $f = g^{2} .$ $B_{g^{2}} = {(B_{g})}^{2} = {(\begin{matrix} \sqrt{d_{1}} & 0 \\ ⋱ \\ 0 & \sqrt{d_{k}} \end{matrix})}^{2} = (\begin{matrix} d_{1} & 0 \\ ⋱ \\ 0 & d_{k} \end{matrix}) = B_{f} .$ So $g^{2} = f .$

(b) $\Rightarrow$ (c) Assume there exists $g : V \to V$ such that $g$ is self adjoint and $f = g^{2} .$
To show: There exists $h : V \to V$ such that $f = h h^{*} .$
Let $h = g .$
To show: $f = h h^{*} .$ $h h^{*} = g g^{*} = g g = g^{2} = f .$ $(g = g^{*}$ since $f$ is self adjoint.)

To show:

(a)	$f$ is self adjoint.
(b)	If $v \in V$ then $⟨ f (v), v ⟩ \geq 0 .$

(1)	To show: $f = f^{} .$ $\begin{matrix} f^{} & = & {(h h^{})}^{} = {(h^{})}^{} h^{}, {(a b)}^{} = b^{} a^{}, \\ = & h h^{} = f, since {(a^{})}^{*} = a . \end{matrix}$
(2)	Assume $v \in V .$ To show: $⟨ f (v), v ⟩ \in ℝ_{\geq 0} .$ $⟨ f (v), v ⟩ = ⟨ h h^{} v, v ⟩ = ⟨ h^{} v, h^{*} v ⟩ \in ℝ_{\geq 0},$ since $⟨, ⟩$ is positive definite.

(d) $\Rightarrow$ (a) Assume $f$ is self adjoint and if $v \in V$ then $⟨ f (b), v ⟩ \in ℝ_{\geq 0} .$

To show:

(1)	$f$ is self adjoint.
(2)	All eigenvalues of $f$ are positive.

(1)

To show:

f

is self adjoint.
By assumption,

f

is self adjoint.

(2)

To show: All eigenvalues of

f

are positive.
To show: If

λ \in ℂ

and

v \in V

and

f v = λ v

then

λ \in ℝ_{\geq 0} .

Assume

λ \in ℂ

and

v \in V

and

f v = λ v .

To show:

λ \in ℝ_{\geq 0} .

We know:

⟨ f (v), v ⟩ \in ℝ_{\geq 0} .

⟨ f v, v ⟩ = ⟨ λ v, v ⟩ = λ ⟨ v, v ⟩ \in ℝ_{\geq 0} .

Since

⟨ v, v ⟩ \in ℝ_{\geq 0},

because

⟨, ⟩

is positive definite, then

λ \in ℝ_{\geq 0} .

$□$

Let $A \in {G L}_{n} (ℂ) .$ Then there exist $P,$ diagonalisable with positive eigenvalues, and $U,$ unitary, such that $A = P U .$

Idea of proof.

Let $P$ be such that $P^{2} = A {\overline{A}}^{t},$ and $U = P^{- 1} A .$

$□$

Notes and References

These are a typed copy of Lecture 19 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on September 6, 2011.

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