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 17: The Spectral Theorem

Let $V$ be a finite dimensional vector space. Let $⟨,⟩:V\times V\to ℂ$ be a positive definite Hermitian form. Let $f:V\to V$ be a linear transformation. The adjoint to $f$ is $f^{*}:V\to V$ such that if $u,w\in V$ then $⟨f\left(u\right),w⟩=⟨u,f^{*}\left(w\right)⟩\text{.}$

The linear transformation $f$ is

$•$	self adjoint, or Hermitian, if $f$ satisfies $$f=f^{*},$$
$•$	an isometry, or unitary, if $f$ satisfies $$f^{*}f=1,$$
$•$	normal, if $f$ satisfies $$f^{*}f=f{f}^{*}\text{.}$$

Let $A$ be an $n\times n$ matrix. The matrix $A$ is

$•$	self adjoint, or Hermitian, if $A$ satisfies $$A={\bar{A}}^t,$$
$•$	an isometry, or unitary, if $A$ satisfies $$$ {\bar{A}}^t$A=1,$$
$•$	normal, if $A$ satisfies $$A{\bar{A}}^t={\bar{A}}^{t}A\text{.}$$

Let $V$ be an inner product space and let $B=\{b_1,b_2,\dots ,b_k\}$ be an orthonormal basis of $V\text{.}$ Then $$C=\{f\left(b_1\right),f\left(b_2\right),\dots ,f\left(b_k\right)\}$$ is an orthonormal basis of $V$ if and only if $f$ is unitary.

(Spectral Theorem). Let $V$ be an inner product space and let $f:V\to V$ be a normal linear transformation. Let $B=\{b_1,\dots b_k\}$ be an orthonormal basis of $V$ and let $A=B_f$ be the matrix of $f$ with respect to $B\text{.}$ Then, there exists a unitary matrix $P$ such that $$PA{P}^{-1}$$ is diagonal.

		Idea of proof.
	Show that $A$ and ${\bar{A}}^t$ have a common eigenvector $v_1$ (so that $A{v}_1=\lambda v_1$ and ${\bar{A}}^t{v}_1=\mu v_1\text{).}$ Let $U_1=\text{span}\left\{v_1\right\}$ and write $V=U_1\oplus U_1^{\perp }\text{.}$ Show that $A$ and ${\bar{A}}^t$ have a common eigenvector $v_2\in U_1^{\perp }\text{.}$ Let $U_2=\text{span}\left\{v_2\right\}$ and write $V=U_1\oplus (U_2\oplus U_2^{\perp })\text{.}$ Continue to get $C=\{v_1,v_2,\dots ,v_k\}\text{.}$ $\square$

Let $$A=\left(\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right),\zeta =\frac{-1+\sqrt{3}i}{2},{\zeta }^2=\frac{-1-\sqrt{3}i}{2}\text{.}$$ Then $${\bar{A}}^t=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right)\text{and}A{\bar{A}}^t=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)={\bar{A}}^{t}A\text{.}$$ So $A$ is a normal matrix. Then $v_1=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$ is an eigenvector $A{v}_1=v_1\text{.}$ If $U_1=\text{span}\left\{\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)\right\}$ then $$U_1^{\perp }=\left\{\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}{a}_1+a_2+a_3=0\}\text{.}$$ Then $v_2=\left(\begin{array}{c}1\\ \frac{-1-\sqrt{3}i}{2}\\ \frac{-1+\sqrt{3}i}{2}\end{array}\right)$ is an eigenvector $A{v}_2=\zeta v_2\text{.}$ $$U_2=\text{span}\left\{v_2\right\}=\text{span}\left\{\left(\begin{array}{c}1\\ \frac{-1-\sqrt{3}i}{2}\\ \frac{-1+\sqrt{3}i}{2}\end{array}\right)\right\}$$ is a subspace of $U_1^{\perp }$ and its complement in $U_1^{\perp }$ is $$\begin{array}{ccc}{\displaystyle U_2^{\perp }}& =& {\displaystyle \left\{\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}{a}_1+a_2+a_3=0,\overline{a_1}+{\zeta }^2\overline{a_2}+\zeta \overline{a_3}=0\}}\\ & =& {\displaystyle \text{span}\left\{\left(\begin{array}{c}1\\ \zeta \\ {\zeta }^2\end{array}\right)\right\}}\end{array}$$ since $\text{dim}\left(U_2^{\perp }\right)=1$ and $1+{\zeta }^2·{\zeta }^2+\zeta ·\zeta =1=\zeta +{\zeta }^2=0\text{.}$ (Note: $\bar{\zeta }={\zeta }^2$ and $\overline{{\zeta }^2}=\zeta \text{).}$ Let $v_3=\left(\begin{array}{c}1\\ \zeta \\ {\zeta }^2\end{array}\right)$ with respect to the basis $\{v_1,v_2,v_3\}=B,$ $$B_a=\left(\begin{array}{ccc}1& 0& 0\\ 0& \zeta & 0\\ 0& 0& {\zeta }^2\end{array}\right)\text{.}$$ If $$P=\frac{1}{\sqrt{3}}\left(\begin{array}{ccc}1& 1& 1\\ 1& {\zeta }^2& \zeta \\ 1& \zeta & {\zeta }^2\end{array}\right)$$ is the change of basis matrix from $S=\{e_1,e_2,e_3\}$ with $e_1=\left(\begin{array}{c}1\\ 0& 0\end{array}\right),$ $e_2=\left(\begin{array}{c}0\\ 1& 0\end{array}\right),$ $e_3=\left(\begin{array}{c}0\\ 0& 1\end{array}\right)$ to $B=\{v_1^{\prime },v_2^{\prime },v_3^{\prime }\}$ with $v_1^{\prime }=\frac{1}{\sqrt{3}}\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right),$ $v_2^{\prime }=\frac{1}{\sqrt{3}}\left(\begin{array}{c}1\\ {\zeta }^2\\ \zeta \end{array}\right),$ $v_3^{\prime }=\frac{1}{\sqrt{3}}\left(\begin{array}{c}1\\ \zeta \\ {\zeta }^2\end{array}\right)$ so that both $S$ and $B$ are orthonormal, then $$P^{-1}={\bar{P}}^t=\frac{1}{\sqrt{3}}\left(\begin{array}{ccc}1& 1& 1\\ 1& \zeta & {\zeta }^2\\ 1& {\zeta }^2& \zeta \end{array}\right),$$ since $P$ is unitary, and $$PA{P}^{-1}=\left(\begin{array}{ccc}1& 0& 0\\ 0& \zeta & 0\\ 0& 0& {\zeta }^2\end{array}\right)$$ is diagonal.

Notes and References

These are a typed copy of Lecture 17 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on August 31, 2011.

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