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: 21 September 2014

Lecture 16: Orthogonal complements and adjoints

Let $V$ be a vector space over $ℂ\text{.}$ Let $⟨,⟩:V\times V\to ℂ$ be a positive definite Hermitian form. Let $W$ be a subspace of $V\text{.}$ The orthogonal complement to $W$ is $$W^{\perp }=\{v\in V\hspace{0.17em}|\hspace{0.17em}\text{if}\hspace{0.17em}w\in W\hspace{0.17em}\text{then}\hspace{0.17em}⟨v,w⟩=0\}\text{.}$$

(a)	$W^{\perp }$ is a subspace of $V\text{.}$
(b)	$V=W\oplus W^{\perp }\text{.}$

		Proof.
	(a) To show: (aa) If $u_1,u^2\in W^{\perp }$ then $u_1+u^2\in W^{\perp }\text{.}$ (ab) If $u\in W^{\perp }$ and $c\in ℂ$ then $cu\in W^{\perp }\text{.}$ (aa) Assume $u_1,u^2\in W^{\perp }\text{.}$ To show: $u_1+u^2\in W^{\perp }\text{.}$ To show: If $w\in W$ then $⟨u_1+u^2,w⟩=0\text{.}$ Assume $w\in W\text{.}$ To show: $⟨u_1+u^2,w⟩=0\text{.}$ $$\begin{array}{ccc}{\displaystyle ⟨u_1+u^2,w⟩}& =& {\displaystyle ⟨u_1,w⟩+⟨u^2,w⟩}\\ & =& {\displaystyle 0+0,\text{since}\hspace{0.17em}{u}_1,u^2\in W^{\perp }}\\ & =& {\displaystyle 0\text{.}}\end{array}$$ (ab) Assume $u\in W^{\perp }$ and $c\in ℂ\text{.}$ To show: $cu\in W^{\perp }\text{.}$ To show: If $w\in W$ then $⟨cu,w⟩=0\text{.}$ Assume $w\in W\text{.}$ To show: $⟨cu,w⟩=0\text{.}$ $$\begin{array}{ccc}{\displaystyle ⟨cu,w⟩}& =& {\displaystyle c⟨u,w⟩}\\ & =& {\displaystyle c·0,\text{since}\hspace{0.17em}u\in W^{\perp }}\\ & =& {\displaystyle 0\text{.}}\end{array}$$ (b) To show: $V=W\oplus W^{\perp }\text{.}$ To show: (ba) $W\cap W^{\perp }=\left\{0\right\}\text{.}$ (bb) $W+W^{\perp }=V\text{.}$ Choose an orthonormal basis of $W$ (by Gram-Schmidt). Extend this to an orthonormal basis of all $V$ (by more Gram-Schmidt). $$\left\{\underset{\underset{\text{basis of}\hspace{0.17em}V}{⏟}}{\stackrel{\stackrel{\text{basis of}\hspace{0.17em}W}{⏞}}{b_1,b_2,\dots ,b_k},b_{k+1},b_{k+2},\dots ,b_{k+\ell }}\right\}$$ Then $\{b_{k+1},b_{k+2},\dots ,b_{k+\ell }\}$ is an orthonormal basis of $W^{\perp }\text{:}$ If $w=c_1{b}_1+\cdots +c_k{b}_k\in W$ then $$\begin{array}{ccc}{\displaystyle ⟨b_{k+i},w⟩}& =& {\displaystyle ⟨b_{k+i},c_1{b}_1+\cdots +c_k{b}_k⟩}\\ & =& {\displaystyle c_1⟨b_{k+i},b_1⟩+\cdots +c_k⟨b_{k+i},b_k⟩}\\ & =& {\displaystyle c_1·0+\cdots +c_k·0}\\ & =& {\displaystyle 0,}\end{array}$$ so that $b_{k+i}\in W^{\perp }\text{.}$ $\square$

Adjoints

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

The linear transformation $f:V\to V$ 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 $V$ be a finite dimensional vector space over $ℂ$ and $⟨,⟩:V\times V\to ℂ$ a positive definite Hermitian form. Let $f:V\to V$ be a linear transformation and $B=\{b_1,\dots ,{\beta }_k\}$ an orthonormal basis of $V\text{.}$ Then $$B_{f^{*}}={\left(\overline{B_f}\right)}^t\text{.}$$

If $A$ is a matrix with $(i,j)$ entry $A_{ij}$ then $A^t$ is a matrix with $(i,j)$ entry $A_{ji}\text{.}$

Let $A$ be a matrix. The transpose of $A$ is the matrix $A^t$ given by $${\left(A^t\right)}_{ij}=A_{ji}\text{.}$$ The conjugate of $A$ is the matrix $\bar{A}$ given by $${\left(\bar{A}\right)}_{ij}=\overline{A_{ij}}\text{.}$$ The conjugate transpose of $A$ is the matrix ${\bar{A}}^t$ given by $${\left({\bar{A}}^t\right)}_{ij}=\overline{A_{ji}}\text{.}$$

		Proof of the theorem.
	If $$f^{*}\left(b_j\right)=p_{1j}{b}_1+p_{2j}{b}_2+\cdots +p_{kj}{b}_k$$ then $$\begin{array}{ccc}{\displaystyle p_{ij}}& =& {\displaystyle ⟨f^{*}\left(bj\right),b_i⟩}\\ & =& {\displaystyle \overline{⟨b_i,f^{*}\left(b_j\right)⟩}}\\ & =& {\displaystyle \overline{⟨f\left(b_i\right),b_j⟩}}\\ & =& {\displaystyle \overline{⟨q_{i1}+b_1+q_{2i}{b}_2+\cdots +q_{ki}{b}_k,b_j⟩}}\\ & =& {\displaystyle \overline{q_{ji}}\text{.}}\end{array}$$ So $B_{f^{*}}=\left(p_{ij}\right)$ and $B_f=\left(q_{ij}\right)$ and $${\left(B_{f^{*}}\right)}_{ij}=p_{ij}=\overline{q_{ji}}={\left({\overline{B_f}}^t\right)}_{ij}\text{.}$$ So $B_{f^{*}}={\overline{B_f}}^t\text{.}$ $\square$

Let $V$ be a vector space over $ℂ$ which is finite dimensional and let $⟨,⟩:V\times V\to ℂ$ be a positive definite Hermitian form. Let $f:V\to V$ be a linear transformation. Let $g:V\to V$ be a linear transformation. Then

(a)	$f^{*}V\to V$ is a linear transformation and is unique,
(b)	$f^{*}+g^{*}={(f+g)}^{*},$
(c)	${\left(fg\right)}^{*}=g^{*}{f}^{*}\text{,}$
(d)	If $c\in ℂ$ then ${\left(cf\right)}^{*}=\bar{c}{f}^{*},$
(e)	${\left(f^{*}\right)}^{*}\text{.}$

		Idea of proof.
	(a) $f^{*}$ has matrix $B_{f^{*}}={\left(\overline{B_f}\right)}^t$ as given in the theorem. Let $B=\{b_1,\dots ,b_k\}$ be an orthonormal basis and let $$A=B_{f^{*}}\text{and}C=B_{g^{*}}\text{.}$$ Then show that (b') ${\left(\overline{A+C}\right)}^t={\bar{A}}^t+{\bar{C}}^t,$ (c') ${\left(\overline{AC}\right)}^t={\bar{C}}^t{\bar{A}}^t,$ (d') If $c\in C$ then ${\left(\overline{cA}\right)}^t=\bar{c}{\bar{A}}^t,$ (e') ${\overline{\left({\stackrel{‾}{A}}^t\right)}}^t=A\text{.}$ $\square$

Notes and References

These are a typed copy of Lecture 16 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on August 30, 2011.

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