# Properties of Adjoint of a Linear Map

Aug-Nov 2020

## Recap

• Vector space $V$ over a scalar field $F$
• $F$: real field $\mathbb{R}$ or complex field $\mathbb{C}$ in this course
• $m\times n$ matrix A represents a linear map $T:F^n\to F^m$
• dim null $T$ + dim range $T$ = dim $V$
• Solution to $Ax=b$ (if it exists): $u+$ null$(A)$
• Four fundamental subspaces of a matrix
• Column space, row space, null space, left null space
• Eigenvalue $\lambda$ and Eigenvector $v$: $Tv=\lambda v$
• Some linear maps are diagonalizable
• Inner products, norms, orthogonality and orthonormal basis
• Upper triangular matrix for a linear map over an orthonormal basis
• Orthogonal projection gives closest vector in the subspace
• Least squares solution to a linear equation is orthogonal projection
• Adjoint of a linear map: $\langle Tv,w\rangle=\langle v,T^*w\rangle$

## Properties

$V,W,U$: finite-dimensional inner product spaces over $F=\mathbb{R}$ or $\mathbb{C}$

1. $(S+T)^* = S^* + T^*$, where $S,T:V\to W$ are linear maps.

2. $(\lambda T)^* = \bar{\lambda}T^*$, where $\lambda\in F$.

3. $(T^*)^*=T$.

4. $I^*=I$, where $I$ is the identity operator.

5. $(ST)^* = T^* S^*$, where $S:W\to U$ and $T:V\to W$ are linear maps.

Sample Proofs

1. $\langle v,(S+T)^* w\rangle=\langle (S+T)v,w\rangle=\langle Sv,w\rangle+\langle Tv,w\rangle=\langle v,S^* w\rangle+\langle v,T^* w\rangle=\langle v,(S^* +T^*)w\rangle$

2. $\langle v,(\lambda T)^* w\rangle=\langle \lambda T v, w\rangle=\lambda\langle T v, w\rangle=\lambda\langle v, T^* w\rangle=\langle v, \bar{\lambda} T^* w\rangle$

## Null and range of adjoint

$T:V\to W$, a linear map and $T^*:W\to V$, adjoint of $T$

1. null $T^* =$ $($range $T)^{\perp}$

2. range $T^* =$ $($null $T)^{\perp}$

3. null $T =$ $($range $T^*)^{\perp}$

4. range $T =$ $($null $T^*)^{\perp}$

5. dim range $T=$ dim range $T^*$

Proof

$w\in$ null $T^*$

iff $T^* w=0$, iff $\langle v,T^* w\rangle=0$ for all $v\in V$, iff $\langle Tv,w\rangle=0$ for all $v\in V$

iff $w\in$ $($range $T)^{\perp}$

(4) is complement of (1), (3) is (1) with $T$ set as $T^*$, (2) is complement of (3)

For (5), use fundamental theorem of linear maps on (3)

## Conjugate transpose: Matrix of adjoint

$V,W$: finite-dimensional inner product spaces over $F=\mathbb{R}$ or $\mathbb{C}$

$T:V\to W$, a linear map

Orthonormal basis for $V$: $B_V=\{e_1,\ldots,e_n\}$

Orthonormal basis for $W$: $B_W=\{f_1,\ldots,f_m\}$

Matrix of $T$ w.r.t. $B_V$ and $B_W$ is $M(T,B_V,B_W)$

$M(T^*,B_W,B_V)=\text{conjugate-transpose}(M(T,B_V,B_W))$

Conjugate-transpose of a matrix: transpose and conjugate each element

Proof

$(i,j)$-th entry of $M(T,B_V,B_W)$: $\langle Te_j,f_i\rangle=\langle e_j,T^* f_i\rangle=\overline{\langle T^* f_i,e_j\rangle}$

$(j,i)$-th entry of $M(T^* ,B_W,B_V)$: $\langle T^* f_i,e_j\rangle$

## Adjoint and four fundamental spaces of a matrix

$A$: $m\times n$ matrix over $\mathbb{R}$

Represents $T:\mathbb{R}^n\to \mathbb{R}^m$ w.r.t. standard basis

$T^*$: represented by the conjugate-transpose$(A)$ or $\overline{A^T}$

Since entries of $A$ are real: $\overline{A^T}=A^T$

1. range $T^* =$ colspace $A^T =$ rowspace $A =$ $($null $A)^{\perp}$

2. null $T^* =$ null $A^T =$ left-null $A =$ $($colspace $A)^{\perp}$

3. range $T =$ colspace $A =$ rowspace $A^T =$ $($left-null $A)^{\perp}$

4. null $T =$ null $A =$ left-null $A^T =$ $($rowspace $A)^{\perp}$

## Composition of operators and their null spaces

$T:V\to W$ and $S:W\to U$ are linear maps

$ST:V\to U$ is a linear map

Suppose $N_W=$ range $T\cap$ null $S$ (a subspace of $W$)

dim null $ST=$ dim null $T+$ dim $N_W$

Proof

$v\in$ null $ST$ iff $STv=0$ iff $Tv\in$ null $S$ iff $Tv\in N_W$

null $ST=\{v:Tv\in N_W\}$

Basis for null $T$: $\{v_1,\ldots,v_k\}$

Basis for $N_W$: $\{w_1,\ldots,w_r\}$, $r=$ dim $R$

Let $v_{k+i}\in V$ be s.t. $Tv_{k+i}=w_i$, $i=1,\ldots,r$

Basis for null $ST$: $\{v_1,\ldots,v_k,v_{k+1},\ldots,v_{k+r}\}$

## Composition of operators: special case

$T:V\to W$ and $S:W\to U$ are linear maps

$ST:V\to U$ is a linear map

if range $T$ and null $S$ intersect only at 0, null $ST=$ null $T$

• $S$ retains all “details” of $T$

• $S$ restricted to range $T$ is one-to-one
• By fundamental theorem of linear maps, dim range $ST=$ dim range $T$

## The operators $TT^*$ and $T^*T$

$T:V\to W$ and $T^*: W\to V$

$T^*T: V\to V$

null $T^* =$ $($range $T)^{\perp}$

So, null $T^*$ and range $T$ intersect only at $0$

Therefore, null $T^* T=$ null $T$ and dim range $T=$ dim range $T^* T=$ dim range $T^*$

$TT^*: W\to W$

null $T =$ $($range $T^*)^{\perp}$

So, null $T$ and range $T^*$ intersect only at $0$

Therefore, null $TT^* =$ null $T^*$ and dim range $T=$ dim range $TT^*=$ dim range $T^*$