Adjoint Operators and Matrices

Subjects: Linear Algebra
Links: Space of Linear Transformations, Matrix Representation of Linear Transformations, Inner Products and Norms, Orthogonal Complements
Def: Suppose $T\in \mathcal{L}(V,W)$. The adjoint of $T$ is the function $T^{\ast }:W\to V$ such that

for every $v\in V$ and every $w\in W$.

If $T\in \mathcal{L}(V,W)$, then $T^{\ast }\in \mathcal{L}(W,V)$.

• Properties
  $(S+T{)}^{\ast }=S^{\ast }+T^{\ast }$, for all $S,T\in \mathcal{L}(V,W)$
  $(\lambda T{)}^{\ast }=\bar{\lambda }{T}^{\ast }$, for all $\lambda \in \mathbb{F}$ and $T\in \mathcal{L}(V,W)$
  $(T^{\ast }{)}^{\ast }=T$ for all $T\in \mathcal{L}(V,W)$
  $I^{\ast }=I$, where $I$ is the identity operator on $V$
  $(ST{)}^{\ast }=T^{\ast }{S}^{\ast }$ for all $T\in \mathcal{L}(V,W)$ and $S\in \mathcal{L}(W,U)$

Th: For arbitrary vector spaces $V$ and $W$, and $T\in \mathcal{L}(V,W)$ then

• $N(T^{\ast })=R(T{)}^{\mathrm{\perp }}$
• $N(T)=R(T^{\ast }{)}^{\mathrm{\perp }}$
• $(\ast ){\mathrm{cl}}_V(R(T^{\ast }))=N(T{)}^{\mathrm{\perp }}$
• $(\ast ){\mathrm{cl}}_W(R(T))=N(T^{\ast }{)}^{\mathrm{\perp }}$
• $R(T^{\ast })\subseteq N(T{)}^{\mathrm{\perp }}$
• $R(T)\subseteq N(T^{\ast }{)}^{\mathrm{\perp }}$

Some special cases are:

• $R(T^{\ast })=N(T{)}^{\mathrm{\perp }}$ if $V$ is finite dimensional
• $R(T)=N(T^{\ast }{)}^{\mathrm{\perp }}$ if $W$ is finite dimensional

Th: Let $V$ be a finite dimensional inner product space, and $\beta$ be an orthonormal basis for $V$. If $T\in \mathcal{L}(V)$, then

Where $[T{]}_{\beta }^{\ast }$ is the conjugate transpose of $[T{]}_{\beta }$.
Cor: Let $A\in {\mathcal{M}}_n(\mathbb{C})$, then $L_{A^{\ast }}=(L_A{)}^{\ast }$

Cor: If $A\in {\mathcal{M}}_{m\times n}(\mathbb{F})$, $x\in {\mathbb{F}}^n$, and $y\in {\mathbb{F}}^m$. Then

Th: Let $A\in {\mathcal{M}}_{m\times n}(\mathbb{F})$. Then $\mathrm{rank}(A^{\ast }A)=\mathrm{rank}(A)$

Cor: If $A\in {\mathcal{M}}_{m\times n}(\mathbb{F})$ such that $\mathrm{rank}(A)=n$, then $A^{\ast }A$ is invertible.

Th: Let $V$ be finite dimensional and $T\in \mathcal{L}(V)$. If $T$ has eigenvector, then so does $T^{\ast }$. Additionally, if $\lambda$ is an eigenvalue of $T$, then $\bar{\lambda }$ is an eigenvalue of $T^{\ast }$.

Th: Suppose $V$is a complex inner product space and $T\in \mathcal{L}(V)$. If for all $v\in V$

Then $T=0$.