Answer to Question #154539 in Linear Algebra for Ashweta Padhan

Question

Find the adjoint of F:C³ tends to C³ defined by F(z_1, z_2, z₃)=(2z₁+(1-i)z₂ , (3+2i)z₁ -4iz₃ , 2iz₁+(4-3i)z₂ - 3z₃ )

Answer 1

Answer:

Step 1

F:4³$\rarr$4³ by

F$\lparen$z₁,z₂,z₃$\rparen$= $\lparen$2 z₁+$\lparen$1-i$\rparen$z₂, $\lparen$3+2i$\rparen$ z₁-4iz₃, 2iz₁+ $\lparen$4-3i$)$z₂-3z₃$\rparen$

Step 2

Way to find the adjoint of F

$($I) Find matrix of f with respect to basis $\lbrace$e_1,e₂, e₃$\rbrace$

where e₁=$($1,0,0$)$ , e₂=$($0,1,0$)$ , e₃=$($0,0,1$)$

Let A be the matrix corresponding to F.

 $($II) Compute $($ $\overline{ A^T}$$)$

 $($II) Hence adjoint F^*: 4³$\rarr$4³ is

 F^* $\lparen$z₁,z₂,z₃$\rparen$= $($ $\overline{ A^T}$$)$ $\begin{bmatrix} z1\\ z2\\ z3\\ \end{bmatrix}$

Now

F$($e₁$)$ = F$($1,0,0$)$ = $($ 2,$\lparen$3+2i$\rparen$, 2i$)$

F$($e₂$)$ = F$($0,1,0$)$ = $($ 1-i, 0, 4-3i$)$

F$($e₃$)$ = F$($0,0,1$)$ = $($ 0,-4i, -3$)$

Step 3

Hence

A= $\begin{bmatrix} 2 & 1-i & 0 \\ 3+2i & 0 & -4i\\ 2i & 4-3i & -3 \\ \end{bmatrix}$

A^T= $\begin{bmatrix} 2 & 3+2i & 2i \\ 1-i & 0 & 4-3i \\ 0 & -4i & -3 \\ \end{bmatrix}$

 $\overline{ A^T}$ =$\begin{bmatrix} 2 & 3-2i & -2i \\ 1+i & 0 & 4+3i \\ 0 & 4i & -3 \\ \end{bmatrix}$

Hence  F^*: 4³$\rarr$4³ is

 F^* $\lparen$z₁,z₂,z₃$\rparen$= $($ $\overline{ A^T}$$)$ $\begin{bmatrix} z1\\ z2\\ z3\\ \end{bmatrix}$

=$\begin{bmatrix} 2 & 3-2i & -2i \\ 1+i & 0 & 4+3i \\ 0 & 4i & -3 \\ \end{bmatrix}$$\begin{bmatrix} z1\\ z2\\ z3\\ \end{bmatrix}$

=$\lparen$2 z₁+$\lparen$3+2i$\rparen$z₂- 2iz_3, $\lparen$1+i$\rparen$z₁+4+3iz₃, iz₂-3z₃$\rparen$

Therefore,

F $\lparen$z₁,z₂,z₃$\rparen$=$\lparen$2 z₁+$\lparen$3+2i$\rparen$z₂- 2iz_3, $\lparen$1+i$\rparen$z₁+4+3iz₃, iz₂-3z₃$\rparen$