Question

1. which sets are a basis for the null space of [(1,1,-1,1),(2,1,1,4),(1,0,0,1)].
2. let T:R^3 to R^3 be defined as T(x,y,z)={x+y, x-y,x+2z). then the basis of range T is...?
3. let R^3 to R^3 defined as T(x,y,z)=(2x,x+y,x-z). then the adjoint operator T*(u,v,w) is (i) (2u+v+w,v,-w), (ii) (2u,v+w,u-w), (iii) (u,v,-w)

Answer 1

1.

$A=\begin{pmatrix} 1 & 1&-1&1 \\ 2 & 1&1&4\\ 1 & 0&0&1\\ \end{pmatrix}$

for null space: $Ax=0$

so, we have:

$x_1=-x_4$

$x_2=x_3$

$-2x_1+2x_2=0\implies x_1=x_2$

solution:

$x=\begin{pmatrix} x_1 \\ x_2\\ x_3\\ x_4 \end{pmatrix}=\begin{pmatrix} x_1 \\ x_1\\ x_1\\ -x_1 \end{pmatrix}=x_1\begin{pmatrix} 1 \\ 1\\ 1\\ -1 \end{pmatrix}$

then basis for the null space:

$\begin{pmatrix} 1 \\ 1\\ 1\\ -1 \end{pmatrix}$

2.

$\begin{pmatrix} 1 &1&0 \\ 1 &-1&0\\ 1 &0&2 \end{pmatrix}\begin{pmatrix} x \\ y\\ z \end{pmatrix}=\begin{pmatrix} x+y \\ x-y\\ x+2z \end{pmatrix}$

$\begin{pmatrix} 1 &1&0 \\ 1 &-1&0\\ 1 &0&2 \end{pmatrix}\to \begin{pmatrix} 1 &1&0 \\ 2 &0&0\\ 1 &0&2 \end{pmatrix}$

basis of range T:

$\begin{pmatrix} 1 \\ 2\\ 1 \end{pmatrix},\begin{pmatrix} 1 \\ 0\\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 0\\ 2 \end{pmatrix}$

3.

i)

Transforming matrix $T$ ;

$\begin{pmatrix} 2x \\ x+y \\ x-z \end{pmatrix}$ =$x\begin{pmatrix} 2\\ 1\\ 1 \end{pmatrix}$ $+y\begin{pmatrix} 0 \\ 1\\ 0 \end{pmatrix}$ +$z\begin{pmatrix} 0 \\ 0\\ -1 \end{pmatrix}$

$T=\begin{pmatrix} 2& 0&0\\ 1&1 & 0\\ 1&0&-1 \end{pmatrix}$

$T^*=T^T=\begin{pmatrix} 2&1 & 1 \\ 0&1& 0\\ 0&0&-1 \end{pmatrix}$

$\begin{pmatrix} 2&1& 1\\ 0&1 & 0\\ 0&0&-1 \end{pmatrix}\begin{pmatrix} u \\ v\\ w \end{pmatrix}=\begin{pmatrix} 2u+v+w \\ v \\ -w \end{pmatrix}$ 

Therefore the adjoint operator $T^*(u,v,w)$

$=(2u+v+w,v,-w)$