Question #283966
  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)
1
Expert's answer
2022-01-03T11:47:07-0500

1.

A=(111121141001)A=\begin{pmatrix} 1 & 1&-1&1 \\ 2 & 1&1&4\\ 1 & 0&0&1\\ \end{pmatrix}


for null space: Ax=0Ax=0

so, we have:

x1=x4x_1=-x_4

x2=x3x_2=x_3

2x1+2x2=0    x1=x2-2x_1+2x_2=0\implies x_1=x_2

solution:

x=(x1x2x3x4)=(x1x1x1x1)=x1(1111)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:

(1111)\begin{pmatrix} 1 \\ 1\\ 1\\ -1 \end{pmatrix}


2.

(110110102)(xyz)=(x+yxyx+2z)\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}


(110110102)(110200102)\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:

(121),(100),(002)\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 TT ;

(2xx+yxz)\begin{pmatrix} 2x \\ x+y \\ x-z \end{pmatrix} =x(211)x\begin{pmatrix} 2\\ 1\\ 1 \end{pmatrix} +y(010)+y\begin{pmatrix} 0 \\ 1\\ 0 \end{pmatrix} +z(001)z\begin{pmatrix} 0 \\ 0\\ -1 \end{pmatrix}


T=(200110101)T=\begin{pmatrix} 2& 0&0\\ 1&1 & 0\\ 1&0&-1 \end{pmatrix}


T=TT=(211010001)T^*=T^T=\begin{pmatrix} 2&1 & 1 \\ 0&1& 0\\ 0&0&-1 \end{pmatrix}


(211010001)(uvw)=(2u+v+wvw)\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)T^*(u,v,w)

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



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