1.
A = ( 1 1 − 1 1 2 1 1 4 1 0 0 1 ) A=\begin{pmatrix}
1 & 1&-1&1 \\
2 & 1&1&4\\
1 & 0&0&1\\
\end{pmatrix} A = ⎝ ⎛ 1 2 1 1 1 0 − 1 1 0 1 4 1 ⎠ ⎞
for null space: A x = 0 Ax=0 A x = 0
so, we have:
x 1 = − x 4 x_1=-x_4 x 1 = − x 4
x 2 = x 3 x_2=x_3 x 2 = x 3
− 2 x 1 + 2 x 2 = 0 ⟹ x 1 = x 2 -2x_1+2x_2=0\implies x_1=x_2 − 2 x 1 + 2 x 2 = 0 ⟹ x 1 = x 2
solution:
x = ( x 1 x 2 x 3 x 4 ) = ( x 1 x 1 x 1 − x 1 ) = x 1 ( 1 1 1 − 1 ) 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} x = ⎝ ⎛ x 1 x 2 x 3 x 4 ⎠ ⎞ = ⎝ ⎛ x 1 x 1 x 1 − x 1 ⎠ ⎞ = x 1 ⎝ ⎛ 1 1 1 − 1 ⎠ ⎞
then basis for the null space:
( 1 1 1 − 1 ) \begin{pmatrix}
1 \\
1\\
1\\
-1
\end{pmatrix} ⎝ ⎛ 1 1 1 − 1 ⎠ ⎞
2.
( 1 1 0 1 − 1 0 1 0 2 ) ( x y z ) = ( x + y x − y x + 2 z ) \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} ⎝ ⎛ 1 1 1 1 − 1 0 0 0 2 ⎠ ⎞ ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ x + y x − y x + 2 z ⎠ ⎞
( 1 1 0 1 − 1 0 1 0 2 ) → ( 1 1 0 2 0 0 1 0 2 ) \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} ⎝ ⎛ 1 1 1 1 − 1 0 0 0 2 ⎠ ⎞ → ⎝ ⎛ 1 2 1 1 0 0 0 0 2 ⎠ ⎞
basis of range T:
( 1 2 1 ) , ( 1 0 0 ) , ( 0 0 2 ) \begin{pmatrix}
1 \\
2\\
1
\end{pmatrix},\begin{pmatrix}
1 \\
0\\
0
\end{pmatrix},\begin{pmatrix}
0 \\
0\\
2
\end{pmatrix} ⎝ ⎛ 1 2 1 ⎠ ⎞ , ⎝ ⎛ 1 0 0 ⎠ ⎞ , ⎝ ⎛ 0 0 2 ⎠ ⎞
3.
i)
Transforming matrix T T T ;
( 2 x x + y x − z ) \begin{pmatrix}
2x \\
x+y \\
x-z
\end{pmatrix} ⎝ ⎛ 2 x x + y x − z ⎠ ⎞ =x ( 2 1 1 ) x\begin{pmatrix}
2\\
1\\
1
\end{pmatrix} x ⎝ ⎛ 2 1 1 ⎠ ⎞ + y ( 0 1 0 ) +y\begin{pmatrix}
0 \\
1\\
0
\end{pmatrix} + y ⎝ ⎛ 0 1 0 ⎠ ⎞ +z ( 0 0 − 1 ) z\begin{pmatrix}
0 \\
0\\
-1
\end{pmatrix} z ⎝ ⎛ 0 0 − 1 ⎠ ⎞
T = ( 2 0 0 1 1 0 1 0 − 1 ) T=\begin{pmatrix}
2& 0&0\\
1&1 & 0\\
1&0&-1
\end{pmatrix} T = ⎝ ⎛ 2 1 1 0 1 0 0 0 − 1 ⎠ ⎞
T ∗ = T T = ( 2 1 1 0 1 0 0 0 − 1 ) T^*=T^T=\begin{pmatrix}
2&1 & 1 \\
0&1& 0\\
0&0&-1
\end{pmatrix} T ∗ = T T = ⎝ ⎛ 2 0 0 1 1 0 1 0 − 1 ⎠ ⎞
( 2 1 1 0 1 0 0 0 − 1 ) ( u v w ) = ( 2 u + v + w v − w ) \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} ⎝ ⎛ 2 0 0 1 1 0 1 0 − 1 ⎠ ⎞ ⎝ ⎛ u v w ⎠ ⎞ = ⎝ ⎛ 2 u + v + w v − w ⎠ ⎞
Therefore the adjoint operator T ∗ ( u , v , w ) T^*(u,v,w) T ∗ ( u , v , w )
= ( 2 u + v + w , v , − w ) =(2u+v+w,v,-w) = ( 2 u + v + w , v , − w )
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