11.22 Null space

For the following linear maps, find \(\text{Null}(T)\) and \(\text{Range}(T)\), give a basis for these subspaces, and verify the dimension theorem.
1. \(T : \mathbb{R}^3 \to \mathbb{R}^5, \, T(x, y, z) = (x - y + z, \, x + y + 2z, \, 2x + 3y - 5z, \, 2x - y + z, \, 4x + 3y - z).\)
  
  \(T(x,y,z)=(x-y+z,\,x+y+2z,\,2x+3y-5z,\,2x-y+z,\,4x+3y-z)=0\)
  
  Then \(\begin{cases}x-y+z=0\\ x+y+2z=0\\ 2x+3y-5z=0\\ 2x-y+z=0\\ 4x+3y-z=0\end{cases}\Rightarrow\begin{pmatrix}1 & -1 & 1\\ 1 & 1 & 2\\ 2 & 3 & -5\\ 2 & -1 & 1\\ 4 & 3 & -1\end{pmatrix}\Rightarrow\begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0\end{pmatrix}\Rightarrow\begin{cases}x=0\\y=0\\z=0\end{cases}\)
  
  \(\text{Null}(T)=\{(0,0,0)\},\text{Range}(T)=\langle\begin{pmatrix}1\\ 1\\ 2\\ 2\\ 4\end{pmatrix},\begin{pmatrix}-1\\ 1\\ 3\\ -1\\ 3\end{pmatrix},\begin{pmatrix}1\\ 2\\ -5\\ 1\\ -1\end{pmatrix}\rangle,\) Yes 2. \(T : M_2(\mathbb{R}) \to M_2(\mathbb{R}), \, T \left( \begin{pmatrix} x & y \\ z & w \end{pmatrix} \right) = \begin{pmatrix} -x + 3y - 2z + w & 7x - 6y + 2z - w \\ -2x + 6y - 4z + 2w & -11x + 13y - 6z + 3w \end{pmatrix}.\)\(T : M_2(\mathbb{R}) \to M_2(\mathbb{R}), \, T \left( \begin{pmatrix} x & y \\ z & w \end{pmatrix} \right) = \begin{pmatrix} -x + 3y - 2z + w & 7x - 6y + 2z - w \\ -2x + 6y - 4z + 2w & -11x + 13y - 6z + 3w \end{pmatrix}.\)
  
  Then \(\begin{cases}-x+3y-2z+w=0\\ 7x-6y+2z-w=0\\ -2x+6y-4z+2w=0\\ -11x+13y-6z+3w=0\end{cases}\Rightarrow\begin{pmatrix}-1 & 3 & -2 & 1\\ 7 & -6 & 2 & -1\\ -2 & 6 & -4 & 2\\ -11 & 13 & -6 & 3\end{pmatrix}\Rightarrow\begin{pmatrix}1 & 0 & -\frac25 & -\frac15\\ 0 & 1 & -\frac45 & \frac25\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}\Rightarrow\begin{cases}x-\frac25z-\frac15w=0\\ y-\frac45z+\frac25w=0\end{cases}\)
  
  Then \(\begin{pmatrix}\frac25z+\frac15w & \frac45z-\frac25w\\ z & w\end{pmatrix}\), thus \(\text{Null}(T)=\langle\begin{pmatrix}\frac25 & \frac45\\ 1 & 0\end{pmatrix},\begin{pmatrix}\frac15 & -\frac25\\ 0 & 1\end{pmatrix}\rangle\)
  
  \(\text{Range}(T)\begin{pmatrix}x & 2x\\ z & 2z-5x\end{pmatrix}=\langle\begin{pmatrix}1 & 2\\ 0 & -5\end{pmatrix},\begin{pmatrix}0 & 0\\ 1 & 2\end{pmatrix}\rangle\)
  
  Yes
In each case, decide if it is possible to define a linear transformation \(T : \mathbb{F}^m \to \mathbb{F}^n\) which satisfies the required conditions. If it is not possible, explain why.

The main idea of these problem is the fundamental theorem of linear map

If we have a basis and a list, then there exist a unique linear map from \(V\) to \(W\)

Thus if the list in \(V\) is linearly independent, thus we can extend to the basis then exists

If the list in \(V\) is linearly dependent, then we can examine whether the \(w\) is same

If yes also exists because we can throw this and do the same thing above
1. \(T(1, -1, 1) = (1, 0)\) and \(T(1, 1, 1) = (0, 1).\)
  
  \(T(x,y,z)=\left(a_1^1x+a_2^1y+a_3^1z,a_1^2x+a_2^2y+a_3^2z\right)\)
  
  Then we have \(\begin{cases}1=a_1^1-a_2^1+a_3^1\\0=a_1^2-a_2^2+a_3^2\end{cases}\) and \(\begin{cases}0=a_1^1+a_2^1+a_3^1\\ 1=a_1^2+a_2^2+a_3^2\end{cases}\)
  
  Then we have \(\begin{cases}1=a_1^1-a_2^1+a_3^1 \\ 0=a_1^1+a_2^1+a_3^1 \\ 1=a_1^2+a_2^2+a_3^2 \\ 0=a_1^2-a_2^2+a_3^2 \end{cases}\)
  
  Let \(a_3^1=0\Rightarrow a_1^1=\frac12,a_2^1=-\frac12\) and \(a_3^2=0\Rightarrow a_1^2=\frac12,a_2^2=\frac12\)
  
  Thus \(T(x,y,z)=\left(\frac12x-\frac12y,\frac12x+\frac12y\right)\) 2. \(T(0, 1) = (1, 2, 0, 0)\) and \(T(1,0)=(1,1,0,0).\)
  
  \(T(x,y)=(x+y,x+2y)\) 3. \(T(1,1,1)=(0,1,3),\,T(1,2,1)=(1,1,3),\,\text{and}\,T(2,1,1)=(3,1,0).\)
  
  \(T(x,y,z)=(a_{11}x+a_{12}y+a_{13}z,a_{21}x+a_{22}y+a_{23}z,a_{31}x+a_{32}y+a_{33}z)\)
  
  \(\begin{cases}a_{11}+a_{12}+a_{13}=0\\a_{21}+a_{22}+a_{23}=1\\a_{31}+a_{32}+a_{33}=3\end{cases}\) and \(\begin{cases}a_{11}+2a_{12}+a_{13}=1\\ a_{21}+2a_{22}+a_{23}=1\\ a_{31}+2a_{32}+a_{33}=3\end{cases}\) and \(\begin{cases}2a_{11}+a_{12}+a_{13}=3\\ 2a_{21}+a_{22}+a_{23}=1\\ 2a_{31}+a_{32}+a_{33}=0\end{cases}\)
  
  Then \(\begin{pmatrix}\begin{array}{ccc|c}1 & 1 & 1 & 0\\ 1 & 2 & 1 & 1\\ 2 & 1 & 1 & 3\end{array}\end{pmatrix}\Rightarrow\begin{pmatrix}\begin{array}{ccc|c}0 & 0 & 1 & -4\\ 0 & 1 & 0 & 1\\ 1 & 0 & 0 & 3\end{array}\end{pmatrix}\Rightarrow a_{13}=-4,a_{12}=1,a_{11}=3\)
  
  \(\begin{pmatrix}\begin{array}{ccc|c}1 & 1 & 1 & 1\\ 1 & 2 & 1 & 1\\ 2 & 1 & 1 & 1\end{array}\end{pmatrix}\Rightarrow\begin{pmatrix}\begin{array}{ccc|c}0 & 0 & 1 & 1\\ 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\end{array}\end{pmatrix}\Rightarrow a_{23}=1,a_{22}=0,a_{21}=0\)
  
  \(\begin{pmatrix}\begin{array}{ccc|c}1 & 1 & 1 & 3\\ 1 & 2 & 1 & 3\\ 2 & 1 & 1 & 0\end{array}\end{pmatrix}\Rightarrow\begin{pmatrix}\begin{array}{ccc|c}0 & 0 & 1 & 6\\ 0 & 1 & 0 & 0\\ 1 & 0 & 0 & -3\end{array}\end{pmatrix}\Rightarrow a_{33}=6,a_{32}=0,a_{31}=-3\)
  
  Thus \(T(x,y,z)=(3x+y-4z,z,-3x+6z)\)

(d) \(T(1, 1, 1) = (3, 2), \, T(1, 0, 1) = (1, 1), \, \text{and} \, T(0, 1, 0) = (1, 0).\)

(e) \(T(0, 1, 1) = (1, 2, 0, 0) \, \text{and} \, T(1, 0, 0) = (1, 1, 0, 0).\)

Defining linear maps.

(a) Define a linear map \(T : \mathbb{R}^3 \to \mathbb{R}^3\) such that \(\text{Range}(T)\) is spanned by \((1, 0, -1)\) and \((1, 2, 2)\). Find \(T(x, y, z).\)

(b) Define a linear map \(T : \mathbb{R}^3 \to M_2(\mathbb{R})\) such that

\(\text{Null}(T) = \{ (x, y, z) \in \mathbb{R}^3 : z = 2x = y \}\)and

\(\text{Range}(T)=\left\{\begin{pmatrix}a & b\\ c & d\end{pmatrix}\in M_2(\mathbb{R}):b=a-c,b-d=a+c\right\}.\)Find \(T(x, y, z).\)

(c) Prove there does not exist a linear map \(T : \mathbb{R}^3 \to M_2(\mathbb{R})\) such that

\(\text{Null}(T) = \{ (x, y, z) \in \mathbb{R}^3 : z = 2x = y \}\)and

\(\text{Range}(T) = \left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in M_2(\mathbb{R}) : b - d = a + c \right\}.\)---

\(\dim(V)=3\)

\(\dim(null)=1\) \(\dim (range)=3\)

Suppose \(V\) and \(W\) are both finite-dimensional.

(a) Prove that there exists an injective linear map from \(V\) to \(W\) if and only if \(\dim V \leq \dim W.\)

(b) Prove that there exists a surjective linear map from \(V\) to \(W\) if and only if \(\dim V \geq \dim W.\)