11.20 Null space and range

Decide if the following function \(T:\mathbb{R}^{n}\to\mathbb{R}^{m}\) is a linear map.
1. \(T(x, y) = (1 + x, y)\)
  
  No, counterexample: \(\lambda=2\implies T(\lambda(x,y))=T(2x,2y)=(1+2x,2y)\)
  
  \(2T(x, y) = 2(1 + x, y) = (2 + 2x, 2y)\)
  
  So \(T\) is not a linear map. 2. \(T(x, y, z) = 3x - 2y + 7z\).
  
  Let’s check the properties:
  
  \(T((x,y,z)+(\tilde{x},\tilde{y},\tilde{z}))=T(x+\tilde{x},y+\tilde{y},z+\tilde{z})=3(x+\tilde{x})-2(y+\tilde{y})+7(z+\tilde{z})=(3x-2y+7z)+(3\tilde{x}-2\tilde{y}+7\tilde{z})=T(x,y,z)+T(\tilde{x},\tilde{y},\tilde{z})\) \(\begin{aligned}T(\lambda(x, y, z)) &= T(\lambda x, \lambda y, \lambda z) = 3(\lambda x) - 2(\lambda y) + 7(\lambda z) = \lambda T(x, y, z).\end{aligned}\) 3. \(T(x_1,\dots,x_{n})=(x_1,-x_1,x_2,-x_2,\dots,x_{n},-x_{n})\).
  
  It is enough to check:
  
  \(T((x_1,\ldots,x_{n})+\lambda(y_1,\ldots,y_{n}))=T(x_1,\ldots,x_{n})+\lambda T(y_1,\ldots,y_{n})\)
  
  \(T((x_1,\ldots,x_{n})+\lambda(y_1,\ldots,y_{n}))=T(x_1+\lambda y_1,\ldots,x_{n}+\lambda y_{n})=(x_1+\lambda y_1,-(x_1+\lambda y_1),\ldots,x_{n}+\lambda y_{n},-(x_{n}+\lambda y_{n}))=(x_1,-x_1,\ldots,x_{n},-x_{n})+\lambda(y_1,-y_1,\ldots,y_{n},-y_{n})=T(x_1,\ldots,x_{n})+\lambda T(y_1,\ldots,y_{n})\) 4. \(T(x_1, \dots, x_n) = (x_1, x_1x_2, x_1x_2x_3, \dots, x_1x_2\cdots x_n)\)
  
  \(T(0,1,0,\dots,0)=(0,0,\dots,0)~~~~~~~T(1,0,\dots,0)=(1,0,\dots,0)\)
  
  \(T(0,1,0,\ldots,0)+T(1,0,\ldots,0)=T(1,1,0,\ldots,0)=(1,1,0,\ldots,0)\neq\left(0,0,\ldots,0\right)+\left(1,0,\ldots,0\right)=\left(1,0,\ldots,0\right)\)
  
  Thus, \(T\) is not a linear map.
Let \(T: V \to W\) be a linear map and \(\{v_1, \dots, v_n\} \subseteq V\).

If \(\{T(v_1), \dots, T(v_n)\}\) is linearly independent in \(W\)\(\implies \{v_1, \dots, v_n\}\) is linearly independent in \(V\).

If we have \(a_1v_1 + \dots + a_nv_n = 0\) for \(a_1, \dots, a_n \in \mathbb{F}\),

\(\implies T(a_1v_1+\dots+a_{n}v_{n})=T(0)=0\)

Since \(T\) is a linear map, \(0 = a_1T(v_1) + \dots + a_nT(v_n).\)

Since \(\{T(v_1), \dots, T(v_n)\}\) is linearly independent, \(\implies a_1 = \dots = a_n = 0.\)
If \(\dim(V) = 1\) and \(T \in \mathcal{L}(V, V)\), \(\implies \exists \lambda \in \mathbb{F}\) such that \(T(v) = \lambda v \,\, \forall v \in V.\)

Since \(\dim(V) = 1\), we have a basis \(\{w\}\) of one element for \(V\).

Let \(v \in V\), since \(\{w\}\) is a basis, \(\exists\alpha\in\mathbb{F}\) such that \(v=\alpha w\).

Since \(T(w) \in V\), \(\exists \lambda \in \mathbb{F}\) such that \(T(w) = \lambda w\).

Therefore, \(T(v)=T(\alpha w)=\alpha T(w)=\alpha(\lambda w)=\lambda(\alpha w)=\lambda v\)
Suppose \(V\) is finite-dimensional. Prove that every linear map on a subspace of \(V\) can be extended to a linear map on \(V\).

In other words, if \(U\) is a subspace of \(V\) and \(S \in \mathcal{L}(U, W)\)\(\implies \exists T \in \mathcal{L}(V, W)\) such that \(T(u) = S(u) \,\, \forall u \in U\).

Proof

Suppose \(U\) is a subspace of \(V\) and \(S \in \mathcal{L}(U, W)\).
Choose a basis \(u_1, \dots, u_m\) of \(U\). Then \(u_1, \dots, u_m\) is a linearly independent list of vectors in \(V\), and so can be extended to a basis \(u_1, \dots, u_m, v_1, \dots, v_n\) of \(V\).
Using theorem, we know that there exists a unique linear map \(T \in \mathcal{L}(V, W)\) such that

\(T u_i = S u_i\) for all \(i \in \{1, 2, \dots, m\}\)

\(T v_j = 0\) for all \(j \in \{1, 2, \dots, n\}\).

Now we are going to prove \(T u = S u\) for all \(u \in U\).

For any \(u \in U\), \(u\) can be written as \(a_1 u_1 + \dots + a_m u_m\), since \(S \in \mathcal{L}(U, W)\),

\(Su=a_1Su_1+a_2Su_2+\dots+a_{m}Su_{m}\).

Since \(T \in \mathcal{L}(V, W)\), we have \(T u = T(a_1 u_1 + \dots + a_m u_m)\)

\(= a_1 T u_1 + a_2 T u_2 + \dots + a_m T u_m\)

\(= a_1 S u_1 + a_2 S u_2 + \dots + a_m S u_m\)

\(= S u\).

Therefore, we have \(T u = S u\) for all \(u \in U\), so we have proved that every linear map on a subspace of \(V\) can be extended to a linear map on \(V\).
For the following linear maps, find \(\text{Null}(T)\), \(\text{Range}(T)\), give a basis for these subspaces, and verify the dimension theorem.
1. \(T: \mathbb{R}^3 \to \mathbb{R}^2\), \(T(x,y,z)=(x-y+z,-2x+2y-2z)\).
  
  \(\text{Null}(T)\): \((x,y,z)\in\mathbb{R}^3\;\iff\;T(x,y,z)=(0,0)\)
  
  \(\;\iff\;\begin{aligned}(x-y+z,-2x+2y-2z)=(0,0)\end{aligned}\)
  
  \(\;\;\iff\;\;\begin{aligned}\begin{cases}x-y+z=0\\ -2x+2y-2z=0\end{cases}\end{aligned}\)\(\;\;\iff\;\;\begin{pmatrix}1 & -1 & 1\\ -2 & 2 & -2\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}\)
  
  So let us row-reduce the matrix:
  
  \(\begin{pmatrix}1 & -1 & 1\\ -2 & 2 & -2\end{pmatrix}\xrightarrow{R_2+2R_1}\begin{pmatrix}1 & -1 & 1\\ 0 & 0 & 0\end{pmatrix}\)\(\implies x = y - z\).
  
  So \((x, y, z) = (y - z, y, z) = y(1, 1, 0) + z(-1, 0, 1)\).
  
  Therefore, \(\text{Null}(T) = \langle (1, 1, 0), (-1, 0, 1) \rangle\) (basis).
  
  \(\text{Range}(T) = \{T(x, y, z) : x, y, z \in \mathbb{R}\}\)
  
  \(=\left\lbrace(x-y+z,-2x+2y-2z):x,y,z\in\mathbb{R}\right\rbrace\)
  
  \(= \{x(1, -2) + y(-1, 2) + z(1, -2) : x, y, z \in \mathbb{R}\}\).
  
  \(\implies \text{Range}(T) = \langle (1, -2), (-1, 2), (1, -2) \rangle = \langle (1, -2) \rangle\).
  
  Note that we could have done:
  
  \(\begin{pmatrix}1 & -2\\ -1 & 2\\ 1 & -2\end{pmatrix}\xrightarrow{R_2+R_1,R_3-R_1}\begin{pmatrix}1 & -2\\ 0 & 0\\ 0 & 0\end{pmatrix}.\)
  
  Let's verify the dimension theorem:
  
  \(\dim(\mathbb{R}^3) = \dim(\text{Null}(T)) + \dim(\text{Range}(T))\)
  
  \(3 = 2 + 1\). 2. \(T: \mathbb{R}^3 \to \mathbb{R}^3\), \(T(x, y, z) = (3x - 2y - z, 7x - 5y - 3z, -x - z)\).
  - \(\text{Null}(T)\): We need to row-reduce:
  \(\begin{pmatrix}3 & -2 & -1\\ 7 & -5 & -3\\ -1 & 0 & -1\end{pmatrix}\xrightarrow{}\begin{pmatrix}0 & 1 & 2\\ 0 & 0 & 0\\ 1 & 0 & 1\end{pmatrix}.\)
  
  Then \(x = -z\), \(y = -2z\).
  
  So \((x, y, z) = z(-1, -2, 1)\).
  
  Therefore, \(\text{Null}(T) = \langle (-1, -2, 1) \rangle\).
  - \(\text{Range}(T)\):
  \(\begin{pmatrix}3 & 7 & -1\\ -2 & -5 & 0\\ -1 & -3 & -1\end{pmatrix}\xrightarrow{}\begin{pmatrix}0 & 0 & 0\\ 0 & 1 & 2\\ 1 & 0 & -5\end{pmatrix}.\)
  
  So \(\text{Range}(T) = \langle (0, 1, 2), (1, 0, -5) \rangle\).
  - Verify: \(\dim(\mathbb{R}^3) = \dim(\text{Null}(T)) + \dim(\text{Range}(T))\):
  \(3 = 1 + 2 \, \checkmark\). 3. \(T: P_2(\mathbb{R}) \to M_2(\mathbb{R})\), \(T(ax^2 + bx + c) = \begin{pmatrix}a & b+c \\ b+c & a\end{pmatrix}\).
  - \(\text{Null}(T)\):
  \(ax^2 + bx + c \in \text{Null}(T) \iff T(ax^2 + bx + c) = \begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}.\)
  
  \(\begin{pmatrix}a & b+c \\ b+c & a\end{pmatrix} = \begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix} \iff \begin{cases}a = 0 \\ b + c = 0\end{cases} \iff a = 0, c = -b.\)
  
  Therefore, \(\text{Null}(T)=\{bx-b:b\in\mathbb{R}\}=\langle x-1\rangle\).
  - \(\text{Range}(T)\):
  \(\text{Range}(T) = \{T(ax^2 + bx + c) : a, b, c \in \mathbb{R}\}.\)
  
  \(=\langle\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix},\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix},\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}\rangle.\)
  
  \(= \langle \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}, \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix} \rangle.\)
  - Verify: \(\dim(P_2(\mathbb{R})) = \dim(\text{Null}(T)) + \dim(\text{Range}(T))\):
  \(3 = 1 + 2 \, \checkmark.\)
Let \(T: V \to W\) be an injective linear map, and let \(\{v_1, \dots, v_n\} \subseteq V\) be linearly independent in \(V\). Prove that \(\{T(v_1), \dots, T(v_n)\}\) is linearly independent in \(W\).

Suppose we have \(a_1 T(v_1) + \dots + a_n T(v_n) = 0\) for \(a_1, \dots, a_n \in \mathbb{F}\).

\(\implies T(a_1 v_1 + \dots + a_n v_n) = 0.\) Since \(T\) is a linear map

Theorem: \(T \in \mathcal{L}(V, W)\) is injective \(\iff \text{Null}(T) = \{0\}\).

So from the above: \(a_1 v_1 + \dots + a_n v_n \in \text{Null}(T) = \{0\} \implies a_1 v_1 + \dots + a_n v_n = 0.\)

\(\implies a_1 = \dots = a_n = 0\), since \(\{v_1, \dots, v_n\}\) is linearly independent in \(V\).

Thus, \(\{T(v_1), \dots, T(v_n)\}\) is linearly independent in \(W\).

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