Skip to content

11.19 Null space and range

Null space and range

Remark

Both may be the \(\{0\}\) subspace, the total vector space, or a non-zero proper subspace.

  • \(T: \mathbb{R}^2 \to \mathbb{R}^2, \ (x, y) \mapsto (y, x). \ \text{Null } T = \{0\}, \ \text{range } T = W\)
  • \(T: V \to W, \ T = 0. \ \text{range } T = \{0\}, \ \text{null } T = V\)
Example

  1. \(T: \mathbb{R}^2 \to \mathbb{R}^3, \ (x, y) \mapsto (x, -x, y)\) \(\dim(\text{range } T) = 2, \ \text{a basis: } \{(1, -1, 0), (0, 0, 1)\}\) \(\text{null } T = \{0\}\)

  2. \(T: F[x] \to F[x], \ p(x) \mapsto x \cdot p(x)\)

    \(\text{null }T=\{0\},\ \text{range }T=\{q(x)=a_0+a_1x+\cdots+a_{n}x^{n}:a_0=0\}\)

  3. \(\text{tr: } \mathbb{R}_{2 \times 2}(\mathbb{C}) \to \mathbb{C}, \ A \mapsto \text{tr } A = a_{11} + a_{22}\)

    \(\text{null } T = \left\{\begin{pmatrix} a & b \\ c & d \end{pmatrix} : a + d = 0 \right\} = \langle \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \rangle\) (basis)

    \(\text{range } T = \mathbb{C}, \ \lambda = \text{tr} \begin{pmatrix} \lambda & 0 \\ 0 & 0 \end{pmatrix}\)

    Remark: Any \(A\) such that \(\text{tr}(A) = \lambda\) is equivalent to the form \(A = \begin{pmatrix} \lambda & 0 \\ 0 & 0 \end{pmatrix} + B\) with \(B \in \text{null } T\).

    • \(\Leftarrow\)) \(\text{tr}\left\lbrack\begin{pmatrix}\lambda & 0\\ 0 & 0\end{pmatrix}+B\right\rbrack=\text{tr}\begin{pmatrix}\lambda & 0\\ 0 & 0\end{pmatrix}+\text{tr}\left(B\right)=\lambda+0=\lambda\)
    • \(\Rightarrow\)) If \(A\) is such that \(\text{tr}(A) = \lambda\), then we need to \(A=\begin{pmatrix}\lambda & 0\\ 0 & 0\end{pmatrix}+B\;\;\iff\;\;B=A-\begin{pmatrix}\lambda & 0\\ 0 & 0\end{pmatrix}\;\iff\;A-\begin{pmatrix}\lambda & 0\\ 0 & 0\end{pmatrix}\in\text{null}T\)​

    \(\text{tr}(A-\begin{pmatrix}\lambda & 0\\ 0 & 0\end{pmatrix})=\text{tr}(A)-\lambda=\lambda-\lambda=0\implies B=A-\begin{pmatrix}\lambda & 0\\ 0 & 0\end{pmatrix}\in\text{null }T\)

    Conclusion: \(A\) is such that \(\text{tr}(A) = \lambda\) iff \(A = \begin{pmatrix} \lambda & 0 \\ 0 & 0 \end{pmatrix} + a \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} + b \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + c \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}\) for \(a, b, c \in \mathbb{R}\).

Example

\(T: \mathbb{R}^3 \to \mathbb{R}^3, \ T(x, y, z) = (x + 2y + z, \ 2y + 2z, \ -x + y + 2z)\)

\(\text{Null } T\): \(\{(x,y,z)\in\mathbb{R}^3\mid(x,y,z)\in\text{null }T\;\}\;\iff\;\;\begin{cases}x+2y+z=0\\ 2y+2z=0\\ -x+y+2z=0\end{cases}\)

\(\text{Range } T\): \(\{(a,b,c)\in\mathbb{R}^3\mid\exists(x,y,z)\}\text{ with }\begin{cases}x+2y+z=a\\ 2y+2z=b\\ -x+y+2z=c\end{cases}\)

\(\begin{aligned}\begin{pmatrix}1 & 2 & 1 & \vert & a\\ 0 & 2 & 2 & \vert & b\\ -1 & 1 & 2 & \vert & c\end{pmatrix}\sim\begin{pmatrix}1 & 0 & -1 & \vert & -2c-a+2b\\ 0 & 1 & 1 & \vert & c+a-b\\ 0 & 0 & 0 & \vert & 3b-2c-2a\end{pmatrix}\end{aligned}\)​

Answers: \(\text{null } T = \langle (1, -1, 1) \rangle\)

\(\text{range } T = \{(a, b, c) \in \mathbb{R}^3 : -2a + 3b - 2c = 0\}\) implicitly description

What's about explicitly descirption?

\(\text{Let’s solve: } -2a + 3b - 2c = 0 \implies a = \frac{3}{2}b - c\)

(1) \(\text{range } T = \langle (\frac{3}{2}, 1, 0), (-1, 0, 1) \rangle\)

(2) Remark: Another way to compute \(\text{range } T = \langle T(1, 0, 0), T(0, 1, 0), T(0, 0, 1) \rangle\)

\(v \in \mathbb{R}^3 = V, \ v = \alpha(1, 0, 0) + \beta(0, 1, 0) + \gamma(0, 0, 1) \implies T v = \alpha T(1, 0, 0) + \beta T(0, 1, 0) + \gamma T(0, 0, 1)\)

\(\text{range } T = \langle (1, 0, -1), (2, 2, 1) \rangle\) (basis) since we choose the column which has a pivot

image​

Proposition

  1. \(T\) is surjective\(\iff \text{range } T = W.\)

  2. \(T\) is injective \(\iff\) \(\text{null } T = \{0\}\)

    Proof

    \(\Rightarrow\)) \(T(0)=0,v\in\text{null }T\implies T(v)=T(0)=0\implies v=0\)

    \(\Leftarrow\)) Take \(T(v_1)=T(v_2)\), then \(T(v_1)-T(v_2)=T(v_1-v_2)=0\Rightarrow v_1=v_2\)​

Fundamental theorems

Let \(V\) and \(W\) be finite-dimensional vector spaces.

Fundamental theorem of linear maps

Let \(T \in \mathcal{L}(V, W)\). Then \(\dim V = \dim (\text{null } T) + \dim (\text{range } T).\)

image​

Proof

Let \(v_1, \dots, v_k\) be a basis of \(\text{null } T\) (if \(\text{null } T \neq 0\), if \(\text{null } T = 0\) do nothing).

Complete it to a basis of \(V\): \(v_1, \dots, v_k, v_{k+1}, \dots, v_n\).

Claim: \(T(v_{k+1}), \dots, T(v_n)\) is a basis of \(\text{range } T\).

  1. \(T(v_{k+1}), \dots, T(v_n)\) spans \(\text{range } T\)​

    Take any \(w\in \text{range}T\), then \(\exists v\in V:T(v)=w\)​

    Since \(v=a_1v_1+...+a_{n}v_{n}\), then \(T(v)=T\left(a_1v_1+...+a_{n}v_{n}\right)=a_1T\left(v_1\right)+\cdots+a_{n}T\left(v_{n}\right)=a_{k+1}T\left(v_{k+1}\right)+\cdots+a_{n}T\left(v_{n}\right)\)​

    Thus \(w=a_{k+1}T\left(v_{k+1}\right)+\cdots+a_{n}T\left(v_{n}\right)\)​

  2. \(T(v_{k+1}), \dots, T(v_n)\) are linearly independent

    Consider \(0=a_{k+1}T(v_{k+1})+\ldots+a_{n}T(v_{n})=0\cdot T\left(v_1\right)+\cdots+a_{k+1}T(v_{k+1})+\ldots+a_{n}T(v_{n})\)​

    Then \(0=T\left(0\cdot v_1+\cdots+a_{n}v_{n}\right)\Rightarrow0=0\cdot v_1+\cdots+a_{n}v_{n}\)​

    Since \(v_1, \dots, v_k, v_{k+1}, \dots, v_n\) is a basis, then \(a_1=...=a_n=0\)​

Corollary

Let \(A \in M_{m \times n}(\mathbb{F})\). Then \(\text{rank } A = \text{row rank } A = \text{column rank } A.\)

Proof. Consider \(T_A: \mathbb{F}^n \to \mathbb{F}^m.\) \(r=\text{row rank }A\)​

Since \(n = \dim (\text{null } T_A) + \dim (\text{range } T_A)\) and \(\dim (\text{range } T_A) = \text{column rank } A.\)

Thus \(\dim(\text{null }T_{A})=n-\dim(\text{range }T_{A})=n-\text{column rank }A\)

Since in the row-reduced echelon form, the \(\text{row rank }=r\) where the pivots are in the column, then the remaining column which has not pivot is null space and their sum is \(n\)

\(\dim(\text{null }T_{A})=n-r\implies r=\text{column rank }A\)​

Fundamental Theorem of Linear Algebra

Let \(v_1, \dots, v_n\) be a basis of \(V\) and let \(w_1, \dots, w_n\) be a list of vectors in \(W\). Then there exists a unique linear transformation \(T: V \to W\) such that \(T(v_i) = w_i\), \(i = 1, \dots, n\).

KEY: the length of basis is same with the length of list of vectors

image​

Proof. Given \(v \in V\), \(v = a_1 v_1 + \cdots + a_n v_n\), for unique \(a_1, \dots, a_n\).

Define \(T\) by: \(T(v) = a_1 w_1 + \cdots + a_n w_n = a_1 T(v_1) + \cdots + a_n T(v_n)\). This \(T\) satisfies \(T(v_i) = w_i\).

  1. \(T\) is linear: If \(v = a_1 v_1 + \cdots + a_n v_n\) and \(w = b_1 v_1 + \cdots + b_n v_n\),

    \(v+w=(a_1+b_1)v_1+\cdots+(a_{n}+b_{n})v_{n}\)

    \(\lambda v=(\lambda a_1)v_1+\cdots+(\lambda a_{n})v_{n}\)

    So that \(T(v+w)=(a_1+b_1)w_1+\cdots+(a_{n}+b_{n})w_{n}=a_1w_1+\cdots+a_{n}w_{n}+b_1w_1+\cdots+b_{n}w_{n}=T(v)+T(w)\).

    \(T(\lambda v) = T((\lambda a_1)v_1 + \cdots + (\lambda a_n)v_n) = (\lambda a_1)w_1 + \cdots + (\lambda a_n)w_n\) \(= \lambda(a_1 w_1 + \cdots + a_n w_n) = \lambda T(v)\).

  2. \(T\) is unique: If \(S: V \to W\) is linear and \(S(v_i) = w_i\), then \(S(v) = S(a_1 v_1 + \cdots + a_n v_n) = a_1 S(v_1) + \cdots + a_n S(v_n)\)

    \(= a_1 w_1 + \cdots + a_n w_n = T(v)\).

‍