Column space, null space and rank of a matrix
Andrew Thangaraj
Aug-Nov 2020
Recap
- Vector space \(V\) over a scalar field \(F\)
- \(F\): real field \(\mathbb{R}\) or complex field \(\mathbb{C}\) in this course
- Linear map \(T:V\to W\)
- Matrix of linear map with respect to bases for \(V\) and \(W\)
- Basis for \(V\): \(\{v_1,\ldots,v_n\}\)
- Column \(j\): coordinates of \(T(v_j)\) with respect to basis of \(W\)
- null \(T=\{v\in V:Tv=0\}\), range \(T=\{Tv:v\in V\}\)
- Fundamental theorem: dim null \(T\) + dim range \(T\) = dim \(V\)
From \(T:F^n\to F^m\) to an \(m\times n\) matrix
Basis for \(F^n\): \(\{v_1,\ldots,v_n\}\), basis for \(F^m\): \(\{w_1,\ldots,w_m\}\)
Matrix for \(T\): \(A=\begin{bmatrix} A_{11}&A_{12}&\cdots&A_{1n}\\ A_{21}&A_{22}&\cdots&A_{2n}\\ \vdots&\vdots&\vdots&\vdots\\ A_{m1}&A_{m2}&\cdots&A_{mn} \end{bmatrix}\)
For input \(v=a_1v_1+\cdots+a_nv_n\leftrightarrow\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix}\),
\[\begin{align}
Tv&=a_1\quad Tv_1\ +\cdots+a_n\quad Tv_n\\[5pt]
&=a_1\begin{bmatrix}A_{11}\\A_{21}\\\vdots\\A_{m1}\end{bmatrix}+\cdots+a_n\begin{bmatrix}A_{1n}\\A_{2n}\\\vdots\\A_{mn}\end{bmatrix}
\quad\leftrightarrow A\begin{bmatrix}a_1\\a_2\\\vdots\\a_n\end{bmatrix}
\end{align}\]
Leads to definition of matrix-vector product
From an \(m\times n\) matrix to \(T:F^n\to F^m\)
\(A=\begin{bmatrix} A_{11}&A_{12}&\cdots&A_{1n}\\ A_{21}&A_{22}&\cdots&A_{2n}\\ \vdots&\vdots&\vdots&\vdots\\ A_{m1}&A_{m2}&\cdots&A_{mn} \end{bmatrix}\)
Given an \(m\times n\) matrix \(A\), fix bases for \(F^n\) and \(F^m\)
\(\{v_1,\ldots,v_n\}\) and \(\{w_1,\ldots,w_m\}\)
say, standard bases
\(j\)-th column of \(A\): \(Tv_j\) expressed in \(\{w_1,\ldots,w_m\}\)
Input vector \(v\): expressed in \(\{v_1,\ldots,v_n\}\) as a column vector
Output \(Tv\): matrix-vector product
Column space and null space of a matrix
\(m\times n\) matrix \(A=\begin{bmatrix} A_{11}&A_{12}&\cdots&A_{1n}\\ A_{21}&A_{22}&\cdots&A_{2n}\\ \vdots&\vdots&\vdots&\vdots\\ A_{m1}&A_{m2}&\cdots&A_{mn} \end{bmatrix}\) \(\leftrightarrow\) \(T:F^n\to F^m\)
- null \(T\)
- \(\{x\in F^n: Ax=0\}\)
- solutions to homogeneous equations \(Ax=0\)
- called null space of matrix \(A\)
- nullity: dimension of null space
- range \(T\)
- \(\{Ax: x\in F^n\}\)
- span of columns of \(A\)
- called column space of \(A\), denoted colspace \(A\)
- column rank or rank: dimension of column space
By fundamental theorem of linear maps, \(n=\text{rank}(A)+\text{nullity}(A)\).
Examples: \(2\times 2\), represents \(T:F^2\to F^2\)
\(A=\begin{bmatrix}0&0\\0&0\end{bmatrix}\)
- colspace \(A = \{0\}\), rank = 0
- null \(A = F^2\), nullity = 2
\(A=\begin{bmatrix}1&0\\2&0\end{bmatrix}\)
- colspace \(A = \text{span}\{(1,2)\}\), rank = 1
- null \(A = \text{span}\{(0,1)\}\), nullity = 1
\(A=\begin{bmatrix}1&3\\2&6\end{bmatrix}\)
- colspace \(A = \text{span}\{(1,2)\}\), rank = 1
- null \(A = \text{span}\{(-3,1)\}\), nullity = 1
Rank 1
- At least one non-zero column
- One column is a multiple of other column
Examples: \(2\times 2\), represents \(T:F^2\to F^2\)
\(A=\begin{bmatrix}1&0\\0&1\end{bmatrix}\)
- colspace \(A = F^2\), rank = 2
- null \(A = \{0\}\), nullity = 0
\(A=\begin{bmatrix}1&3\\2&4\end{bmatrix}\)
- colspace \(A = \text{span}\{(1,2),(3,4)\}\), rank = 2
- null \(A = \{0\}\), nullity = 0
Rank 2
- colspace = \(F^2\), null space = \(\{0\}\)
- columns are linearly independent
- called full rank or full column rank
Examples: \(3\times 2\), represents \(T:F^2\to F^3\)
\(A=\begin{bmatrix}1&3\\2&6\\3&9\end{bmatrix}\)
- colspace \(A = \text{span}\{(1,2,3)\}\), rank = 1
- null \(A=\text{span}\{(-3,1)\}\), nullity = 1
Rank 1
- At least one non-zero column
- One column is a multiple of other column
\(A=\begin{bmatrix}1&3\\2&4\\3&5\end{bmatrix}\)
- colspace \(A = \text{span}\{(1,2,3),(3,4,5)\}\), rank = 2
- null \(A=\{0\}\), nullity = 0
Rank 2
- null space = \(\{0\}\)
- col space = dim 2 subspace of \(F^3\)
Examples: \(2\times 3\), represents \(T:F^3\to F^2\)
\(A=\begin{bmatrix}1&3&5\\2&6&10\end{bmatrix}\)
- colspace \(A = \text{span}\{(1,2)\}\), rank = 1
- null \(A=\text{span}\{(-3,1,0),(-5,0,1)\}\), nullity = 2
Rank 1
- At least one non-zero column
- Columns are multiples of non-zero column
\(A=\begin{bmatrix}1&3&5\\2&4&6\end{bmatrix}\)
- colspace \(A = F^2\), rank = 2
- null \(A=\text{span}\{(1,-2,1)\}\), nullity = 1
Rank 2
- col space = \(F^2\)
- null space, rank 1
\(m\times n\) matrix representing \(T:F^n\to F^m\)
col space = span{columns}
- use Gaussian elimination to reduce and find dimension
null space
- use fundamental theorem to find dimension
- solve homogeneous linear equation to find basis