$$

How to find a basis of the kernel?

Problem

Let

$$A=\left[\begin{array}{cccccc}1& -5& -4& -4& 5& 8\\ -2& 10& 8& 8& -9& -14\\ -5& 25& 20& 20& -23& -36\\ -5& 25& 20& 20& -23& -36\end{array}\right]\text{ and }R=\left[\begin{array}{cccccc}1& -5& -4& -4& 0& -2\\ 0& 0& 0& 0& 1& 2\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right].$$

It is known that

$R$ is the reduced echelon form of

$A$. Find a basis for the kernel

$\ker (A)$.

Thought

According to the article "How to solve a system of linear equations?", the kernel of

$A$ can be written as

$\mathrm{span}(\{{\mathbf{h}}_1,\dots ,{\mathbf{h}}_k\})$ for some homogeneous solutions

${\mathbf{h}}_i$'s, where

$k$ is the number of free variables. Let

$S=\{{\mathbf{h}}_1,\dots ,{\mathbf{h}}_k\}$. Then we have

$\ker (A)=\mathrm{span}(S)$, which is one of the conditions for

$S$ being a basis of

$\ker (A)$. As long as we can show that

$S$ is linearly independent, then

$S$ is a basis of

$\ker (A)$.

Sample answer

By observing the reduced echelon form, we know the indices

$2$,

$3$,

$4$, and

$6$ correspond to the free variables. By setting one of them as

$1$ and others as

$0$, we get the set of homogeneous solutions

$$\begin{array}{rl}S& =\{{\mathbf{h}}_1,{\mathbf{h}}_2,{\mathbf{h}}_3,{\mathbf{h}}_4\}\\ & =\{\left[\begin{array}{c}5\\ 1\\ 0\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}4\\ 0\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}4\\ 0\\ 0\\ 1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}2\\ 0\\ 0\\ 0\\ -2\\ 1\end{array}\right]\}.\end{array}$$

By the way we solve the solutions, we know

$\ker (A)=\mathrm{span}(S)$. Now we may check the linearly independency of

$S$. Suppose

$$c_1{\mathbf{h}}_1+c_2{\mathbf{h}}_2+c_3{\mathbf{h}}_3+c_4{\mathbf{h}}_4=\mathbf{0}.$$

By comparing the second entries on the both sides, we know

$c_1=0$. Similarly, the equation on the third entries gives

$c_2=0$. The fourth entries give

$c_3=0$, and the sixth entries give

$c_4=0$. Therefore,

$S$ is linearly independent.

Combining all above results, we know

$S$ is a basis of

$\ker (A)$.

Note

It turns out that by the way we found the homogeneous solutions, there is a "hidden identity matrix" in

$S$. (Use the electronic version of this note to see the red part above.) Therefore, The set

$S$ of homogeneous solutions is always linearly independent by our choice (why?).

This means the dimension of

$\ker (A)$ is the number of free variables. This number is called the nullity of

$A$, denoted as

$\mathrm{null}(A)$.

Run the SageMath code below or simply click here. The code generates some random matrices. Use the code to do some practices by yourself, and then check your answer by modifying print_ans in the code.

load("https://raw.githubusercontent.com/jephianlin/LA-notebook/main/lingeo.py")

set_random_seed(None) # replace None with your favorite number
m,n,r = 4,6,2 # m x n matrix with rank r
A, R, pivots = random_good_matrix(m,n,r, return_answer=True)
pretty_print(LatexExpr("A="), A, LatexExpr("\\rightsquigarrow R="), R)

Kb = kernel_matrix(A)
Cb = column_space_matrix(A)
Rb = row_space_matrix(A)

print_ans = "No" # replace No with K, C, R, or All
if print_ans in ["K", "All"]:
    K_basis = [(Kb[:,i//2] if i % 2 == 0 else ",") for i in range(2*(n-r)-1)]
    pretty_print("a basis of ker(A) can be", LatexExpr("\\ \\Big\\{"), *K_basis, LatexExpr("\\Big\\}"))
if print_ans in ["C", "All"]:
    K_basis = [(Cb[:,i//2] if i % 2 == 0 else ",") for i in range(2*r-1)]
    pretty_print("a basis of Col(A) can be", LatexExpr("\\ \\Big\\{"), *K_basis, LatexExpr("\\Big\\}"))
if print_ans in ["R", "All"]:
    K_basis = [(Rb[i//2,:] if i % 2 == 0 else ",") for i in range(2*r-1)]
    pretty_print("a basis of Row(A) can be", LatexExpr("\\ \\Big\\{"), *K_basis, LatexExpr("\\Big\\}"))

This note can be found at Course website > Learning resources.