[Linear_Algebra] Null space of Matrix and Linear system equations

 

In this post, we are going to talk about the Null space and solving linear system equations by using the null space.

Null space

Let A be a matrix whose columns are [a1, a2, ..., an].

The equation is as follows:

AX = 0 ; N(A) = {X | AX = 0}.

We can interpret this intuitively: when matrix A multiplies a vector X and the result is zero, those X vectors form the null space. This means that the null space contains all X that are mapped to the zero vector by A.

From previous posts, a vector space is defined when 3 conditions are satisfied. So let’s check:

If A(x1) = 0 and A(x2) = 0, then A(k1x1 + k2x2) = k1A(x1) + k2A(x2) = 0.
This shows that the null space is a subspace.

Example 1. A = [[1,5,2],[0,4,4]]. Then the null space is N(A) = {0}, and the column space C(A) is all of R^2.

Example 2. A = [[1,5,2],[0,4,4],[1,4,4]]. Then the null space is N(A) = {k(1,1,-1)}, which is a line through the origin. The column space C(A) is a plane in R^3.

Example 3. A = [[1,1],[2,2]]. In this case, the two columns are the same.

• The null space is N(A) = {k(1,-1)}, which is a line through the origin.
• The column space is C(A) = span{(1,2)}, which is also a line in R^2.

Linear system equations with Null space

1. Unique solution: Ax = b, where A is an invertible n×n matrix.

If A^-1 exists, it means all the column vectors of A are linearly independent. So the rank of A is n, the column space C(A) = R^n, and the null space N(A) = {0}.
Because Ax = 0 → A^-1Ax = A^-1(0) → x = 0.

1. Non-invertible case: Ax = b.

Suppose Ax = b has infinite solutions.
In this case, we can write the general solution as x = xp + xn,
where xp is a particular solution of Ax = b, and xn is any solution of Ax = 0 (element of the null space).

Example: A = [[1,1],[2,2]].

• If b is not in span{(1,2)}, i.e. b2 ≠ 2b1 → no solution.
• If b2 = 2b1 → infinite solutions.

First, solve Ax = 0:
Let X = [y,z]. Then the equation gives y + z = 0 → y = -z.
So the null space is span{(1,-1)}.

Next, find a particular solution xp.
Let’s say b = [1,2]. One valid solution is y=1, z=0 → xp = (1,0).

Therefore, the general solution is
X = xp + k·(1,-1) = (1,0) + k(1,-1) = [1+k, -k].