# How to find dimension of null space of a given matrix?

Let $$A$$ be a $$5×4$$ matrix with real entries such that the space of all solutions of linear system $$AX^t = \begin{pmatrix}{1,2,3,4,5}\end{pmatrix}$$ is given by $$\{[1+2s,2+3s,3+4s,4+5s]^t : s \in \Bbb R \}$$ (Here $$M^t$$ denotes the transpose of a matrix $$M$$). Then how to find dimension of null space.

My attempt

I saw somewhere that

$$\begin{pmatrix}{1+2s \\ 2+3s \\ 3+4s \\ 4+5s}\end{pmatrix} = \begin{pmatrix}{1 \\ 2 \\ 3 \\ 4}\end{pmatrix} + s \begin{pmatrix}{2 \\ 3 \\ 4 \\ 5}\end{pmatrix}$$

where $$s$$ is a free variable. So $$\dim(\mbox{null}(A)) = 1$$. But don't know concept that is hidden behind it. Please help me.

• Shouldn't you be solving $AX=0$ if you're looking for the nullspace?
– fwd
Aug 24 at 7:10
• I know that dimension of null space = $n-r$ for $A_{m×n}X_{n×1} = O_{m×1}$. Where $n$ is number of columns of matrix $A_{m×n}$ and $r$ is rank of $A_{m×n}$ Aug 24 at 7:19

I'd say the fastest way to get there is by the Rank-Nullity theorem from which you can deduce:

$$\mathrm{dim}(\mathrm{Nul}\, A) = \text{ the number of free variables}$$

I believe that you referred to it in a comment. For an explanation of its validity in the context of the echelon form of a matrix, see this answer.

Say, we were unaware of the Rank-Nullity theorem and wanted to find the dimension of $$\mathrm{Nul}\, A$$. One way to find the dimension of the null space of a matrix is to find a basis for the null space. The number of vectors in this basis is the dimension of the null space. As I will show for the case of one free variable,$$^1$$ the number of vectors in the basis corresponds to the number of free variables.

We are told that all solutions $$\mathbf{x}$$ to the given matrix equation

$$A\mathbf{x} = \mathbf{b}\tag1$$ where

$$\mathbf{b} = \begin{bmatrix}1\\2\\3\\4\\5\end{bmatrix}$$

are of the form $$\mathbf{x} = \mathbf{p} + s\mathbf{q}$$ for $$s \in \mathbb{R}$$ where

$$\mathbf{p} = \begin{bmatrix}1\\2\\3\\4\end{bmatrix} \qquad \text{and} \qquad \mathbf{q} = \begin{bmatrix}2\\3\\4\\5\end{bmatrix} \tag2$$

[The vector $$\mathbf{p}$$ is called a particular solution of (1).]

Let $$X$$ denote the solution set of (1)–that is, $$X = \{\mathbf{x} : \mathbf{x} = \mathbf{p} + s\mathbf{q} \text{ for } s \in \mathbb{R}\}$$, and recall that the null space of $$A$$ is the set of all vectors $$\mathbf{x}$$ which satisfy the equation

$$A\mathbf{x} = \mathbf{0} \tag3$$

There exists an intimate relationship between the vectors in $$X$$ and those in $$\mathrm{Nul}\, A$$. Denote the set $$Z = X - X = \{\mathbf{z} : \mathbf{z} = \mathbf{x} - \mathbf{y} \text{ for }\mathbf{x}, \mathbf{y} \in X\}$$, and observe that any $$\mathbf{z} \in Z$$ is a solution to (3).

$$A\mathbf{z} = A(\mathbf{x} - \mathbf{y}) = A\mathbf{x} - A\mathbf{y} = \mathbf{b} - \mathbf{b} = \mathbf{0} \qquad \text{where \mathbf{x},\mathbf{y} \in X}$$

Thus, $$Z \subset \mathrm{Nul}\, A$$.

Since $$\mathbf{x} = \mathbf{p} + r\mathbf{q}$$ and $$\mathbf{y} = \mathbf{p} + s\mathbf{q}$$ for some $$r,s \in \mathbb{R}$$, we have that $$\mathbf{z} = \mathbf{x} - \mathbf{y} = (r-s)\mathbf{q}$$. Thus, any $$\mathbf{z} \in Z$$ can be written in the form $$t\mathbf{q}$$ (choose $$t = r -s$$), and every vector $$t\mathbf{q}$$ corresponds to a vector $$\mathbf{z} \in Z$$ (choose $$r-s =t$$). In other words, $$Z = \mathrm{Span}\{\mathbf{q}\}$$, so similarly, we have $$\mathrm{Span}\{\mathbf{q}\} \subset \mathrm{Nul}\, A$$.

Next, we will show that every vector in the null space of $$A$$ is in $$\mathrm{Span}\, \{\mathbf{q}\}$$.

Let $$\mathbf{x} = \mathbf{p} + r\mathbf{q}$$ where $$r \in \mathbb{R}$$ (that is, $$\mathbf{x} \in X$$), and suppose $$\mathbf{u} \in \mathrm{Nul}\, A$$. Then, we have

$$A(\mathbf{x} + \mathbf{u}) = A(\mathbf{p} + r\mathbf{q} + \mathbf{u}) = A(\mathbf{p} + \mathbf{u}) + A(r\mathbf{q}) = A(\mathbf{p} + \mathbf{u}) = \mathbf{b} + \mathbf{0} = \mathbf{b}$$

Thus, $$\mathbf{p} + \mathbf{u}$$ is a solution to (1) and must be of the form $$\mathbf{p} + s\mathbf{q}$$. That is,

$$\mathbf{p} + \mathbf{u} = \mathbf{p} + s\mathbf{q} \quad \text{or equivalently} \quad \mathbf{u} = s\mathbf{q} \quad \text{for some s \in \mathbb{R}}$$

Thus, if $$\mathbf{u} \in \mathrm{Nul}\, A$$, $$\mathbf{u}$$ is a scalar multiple of $$\mathbf{q}$$, which by definition means that $$\mathbf{u} \in \mathrm{Span}\{\mathbf{q}\}$$.

As we have shown that $$\mathrm{Nul}\, A \subset \mathrm{Span} \{\mathbf{q}\}$$ and $$\mathrm{Span} \{\mathbf{q}\} \subset \mathrm{Nul}\, A$$, we have that $$\mathrm{Span} \{\mathbf{q}\} = \mathrm{Nul}\, A$$.

Since $$\{\mathbf{q}\}$$ is a linearly independent set which spans $$\mathrm{Nul}\, A$$, $$\{\mathbf{q}\}$$ is a basis for $$\mathrm{Nul}\, A$$, and $$\mathrm{dim}(\mathrm{Nul})\, A = 1$$.

$$^1$$ For the case of $$n$$ free variables, this relationship holds as well. Any solution $$\mathbf{x}$$ to a nonhomogeneous system can be written as

$$\mathbf{x} =\mathbf{p} + \sum_{i = 1}^n s_i \mathbf{q}_i$$

where $$\mathbf{p} \ne \mathbf{0}$$ denotes a particular solution to the nonhomogeneous system, $$s_i$$ denotes a free variable, and each $$\mathbf{q}_i$$ is analogous to $$\mathbf{q}$$ from (2). From here, we would show that each $$\mathbf{q}_i \in \mathrm{Nul}\, A$$ and that if $$\mathbf{x} \in \mathrm{Nul}\, A$$, then $$\mathbf{x} \in \mathrm{Span}\{\mathbf{q}_1,\dotsc,\mathbf{q}_n\}$$. Confirming the linear independence of the set $$\{\mathbf{q}_1,\dotsc,\mathbf{q}_n\}$$ establishes $$\{\mathbf{q}_1,\dotsc,\mathbf{q}_n\}$$ as a basis for $$\mathrm{Nul}\, A$$ with $$n$$ vectors, so that we may conclude $$\mathrm{dim}(\mathrm{Nul}\, A) = n$$.

The general solution for a linear system $$Ax=b$$ is given by two terms:

$$x=x_p+x_h$$

with

• particular solution $$Ax_p =b$$
• homogeneous solution $$Ax_h =0$$

such that

$$Ax=A(x_p+x_h)=Ax_p+Ax_h=b+0=b$$

In your case we have

$$x_p=\begin{pmatrix}{1 \\ 2 \\ 3 \\ 4}\end{pmatrix}\quad x_h= s \begin{pmatrix}{2 \\ 3 \\ 4 \\ 5}\end{pmatrix}$$

and since $$x_h$$ is a subspace with dimension $$1$$ we have that $$\dim(\mbox{null}(A)) = 1$$.