# Day 6:

Finding bases for the Nullspace

Example. Find a basis for the subspace

$V = \left\{\begin{bmatrix} a\\ b\\ c\end{bmatrix} : a + b + c=0\right\}$

Note that if $$x\in V$$, then

$x=\begin{bmatrix} a\\ b\\ c\end{bmatrix} = \begin{bmatrix} -b-c\\ b\\ c\end{bmatrix} = b\begin{bmatrix} -1\\ 1\\ 0\end{bmatrix} + c\begin{bmatrix} -1\\ 0\\ 1\end{bmatrix}$

From this we see that

$\left\{\begin{bmatrix} -1\\ 1\\ 0\end{bmatrix}, \begin{bmatrix} -1\\ 0\\ 1\end{bmatrix}\right\}$

spans $$V$$. We also see that this set is independent, since

$x_{1}\begin{bmatrix} -1\\ 1\\ 0\end{bmatrix}+x_{2}\begin{bmatrix} -1\\ 0\\ 1\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}\ \Rightarrow \begin{bmatrix} -x_{1}-x_{2}\\ x_{1}\\ x_{2}\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}\ \Rightarrow x_{1}=x_{2}=0$

Example. Find a basis for the subspace

$V = \left\{\begin{bmatrix} a\\ b\\ c\end{bmatrix} : a + b + c=0\right\} = \left\{\begin{bmatrix} a\\ b\\ c\end{bmatrix} : \begin{bmatrix} 1 & 1 & 1\end{bmatrix}\begin{bmatrix} a\\ b\\ c\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix} \right\}$

Note that if $$x\in V$$, then

$x=\begin{bmatrix} a\\ b\\ c\end{bmatrix} = \begin{bmatrix} -b-c\\ b\\ c\end{bmatrix} = b\begin{bmatrix} -1\\ 1\\ 0\end{bmatrix} + c\begin{bmatrix} -1\\ 0\\ 1\end{bmatrix}$

From this we see that

$\left\{\begin{bmatrix} -1\\ 1\\ 0\end{bmatrix}, \begin{bmatrix} -1\\ 0\\ 1\end{bmatrix}\right\}$

spans $$V$$. We also see that this set is independent, since

$x_{1}\begin{bmatrix} -1\\ 1\\ 0\end{bmatrix}+x_{2}\begin{bmatrix} -1\\ 0\\ 1\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}\ \Rightarrow \begin{bmatrix} -x_{1}-x_{2}\\ x_{1}\\ x_{2}\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}\ \Rightarrow x_{1}=x_{2}=0$

Definition. If $$A$$ is an $$m\times n$$ matrix, then the null space of $$A$$, denoted $$N(A)$$, is the set

$N(A) = \{x\in\mathbb{R}^{n} : Ax=\mathbf{0}\}.$

Example. From the previous example

$V = \left\{\begin{bmatrix} a\\ b\\ c\end{bmatrix} : a + b + c=0\right\} = \left\{\begin{bmatrix} a\\ b\\ c\end{bmatrix} : \begin{bmatrix} 1 & 1 & 1\end{bmatrix}\begin{bmatrix} a\\ b\\ c\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix} \right\}$

$= N(\begin{bmatrix} 1 & 1 & 1\end{bmatrix}) = \text{span}\left\{\begin{bmatrix} -1\\ 1\\ 0\end{bmatrix}, \begin{bmatrix} -1\\ 0\\ 1\end{bmatrix}\right\}$

Theorem. If $$A$$ is an $$m\times n$$ matrix, then $$N(A)$$ is a subspace of $$\R^{n}$$.

Proof. We must show that $$N(A)$$ is closed under addition and scalar multiplication. Let $$x,y\in N(A)$$ and let $$a$$ be a scalar. Observe that

$A(ax) = aAx = a\mathbf{0} = \mathbf{0}.$

This shows that $$ax\in N(A)$$, and hence $$N(A)$$ is closed under scalar multiplication. Next, observe that

$A(x+y) = Ax+Ay = \mathbf{0} + \mathbf{0} = \mathbf{0}.$

This shows that $$x+y\in N(A)$$, and hence $$N(A)$$ is closed under addition. $$\Box$$

Example. Consider the matrix

$C=\begin{bmatrix} 1 & 0 & 1 & -1 & 0\\ 2 & 1 & 1 & 0 & 1\\ 0 & 2 & 0 & 0 & 0\end{bmatrix}$

Find a basis for $$N(C)$$.

By a theorem from last time we see that $$N(C) = N(\text{rref}(C))$$ and

$\text{rref}(C)=\begin{bmatrix} 1 & 0 & 0 & 1 & 1\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & -2 & -1\end{bmatrix}$

We need to solve

$\begin{bmatrix} 1 & 0 & 0 & 1 & 1\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & -2 & -1\end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5}\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}$

We need to solve

$\begin{bmatrix} 1 & 0 & 0 & 1 & 1\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & -2 & -1\end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5}\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}$

We're going to write down one solution for each non-pivot column:

$\begin{bmatrix} -1\\ 0\\ 2\\ 1\\ 0\end{bmatrix}\quad\quad\quad\quad$

$,\ \begin{bmatrix} -1\\ 0\\ 1\\ 0\\ 1\end{bmatrix}$

We're going to write down one solution for each non-pivot column:

I claim that this is a basis for the null space of $$C$$.

It is quick to show that these vectors are independent

$\begin{bmatrix} -1\\ 0\\ 2\\ 1\\ 0\end{bmatrix},\ \begin{bmatrix} -1\\ 0\\ 1\\ 0\\ 1\end{bmatrix}$

Since

$a\begin{bmatrix} -1\\ 0\\ 2\\ 1\\ 0\end{bmatrix}+b\begin{bmatrix} -1\\ 0\\ 1\\ 0\\ 1\end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0\\ 0\\ 0\end{bmatrix}\ \Leftrightarrow\ \begin{bmatrix} -a-b\\ 0\\ 2a+b\\ a\\ b\end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0\\ 0\\ 0\end{bmatrix}\ \Leftrightarrow\ a=b=0$

Now, we must show that anything in $$N(C)$$ is in the span of

$\begin{bmatrix} -1\\ 0\\ 2\\ 1\\ 0\end{bmatrix},\ \begin{bmatrix} -1\\ 0\\ 1\\ 0\\ 1\end{bmatrix}$

Assume

$\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5}\end{bmatrix}\in N(C).$

Since $$N(C)$$ is a subspace, we have

$\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5}\end{bmatrix}-x_{4}\begin{bmatrix} -1\\ 0\\ 2\\ 1\\ 0\end{bmatrix}-x_{5}\begin{bmatrix} -1\\ 0\\ 1\\ 0\\ 1\end{bmatrix} = \begin{bmatrix} x_{1}+x_{4}+x_{5}\\ x_{2}\\ x_{3}-2x_{4}-x_{5}\\ 0\\ 0\end{bmatrix}\in N(C)$

$\Rightarrow \begin{bmatrix} 1 & 0 & 0 & 1 & 1\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & -2 & -1\end{bmatrix}\begin{bmatrix} x_{1}+x_{4}+x_{5}\\ x_{2}\\ x_{3}-2x_{4}-x_{5}\\ 0\\ 0\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}$

$\Rightarrow \begin{bmatrix} x_{1}+x_{4}+x_{5}\\ x_{2}\\ x_{3}-2x_{4}-x_{5}\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}$

$\Rightarrow \left\{\begin{matrix} x_{1}+x_{4}+x_{5} & = 0\\ x_{2} & = 0\\ x_{3}-2x_{4}-x_{5} & = 0\end{matrix}\right.$

$\Rightarrow\ \begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5}\end{bmatrix}-x_{4}\begin{bmatrix} -1\\ 0\\ 2\\ 1\\ 0\end{bmatrix}-x_{5}\begin{bmatrix} -1\\ 0\\ 1\\ 0\\ 1\end{bmatrix} = \begin{bmatrix} x_{1}+x_{4}+x_{5}\\ x_{2}\\ x_{3}-2x_{4}-x_{5}\\ 0\\ 0\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ 0\end{bmatrix}$

$\Rightarrow\ \begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5}\end{bmatrix}=x_{4}\begin{bmatrix} -1\\ 0\\ 2\\ 1\\ 0\end{bmatrix}+x_{5}\begin{bmatrix} -1\\ 0\\ 1\\ 0\\ 1\end{bmatrix}\in\text{span}\left\{\begin{bmatrix} -1\\ 0\\ 2\\ 1\\ 0\end{bmatrix}, \begin{bmatrix} -1\\ 0\\ 1\\ 0\\ 1\end{bmatrix}\right\}$

Let $$A$$ be an $$m\times n$$ matrix and look at

$\text{rref}(A) = \begin{bmatrix} 0 & 1 & \cdots & 0 & \cdots & 0 & \cdots & b_{1} & \cdots\\ 0 & 0 & \cdots & 1 & \cdots & 0 & \cdots & b_{2} & \cdots\\ \vdots & \vdots & & \vdots & \ddots & \vdots & & \vdots & \\ 0 & 0 & \cdots & 0 & \cdots & 1 & \cdots & b_{r} & \cdots\\ 0 & 0 & \cdots & 0 & \cdots & 0 & \cdots & 0 & \cdots\\ \vdots & \vdots & & \vdots & & \vdots & & \vdots & \\ 0 & 0 & \cdots & 0 & \cdots & 0 & \cdots & 0 & \cdots\end{bmatrix}$

Non-pivot column

Pivot columns to the left of the non-pivot column

# ...

### Finding a basis for $$N(A)$$

Let $$A$$ be an $$m\times n$$ matrix and look at

$\text{rref}(A) = \begin{bmatrix} 0 & 1 & \cdots & 0 & \cdots & 0 & \cdots & b_{1} & \cdots\\ 0 & 0 & \cdots & 1 & \cdots & 0 & \cdots & b_{2} & \cdots\\ \vdots & \vdots & & \vdots & \ddots & \vdots & & \vdots & \\ 0 & 0 & \cdots & 0 & \cdots & 1 & \cdots & b_{r} & \cdots\\ 0 & 0 & \cdots & 0 & \cdots & 0 & \cdots & 0 & \cdots\\ \vdots & \vdots & & \vdots & & \vdots & & \vdots & \\ 0 & 0 & \cdots & 0 & \cdots & 0 & \cdots & 0 & \cdots\end{bmatrix}$

### Finding a basis for $$N(A)$$

$x=\left[\quad -b_{1} \quad\ \ -b_{2}\ \cdots\ -b_{r} \quad\quad\ \ 1\quad\quad\quad\right]^{\top}$

Zeros elsewhere

For each non-pivot column we make such a vector.

Example. Consider a matrix $$A$$ such that

$\text{rref}(A) = \begin{bmatrix} 0 & 1 & -2 & 0 & 4 & 0 & 1\\ 0 & 0 & 0 & 1 & 2 & 0 & -3\\ 0 & 0 & 0 & 0 & 0 & 1 & 0\end{bmatrix}$

### Finding a basis for $$N(A)$$

$\left\{\begin{bmatrix} 1\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\end{bmatrix}\right.$

Consider a matrix $$A$$ such that

$\text{rref}(A) = \begin{bmatrix} 0 & 1 & -2 & 0 & 4 & 0 & 1\\ 0 & 0 & 0 & 1 & 2 & 0 & -3\\ 0 & 0 & 0 & 0 & 0 & 1 & 0\end{bmatrix}$

### Finding a basis for $$N(A)$$

$\left\{\begin{bmatrix} 1\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\end{bmatrix}\right.$

$,\ \begin{bmatrix} 0\\ 2\\ 1\\ 0\\ 0\\ 0\\ 0\end{bmatrix}$

Consider a matrix $$A$$ such that

$\text{rref}(A) = \begin{bmatrix} 0 & 1 & -2 & 0 & 4 & 0 & 1\\ 0 & 0 & 0 & 1 & 2 & 0 & -3\\ 0 & 0 & 0 & 0 & 0 & 1 & 0\end{bmatrix}$

### Finding a basis for $$N(A)$$

$\left\{\begin{bmatrix} 1\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\end{bmatrix}\right.$

$,\ \begin{bmatrix} 0\\ 2\\ 1\\ 0\\ 0\\ 0\\ 0\end{bmatrix}$

$,\ \begin{bmatrix} 0\\ -4\\ 0\\ -2\\ 1\\ 0\\ 0\end{bmatrix}$

Consider a matrix $$A$$ such that

$\text{rref}(A) = \begin{bmatrix} 0 & 1 & -2 & 0 & 4 & 0 & 1\\ 0 & 0 & 0 & 1 & 2 & 0 & -3\\ 0 & 0 & 0 & 0 & 0 & 1 & 0\end{bmatrix}$

### Finding a basis for $$N(A)$$

$\left\{\begin{bmatrix} 1\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\end{bmatrix}\right.$

$,\ \begin{bmatrix} 0\\ 2\\ 1\\ 0\\ 0\\ 0\\ 0\end{bmatrix}$

$,\ \begin{bmatrix} 0\\ -4\\ 0\\ -2\\ 1\\ 0\\ 0\end{bmatrix}$

$,\left.\ \begin{bmatrix} 0\\ -1\\ 0\\ 3\\ 0\\ 0\\ 1\end{bmatrix}\right\}$

Consider a matrix $$A$$ such that

$\text{rref}(A) = \begin{bmatrix} 0 & 1 & -2 & 0 & 4 & 0 & 1\\ 0 & 0 & 0 & 1 & 2 & 0 & -3\\ 0 & 0 & 0 & 0 & 0 & 1 & 0\end{bmatrix}$

### Finding a basis for $$N(A)$$

$\left\{\begin{bmatrix} 1\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\end{bmatrix}\right.$

$,\ \begin{bmatrix} 0\\ 2\\ 1\\ 0\\ 0\\ 0\\ 0\end{bmatrix}$

$,\ \begin{bmatrix} 0\\ -4\\ 0\\ -2\\ 1\\ 0\\ 0\end{bmatrix}$

$,\left.\ \begin{bmatrix} 0\\ -1\\ 0\\ 3\\ 0\\ 0\\ 1\end{bmatrix}\right\}$

This set is clearly independent (Why?), but how do we know that it spans $$N(A)$$? (We will find out in the upcoming lectures.)

By John Jasper

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