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.)

End Day 6

Linear Algebra Day 6

By John Jasper

Linear Algebra Day 6

  • 442