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
- 478
