Lab 4. Bases, Dimension, Coordinates, and Rank–Nullity: Independent Study
This lab accompanies Chapter 4: Coordinates: Bases, Dimension, and Rank–Nullity.
The goal is to connect the story of Chapter 4 with computation:
- Linear independence: find redundancy in a list of vectors.
- Bases and coordinates: express a vector in a chosen coordinate system.
- Dimension: count degrees of freedom.
- Rank–nullity: separate information that survives from information that is lost.
- Change of basis: translate coordinates between two coordinate systems.
This is an independent-study lab. Each main question includes a worked solution and a similar practice question so that students can check their work as they go.
Python practice notebook
You may also use the Jupyter notebook version for longer Python practice:
Interactive lab
Study guide and worked questions
Question 1. Linear independence and redundancy
Let
\[ v_1=\begin{bmatrix}1\\-3\\4\end{bmatrix},\qquad v_2=\begin{bmatrix}2\\-2\\5\end{bmatrix},\qquad v_3=\begin{bmatrix}3\\-1\\6\end{bmatrix}. \]
Determine whether the vectors are linearly independent. If they are dependent, find a nontrivial linear relation.
Solution
Put the vectors into the columns of a matrix:
\[ A=\begin{bmatrix} 1&2&3\\ -3&-2&-1\\ 4&5&6 \end{bmatrix}. \]
Row reduction gives
\[ \operatorname{rref}(A)= \begin{bmatrix} 1&0&-1\\ 0&1&2\\ 0&0&0 \end{bmatrix}. \]
The equation \(Ax=0\) has a free variable \(x_3\). From the RREF,
\[ x_1-x_3=0,\qquad x_2+2x_3=0. \]
Let \(x_3=1\). Then \(x_1=1\) and \(x_2=-2\), so
\[ v_1-2v_2+v_3=0. \]
Therefore the vectors are linearly dependent.
Similar practice
Let
\[ w_1=\begin{bmatrix}1\\0\\1\end{bmatrix},\qquad w_2=\begin{bmatrix}0\\1\\1\end{bmatrix},\qquad w_3=\begin{bmatrix}1\\1\\2\end{bmatrix}. \]
Are these vectors independent?
Answer: They are dependent because \(w_3=w_1+w_2\), so
\[ w_1+w_2-w_3=0. \]
Question 2. Basis and coordinates in \(\mathbb R^2\)
Let
\[ b_1=\begin{bmatrix}1\\1\end{bmatrix},\qquad b_2=\begin{bmatrix}1\\-1\end{bmatrix},\qquad v=\begin{bmatrix}4\\2\end{bmatrix}. \]
- Show that \(\mathcal B=(b_1,b_2)\) is a basis of \(\mathbb R^2\).
- Find the coordinate vector \([v]_{\mathcal B}\).
Solution
Let
\[ P_{\mathcal B}=\begin{bmatrix}1&1\\1&-1\end{bmatrix}. \]
The determinant is
\[ \det(P_{\mathcal B})=(1)(-1)-(1)(1)=-2\neq 0. \]
So the columns are linearly independent and form a basis of \(\mathbb R^2\).
To find coordinates, solve
\[ P_{\mathcal B}c=v. \]
That is,
\[ c_1\begin{bmatrix}1\\1\end{bmatrix} +c_2\begin{bmatrix}1\\-1\end{bmatrix} =\begin{bmatrix}4\\2\end{bmatrix}. \]
This gives
\[ c_1+c_2=4,\qquad c_1-c_2=2. \]
Hence \(c_1=3\) and \(c_2=1\). Therefore
\[ [v]_{\mathcal B}=\begin{bmatrix}3\\1\end{bmatrix}. \]
Similar practice
For the same basis \(\mathcal B\), find the coordinates of \(u=(5,-1)\).
Answer: Solve \(c_1+c_2=5\) and \(c_1-c_2=-1\). Then \(c_1=2\) and \(c_2=3\), so
\[ [u]_{\mathcal B}=\begin{bmatrix}2\\3\end{bmatrix}. \]
Question 3. Finding a basis from a spanning list
Let
\[ S=\left\{ \begin{bmatrix}1\\0\\1\end{bmatrix}, \begin{bmatrix}0\\1\\1\end{bmatrix}, \begin{bmatrix}1\\1\\2\end{bmatrix}, \begin{bmatrix}2\\1\\3\end{bmatrix} \right\}. \]
Find a basis for \(\operatorname{span}(S)\).
Solution
Place the vectors as columns:
\[ A=\begin{bmatrix} 1&0&1&2\\ 0&1&1&1\\ 1&1&2&3 \end{bmatrix}. \]
A row reduction shows that columns \(1\) and \(2\) are pivot columns. Therefore a basis for the span is given by the corresponding original columns:
\[ \left\{ \begin{bmatrix}1\\0\\1\end{bmatrix}, \begin{bmatrix}0\\1\\1\end{bmatrix} \right\}. \]
The other vectors are redundant:
\[ \begin{bmatrix}1\\1\\2\end{bmatrix} = \begin{bmatrix}1\\0\\1\end{bmatrix} + \begin{bmatrix}0\\1\\1\end{bmatrix}, \]
and
\[ \begin{bmatrix}2\\1\\3\end{bmatrix} =2\begin{bmatrix}1\\0\\1\end{bmatrix} + \begin{bmatrix}0\\1\\1\end{bmatrix}. \]
Similar practice
Find a basis for the span of
\[ (1,2,0),\qquad (2,4,0),\qquad (0,1,1),\qquad (1,3,1). \]
Answer: A basis is
\[ \{(1,2,0),(0,1,1)\}. \]
The second vector is \(2(1,2,0)\), and the fourth vector is \((1,2,0)+(0,1,1)\).
Question 4. Kernel, image, rank, and nullity
Let \(T:\mathbb R^4\to\mathbb R^3\) be defined by \(T(x)=Ax\), where
\[ A=\begin{bmatrix} 1&0&2&1\\ 0&1&-1&3\\ 1&1&1&4 \end{bmatrix}. \]
Find bases for \(\operatorname{im}(T)\) and \(\ker(T)\), and verify the rank–nullity theorem.
Solution
Row reducing \(A\) gives
\[ \operatorname{rref}(A)= \begin{bmatrix} 1&0&2&1\\ 0&1&-1&3\\ 0&0&0&0 \end{bmatrix}. \]
The pivot columns are columns \(1\) and \(2\). Therefore
\[ \operatorname{im}(T)=\operatorname{span}\left\{ \begin{bmatrix}1\\0\\1\end{bmatrix}, \begin{bmatrix}0\\1\\1\end{bmatrix} \right\}. \]
The equations from the RREF are
\[ x_1+2x_3+x_4=0,\qquad x_2-x_3+3x_4=0. \]
Let \(x_3=s\) and \(x_4=t\). Then
\[ x_1=-2s-t,\qquad x_2=s-3t. \]
So
\[ x=s\begin{bmatrix}-2\\1\\1\\0\end{bmatrix} +t\begin{bmatrix}-1\\-3\\0\\1\end{bmatrix}. \]
Hence
\[ \ker(T)=\operatorname{span}\left\{ \begin{bmatrix}-2\\1\\1\\0\end{bmatrix}, \begin{bmatrix}-1\\-3\\0\\1\end{bmatrix} \right\}. \]
The rank is \(2\) and the nullity is \(2\). Since the domain is \(\mathbb R^4\),
\[ \operatorname{rank}(T)+\operatorname{nullity}(T)=2+2=4=\dim(\mathbb R^4). \]
Similar practice
Let
\[ B=\begin{bmatrix} 1&0&0&2\\ 0&1&0&-1\\ 0&0&1&3 \end{bmatrix}. \]
Find \(\ker(B)\) and the rank of \(B\).
Answer: The rank is \(3\). The equations are
\[ x_1+2x_4=0,\qquad x_2-x_4=0,\qquad x_3+3x_4=0. \]
Let \(x_4=t\). Then
\[ \ker(B)=\operatorname{span}\left\{ \begin{bmatrix}-2\\1\\-3\\1\end{bmatrix} \right\}. \]
The nullity is \(1\), and \(3+1=4\).
Question 5. Change of basis
Let
\[ \mathcal B=\left(\begin{bmatrix}1\\1\end{bmatrix},\begin{bmatrix}1\\-1\end{bmatrix}\right), \qquad \mathcal C=\left(\begin{bmatrix}2\\0\end{bmatrix},\begin{bmatrix}0\\3\end{bmatrix}\right). \]
Suppose
\[ [v]_{\mathcal B}=\begin{bmatrix}3\\1\end{bmatrix}. \]
Find \([v]_{\mathcal C}\).
Solution
First convert from \(\mathcal B\)-coordinates to standard coordinates:
\[ v=P_{\mathcal B}[v]_{\mathcal B} =\begin{bmatrix}1&1\\1&-1\end{bmatrix} \begin{bmatrix}3\\1\end{bmatrix} =\begin{bmatrix}4\\2\end{bmatrix}. \]
Now solve
\[ P_{\mathcal C}[v]_{\mathcal C}=v, \qquad P_{\mathcal C}=\begin{bmatrix}2&0\\0&3\end{bmatrix}. \]
Thus
\[ [v]_{\mathcal C}=P_{\mathcal C}^{-1}v =\begin{bmatrix}1/2&0\\0&1/3\end{bmatrix} \begin{bmatrix}4\\2\end{bmatrix} =\begin{bmatrix}2\\2/3\end{bmatrix}. \]
Similar practice
Using the same bases, if \([u]_{\mathcal B}=(2,3)\), find \([u]_{\mathcal C}\).
Answer: First \(u=2(1,1)+3(1,-1)=(5,-1)\). Then
\[ [u]_{\mathcal C}=\begin{bmatrix}5/2\\-1/3\end{bmatrix}. \]
Question 6. AI companion critique
Ask an AI tool:
Explain rank–nullity using the metaphor of information preserved and information lost.
Then critique the response. Your critique should answer:
- Does it correctly state that the domain dimension is split into rank and nullity?
- Does it identify the image as the information that survives?
- Does it identify the kernel as the directions that are sent to zero?
- Does it include a concrete matrix example?
- Does it avoid saying that nullity is always bad? In applications, null directions may represent compression, invariance, or constraints.