Section 23 Exam 2 Review

23.1 Overview

  • I will hand it out right away, even a few minutes early, so you can start right on time.
  • You must turn it in by 9:35 (for the 8:30 class) and 10:45 for the 10:50 class). No exceptions. The next class needs to come in and get started.
  • No calculators are allowed, and none are needed.
  • If row reductions are needed, they will be easy (integer) calculations, and there will not be many.
  • It will be closed book but you can bring a 3" x 5" notecard with notes written on both sides. These notes should be hand written by you. You do not need to turn in the note card with the exam.
  • There will be some basic calculations, but the problems will focus more on the ideas than on the calculations.
  • I will ask some problems that are very similar to homework problems, Edfinity problems, and examples from class or the videos.
  • I will ask other problems that are somewhat different from things you have done. On these, you are to apply your knowledge in a slightly new setting to demonstrate an even higher mastery of the material.
  • You will be allowed to re-write one problem to earn back half of the points that you lost. This will be due the class period after I hand the exam back.
  • The exam is worth 12% of your course grade as outlined in the syllabus.

23.1.1 Vocabulary and Concepts

  • Subspaces
    • 4.1: Subspaces of \(\mathbb{R}^n\)
    • 4.2: Null Space and Column Space
    • 4.3: Bases
    • 4.4: Coordinates
    • 4.5: Dimension, Rank, and Nullity
    • 4.6: Row Space

This corresponds to Problem Sets 5 and 6.

The best way to study is to do practice problems. The Exam will have some calculation problems (like Edfinity) and more conceptual problems (like the problem sets). Here are some ways to practice:

  • Make sure that you have mastered the Vocabulary, Skills and Concepts listed below.
  • Look over the Edfinity homework assignments
  • Redo the class exercises problems
  • Try to re-solve the Problem Sets and compare your answers to the solutions.
  • Do the practice problems below. Compare your answers to the solutions.

23.1.2 Vocabulary and Concepts

You should understand these concepts and be able to read and use these terms correctly:

  • all of the Important Definitions found here.
  • subspaces
  • null space and column space of a matrix
  • kernel and image of a linear transformation
  • basis (span and linearly independent)
  • coordinate vector with respect to a basis \(\mathcal{B}\)
  • change-of-coordinates matrix
  • dimension
  • rank
  • row space

23.1.3 Skills

You should be able to perform these linear algebra tasks.

  • show that a subset is a subspace or demonstrate that it is not a subspace
  • describe the null space and the column space of a matrix A, including find a basis for theses spaces.
  • determine if a vector v is in Nul(A), Col(A), or Row(A).
  • find a basis of a subspace, including the null space and column space
  • answer questions about the connections between all these ideas. For example,
    • What is the connection between the column space and null space and solving equations \(A x = b\).
    • What is the connection between the column space and null space and the linear transformation \(T_A: \mathbb{R}^n \to \mathbb{R}^m\)?
    • What is the connection between the column space and null space and the vectors in the matrix of \(A\): linear independence and span?
  • write short proofs of basic statements using the Important Definitions. For example, something like the Getting into a Subspace problem

23.2 Practice Problems

23.2.1

Here are the row reductions of 4 matrices into reduced row echelon form. \[ \begin{array}{ll} A \longrightarrow \begin{bmatrix} 1 & 0 & 5 & -3 & 0\\ 0 & 1 & -2 & 8 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \qquad & B \longrightarrow \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} \\ \\ C \longrightarrow \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} & D \longrightarrow \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix} \end{array} \]

In each case, if \(T_M\) is the linear transformation given by the matrix product \(T_M(x) = M x\), where \(M\) is the given matrix, then \(T_M: \mathbb{R}^n \to \mathbb{R}^m\) is a transformation from domain \(\mathbb{R}^n\) to codomain (aka target) \(\mathbb{R}^m\).

Determine the appropriate values for \(n\) and \(m\), and decide whether \(T_M\) is one-to-one and/or onto. Submit your answers in table form, as shown below. \[ \begin{array} {|c|c|c|c|c|c|} \hline & n & m & \text{one-to-one?} & \text{onto?} & \text{rank}=dim(Col(A)) & \text{nullity}=dim(Nul(A)) \\ \hline T_A &\phantom{\Big\vert XX}&\phantom{\Big\vert XX}&& \\ \hline T_B &\phantom{\Big\vert XX}&&& \\ \hline T_C &\phantom{\Big\vert XX}&&& \\ \hline T_D &\phantom{\Big\vert XX}&&& \\ \hline \end{array} \hskip5in \]

23.2.2

Suppose that \(A\) is an \(n \times n\) matrix with the property that \(A \mathsf{u} = A\mathsf{v}\) for \(\mathsf{u}, \mathsf{v} \in \mathbb{R}^n\) with \(\mathsf{u} \not= \mathsf{v}\). Decide if the following are true or false. Justify each answer, briefly.

  1. \(Nul(A) = \{0\}\)
  2. \(Col(A) = \mathbb{R}^n\).
  3. The columns of \(A\) are linearly independent.
  4. The columns of \(A\) form a basis of \(\mathbb{R}^n\).

23.2.3

Suppose that \(\mathsf{v}_1, \mathsf{v}_2, \mathsf{v}_3, \mathsf{v}_4\) are four vectors in \(\mathbb{R}^4\) and that there is a vector \(\mathsf{v} \in \mathbf{R}^4\) such that \(\mathsf{v}\) can be expressed in two different ways as linear combinations of these vectors: \[ \begin{array}{rrrrrrrrr} \mathsf{v} &=& 2 \mathsf{v}_1 &+& 7 \mathsf{v}_2 &+& 5 \mathsf{v}_3 &+& (-5) \mathsf{v}_4 \end{array} \] and \[ \begin{array}{rrrrrrrrr} \mathsf{v} &=& 3 \mathsf{v}_1 &+& 5 \mathsf{v}_2 &+& (-2) \mathsf{v}_3 &+& \mathsf{v}_4. \end{array} \] Use this information to show that \(\{\mathsf{v}_1,\mathsf{v}_2, \mathsf{v}_3, \mathsf{v_4}\}\) is a linearly dependent set by finding a dependence relation among these vectors.

23.2.4

Let \(U\) and \(W\) be subspaces of a vector space \(\mathbb{R}^n\). Prove or disprove the following statements. Prove them by showing that the conditions are being a subspace are satisfied. Disprove them with a specific counter example.

  1. \(U \cap W = \{ \mathsf{v} \in \mathbb{R}^n \mid \mathsf{v} \in U \mbox{ and } \mathsf{v} \in W \}\) is a subspace

  2. \(U \cup W = \{ \mathsf{v} \in \mathbb{R}^n \mid \mathsf{v} \in U \mbox{ or } \mathsf{v} \in W \}\) is a subspace

  3. \(U+W = \{\mathsf{u} + \mathsf{w} \mid \mathsf{u} \in U \mbox{ and } \mathsf{w} \in W \}\) is a subspace

23.2.5

I have performed some row operations below for you on a matrix \(A\). Find a basis for the column space and the null space of \(A\). \[ A= \left[ \begin{matrix} 1& 2& 0& 2& 0& -1 \\ 1& 2& 1& 1& 0& -2 \\ 2& 4& -2& 6& 1& 2 \\ 1& 2& 0& 2& -1& -3 \\ \end{matrix}\right] \longrightarrow \left[ \begin{matrix} 1& 2& 0& 2& 0& -1\\ 0& 0& 1& -1& 0& -1\\ 0& 0& 0& 0& 1& 2\\ 0& 0& 0& 0& 0& 0\\ \end{matrix}\right] \]

23.2.6

Consider the matrix \[ A = \left[ \begin{array}{cccc} 1 & 5 & 2 & -4 \\ 3 & 10 & 2 & 8 \\ 4 & 15 & 4 & 4 \end{array} \right] \] Find a basis for \(\mathrm{Col}(A)\). Find a basis for \(\mathrm{Nul}(A)\).

23.2.7

Are the vectors in \({\mathcal B}\) a basis of \(\mathbb{R}^3\)? If not, find a basis of the span of those vectors. Explain your reasoning. \[ \mathcal{B}=\left\{ \begin{bmatrix} 1 \\ -1 \\ -2 \end{bmatrix},\begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix},\begin{bmatrix} -1 \\ -1 \\ -8 \end{bmatrix} \right\} \]

23.2.8

Find the coordinates of \(\mathsf{w}\) in the standard basis and of \(\mathsf{v}\) in the \(\mathcal{B}\)-basis. \[ \mathcal{B} = \left\{ \mathsf{v}_1=\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \mathsf{v}_2=\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \mathsf{v}_3=\begin{bmatrix} 1 \\ 1 \\ 1 \\ 0 \end{bmatrix}, \mathsf{v}_4=\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix} \right\}, \] \[ \mathsf{w} = \begin{bmatrix} 3 \\ -2 \\ 0 \\ -1 \end{bmatrix}_{\mathcal{B}}, \qquad \mathsf{v} = \begin{bmatrix} 10 \\ 9 \\ 7 \\ 4 \end{bmatrix}_{\mathcal{S}} \]

23.2.9

The subspace \(S \subset \mathbb{R}^5\) is given by \[ \mathsf{S} = \mathsf{span} \left( \begin{bmatrix}1\\ 1\\ 0\\ -1\\ 2 \end{bmatrix}, \begin{bmatrix} 0\\ 1\\ 1\\ 1\\ 1 \end{bmatrix}, \begin{bmatrix} 3\\ 1\\ -2\\ -5\\ 4 \end{bmatrix}, \begin{bmatrix} 1\\ 0\\ 1\\ 0\\ 1 \end{bmatrix}, \begin{bmatrix} 2\\ -1\\ -1\\ -3\\ 1 \end{bmatrix}, \right)\]

  1. Use the following matrix to find a basis for \(S\). What is the dimension of \(S\)? \[ A=\left[ \begin{array}{ccccc} 1 & 0 & 3 & 1 & 2 \\ 1 & 1 & 1 & 0 & -1 \\ 0 & 1 & -2 & 1 & -1 \\ -1 & 1 & -5 & 0 & -3 \\ 2 & 1 & 4 & 1 & 1 \\ \end{array} \right] \rightarrow \left[ \begin{array}{ccccc} 1 & 0 & 3 & 0 & 1 \\ 0 & 1 & -2 & 0 & -2 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right] \]

  2. Find a basis for \(\mathrm{Nul}(A)\). What is the dimension of this null space?

23.2.10

A \(6 \times 8\) matrix \(A\) has rank 5. For each of \(\mathrm{Col}(A)\) and \(\mathrm{Nul}(A)\),

  • Determine the dimension of the subspace,
  • Indicate whether it is subspace of \(\mathbb{R}^6\) or \(\mathbb{R}^8\), and
  • Describe how you would find a basis of the subspace.

23.2.11

Here is a matrix \(A\) and \(rref(A)\).

##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    1    1    1    1    6    3
## [2,]    1    2   -1   -2    6    2
## [3,]    0    1    1    0    3    2
## [4,]    1    2   -1   -2    6    2
## [5,]   -1    1    1   -1    0    1
## [6,]    2    2   -1   -1    9    3
##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    1    0    0    1    3    1
## [2,]    0    1    0   -1    2    1
## [3,]    0    0    1    1    1    1
## [4,]    0    0    0    0    0    0
## [5,]    0    0    0    0    0    0
## [6,]    0    0    0    0    0    0

Here is \(A^T\) and \(rref(A^T)\).

##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    1    1    0    1   -1    2
## [2,]    1    2    1    2    1    2
## [3,]    1   -1    1   -1    1   -1
## [4,]    1   -2    0   -2   -1   -1
## [5,]    6    6    3    6    0    9
## [6,]    3    2    2    2    1    3
##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    1    0    0    0   -1    1
## [2,]    0    1    0    1    0    1
## [3,]    0    0    1    0    2   -1
## [4,]    0    0    0    0    0    0
## [5,]    0    0    0    0    0    0
## [6,]    0    0    0    0    0    0
  1. Give two bases of \(Col(A)\).
  2. One of the vectors below is in \(Col(A)\) and one is not. For the vector that is in \(Col(A)\) give the coordinates of the vector with respect to one of your bases.

\[ \mathsf{u} = \begin{bmatrix} 5\\ 7\\ 4\\ 7 \\ 3\\ 8 \end{bmatrix}, \qquad \mathsf{v} = \begin{bmatrix} 5\\ 4\\ 3\\ 2 \\ 1\\ 1 \end{bmatrix}. \]

23.2.12

Suppose that \(T: \mathbb{R}^n \to \mathbb{R}^m\) and \(S: \mathbb{R}^n \to \mathbb{R}^m\) are linear transformations. Let \(V \subset \mathbb{R}^n\) be the set \[ V = \{ \mathsf{v} \in \mathbb{R}^n\mid T(\mathsf{v}) = S(\mathsf{v}) \}. \] Prove that the set \(V\) is a subspace.

23.3 Solutions to Practice Problems

23.3.1

\[ \begin{array} {|c|c|c|c|c|} \hline & n & m & \text{one-to-one?} & \text{onto?} & \text{onto?} & \text{rank}=dim(Col(A)) & \text{nullity}=dim(Nul(A)) \\ \hline T_A &5&5& No & No \\ \hline T_B &4&5&Yes& No\\ \hline T_C &4&4&Yes&Yes \\ \hline T_D &4&3&No&Yes\\ \hline \end{array} \hskip5in \]

23.3.2

All are false!

23.3.3

Hint: subtract the two representations of \(\mathsf{v}\).

23.3.4

  1. True

  • Since \(U\) and \(W\) are subspaces, we know that \(\mathbb{0} \in U\) and \(\mathbb{0} \in W\). Therefore \(\mathbb{0} \in U \cap W\).

  • Let \(\mathsf{v}_1 \in U \cap W\) and \(\mathsf{v}_2 \in U \cap W\).

  • We know that \(\mathsf{v}_1 \in U\) and \(\mathsf{v}_2 \in U\). Since \(U\) is a subspace, we have \(\mathsf{v}_1 + \mathsf{v}_2 \in U\).

  • We know that \(\mathsf{v}_1 \in W\) and \(\mathsf{v}_2 \in W\). Since \(W\) is a subspace, we have \(\mathsf{v}_1 + \mathsf{v}_2 \in W\). Therefore \(\mathsf{v}_1 + \mathsf{v}_2 \in U \cap W\).

  • Let \(\mathsf{v} \in U \cap W\) and \(c \in \mathbb{R}\).

  • We know that \(\mathsf{v} \in U\) and \(c \in R\). Since \(U\) is a subspace, we have \(c \mathsf{v} \in U\).

  • We know that \(\mathsf{v} \in W\) and \(c \in R\). Since \(W\) is a subspace, we have \(c \mathsf{v} \in W\).

  • Therefore \(c \mathsf{v} \in U \cap W\).

  1. False. Here is an example that shows this is not always true. Let \(V= \mathbb{R}^2\), \(U = \{ { x \choose 0} \mid x \in \mathbb{R} \}\) and \(W= \{ { 0 \choose y} \mid y \in \mathbb{R} \}\). The set \(U \cup W\) is not closed under addition. For example, \({1 \choose 0} + {0 \choose 1} = { 1 \choose 1} \notin U \cup W\).

  2. True.

  • Since \(U\) and \(W\) are subspaces, we know that \(\mathbb{0} \in U\) and \(\mathbb{0} \in W\). Therefore \(\mathbb{0} = \mathbb{0} + \mathbb{0} \in U + W\).

  • Let \(\mathsf{u}_1 + \mathsf{w}_1 \in U + W\) and \(\mathsf{u}_1 + \mathsf{w}_2 \in U + W\), where \(\mathsf{u}_1, \mathsf{u}_2 \in U\) and \(\mathsf{w}_1, \mathsf{w}_2 \in W\). Then \[ (\mathsf{u}_1 + \mathsf{w}_1) + (\mathsf{u}_2 + \mathsf{w}_2) = (\mathsf{u}_1 + \mathsf{u}_2) + (\mathsf{w}_1 + \mathsf{w}_2) \] and \(\mathsf{u}_3 = (\mathsf{u}_1 + \mathsf{u}_2) \in U\) (because \(U\) is a subspace) and \(\mathsf{w}_3 = (\mathsf{w}_1 + \mathsf{w}_2) \in W\) (because \(W\) is a subspace).

  • Therefore \((\mathsf{u}_1 + \mathsf{v}_1) + (\mathsf{u}_2 + \mathsf{w}_2) = \mathsf{u}_3 + \mathsf{w}_3 \in U + W\).

  • Let \(\mathsf{u} + \mathsf{w} \in U + W\) and \(c \in \mathbb{R}\). Then \(c(\mathsf{u} + \mathsf{w}) = c \mathsf{u} + c \mathsf{w}\). We know that \(c \mathsf{u} \in U\) (since \(U\) is a subspace) and \(c \mathsf{w} \in W\) (since \(W\) is a subspace). Therefore \(c(\mathsf{u} + \mathsf{w}) = c \mathsf{u} + c \mathsf{w} \in U+W\).

23.3.5

A basis for \(\mathrm{Col}(A)\) is \[ \begin{bmatrix} 1 \\1 \\ 2 \\ 1 \end{bmatrix}, \quad \begin{bmatrix} 0 \\1 \\ -2 \\ 0 \end{bmatrix}, \quad \begin{bmatrix} 0 \\0 \\ 1 \\ -1 \end{bmatrix} \] and a basis for \(\mathrm{Nul}(A)\) is \[ \begin{bmatrix} -2 \\1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \quad \begin{bmatrix} -2 \\0 \\ 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \quad \begin{bmatrix} 1 \\0 \\ 1\\ 0 \\ -2 \\ 1 \end{bmatrix}. \]

23.3.6

Using R we find:

##      [,1] [,2] [,3] [,4]
## [1,]    1    5    2   -4
## [2,]    3   10    2    8
## [3,]    4   15    4    4
##      [,1] [,2] [,3] [,4]
## [1,]    1    0 -2.0   16
## [2,]    0    1  0.8   -4
## [3,]    0    0  0.0    0

A basis for \(\mathrm{Col}(A)\) is \[ \begin{bmatrix} 1 \\ 3 \\ 4 \end{bmatrix}, \quad \begin{bmatrix} 5 \\ 10 \\ 15 \end{bmatrix}. \]

A basis for \(\mathrm{Nul}(A)\) is \[ \begin{bmatrix} 2 \\ -0.8 \\ 1 \\ 0 \end{bmatrix}, \quad \begin{bmatrix} -16 \\ 4 \\ 0 \\ 1 \end{bmatrix}. \]

23.3.7 

A =  cbind(c(1,-1,-2),c(2,-1,1),c(-1,-1,-8))
A
##      [,1] [,2] [,3]
## [1,]    1    2   -1
## [2,]   -1   -1   -1
## [3,]   -2    1   -8
rref(A)
##      [,1] [,2] [,3]
## [1,]    1    0    3
## [2,]    0    1   -2
## [3,]    0    0    0

The vectors are linearly dependent. If we choose the first two vectors \(\mathcal{B} = \{ \mathsf{v}_1, \mathsf{v}_2\}\), then \(\mathcal{B}\) is a basis for \(S = \mathsf{span}(\mathsf{v}_1, \mathsf{v}_2, \mathsf{v}_3)\).

23.3.8

We use the change of basis matrix. \[ P_{\cal B} = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix} \] Then, the desired coordinate vectors are \[ \mathsf{w} = \begin{bmatrix} 0 \\ -3 \\ -1 \\ -1 \end{bmatrix}_{\mathcal{S}}, \qquad \mathsf{v} = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix}_{\mathcal{B}} \] You can find these vectors by multiplying by \(P_\mathcal{B}\) and by augmenting and row reducing as seen here.

A =  cbind(c(1,0,0,0),c(1,1,0,0),c(1,1,1,0),c(1,1,1,1))
w = c(3,-2,0,-1)
v = c(10,9,7,4)
A  %*% w
##      [,1]
## [1,]    0
## [2,]   -3
## [3,]   -1
## [4,]   -1
Av = cbind(A,v)
rref(Av)
##              v
## [1,] 1 0 0 0 1
## [2,] 0 1 0 0 2
## [3,] 0 0 1 0 3
## [4,] 0 0 0 1 4

Or we can use the inverse of \(P_\mathcal{B}\). \[ P_{\cal B}^{-1} = \begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 &-1 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]

Ainv = solve(A)
Ainv %*% v
##      [,1]
## [1,]    1
## [2,]    2
## [3,]    3
## [4,]    4

23.3.9

  1. \(\dim(S) = 3\) and a basis for \(S\) is \[ \begin{bmatrix} 1 \\ 1 \\ 0 \\ -1 \\2 \end{bmatrix}, \quad \begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}, \quad \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \\1 \end{bmatrix}. \]

  2. \(\dim(\mathrm{Nul}(A))=2\) and a basis is \[ \begin{bmatrix} -3 \\ 2 \\ 1 \\ 0 \\0\end{bmatrix}, \quad \begin{bmatrix} -1 \\ 2 \\ 0 \\ -1 \\1 \end{bmatrix}. \]

23.3.10

  • \(\mathrm{Col}(A)\) has dimension 5, and it is a subspace of \(\mathbb{R}^6\). You would find a basis by taking the pivot columns of \(A\).
  • \(\mathrm{Nul}(A)\) has dimension 3, and it is a subspace of \(\mathbb{R}^8\). You would find a basis by finding the parametric solution to \(A \mathsf{x}= \mathbb{0}\).