Section 3 Problem Set 3
- Due: Wednesday September 22 by midnight.
- Free Extension: anything turned in by 5PM Friday will be considered on time.
- After 5PM Friday, late assignments are subject to a 15% late penalty.
- After I post the solution on Monday, there is a 50% late penalty.
- Upload your solutions to Moodle in a PDF.
- Please feel free to use RStudio for all row reductions.
- You can download the Rmd source file for this problem set.
The Problem Set covers sections 1.7 Linear Independence and 1.8 Linear Transformations.
3.1 RREF for a set of vectors
Suppose that we have five vectors \(\mathsf{v}_1, \mathsf{v}_2,\mathsf{v}_3,\mathsf{v}_4,\mathsf{v}_5\) in \(\mathbb{R}^4\) and that the matrix \(A\) containing those vectors row reduces as follows \[ A = \left[ \begin{array}{ccc} \mid & \mid & \mid & \mid & \mid \\ \mathsf{v}_1 & \mathsf{v}_2 & \mathsf{v}_3 &\mathsf{v}_4 &\mathsf{v}_5 \\ \mid & \mid & \mid & \mid & \mid \end{array} \right] \longrightarrow \begin{bmatrix} 1 & 0 & -3 & 0 & 2 \\ 0 & 1 & 4 & 0 & 1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}. \]
Do the vectors \(\mathsf{v}_1, \mathsf{v}_2, \mathsf{v}_3, \mathsf{v}_4, \mathsf{v}_5\) span \(\mathbb{R}^4\)? Justify your answer.
Give the solution, in parametric form, to the homogeneous system of equations \(A x = 0\) for this problem.
Give a dependence relation among the vectors \(\mathsf{v_1}, \mathsf{v_2},\mathsf{v_3},\mathsf{v_4}, \mathsf{v_5}\).
Is the vector \(\mathsf{v}_3\) in \(\mathrm{span}(\mathsf{v}_1,\mathsf{v}_2)\)? Justify your answer.
Suppose that \(\mathsf{b} = 5 \mathsf{v}_1 + \mathsf{v}_2 - 3 \mathsf{v}_3 +4 \mathsf{v}_4 - \mathsf{v}_5\). Use what you have done above to write \(\mathsf{b}\) as a different linear combination of \(\mathsf{v_1}, \mathsf{v_2},\mathsf{v_3},\mathsf{v_4}, \mathsf{v_5}\) (i.e., with different weights).
3.2 Linear independence and unique expressions.
It is an important fact that if a set of vectors \(\mathsf{v}_1, \mathsf{v}_2, \mathsf{v}_3, \ldots, \mathsf{v}_n\) is linearly independent then any vector in the span of these vectors can be written as a unique linear combination of those vectors.
- This property has a fairly simple proof. Let’s suppose that n = 4 for simplicity and that a vector v can be written in two ways as a combination of those vectors: \[ \begin{array}{ccccccc} v &=& c_1 \mathsf{v}_1 &+& c_2 \mathsf{v}_2 &+& c_3 \mathsf{v}_3 &+& c_4 \mathsf{v}_4 \\ &=& d_1 \mathsf{v}_1 &+& d_2 \mathsf{v}_2 &+& d_3 \mathsf{v}_3 &+& d_4 \mathsf{v}_4 \end{array} \] Then use the definition of linear independence to prove that \(c_1 = d_1\), \(c_2 = d_2\), \(c_3 = d_3\), and \(c_4 = d_4\).
3.3 Is the transformation linear?
There are three transformations below. If you believe that the transformation is linear, then show it is by showing that the three linear transformation rules apply for arbitrary vectors. If you believe that the transformation is not linear. Then show that one of the rules fails for specific vectors.
\[ T \left( \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \right) = \begin{bmatrix} x_1 + x_2 + x_3 -1 \\ x_1 - x_2 + x_3 + 1 \end{bmatrix} \]
\[ T \left( \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \right) = \begin{bmatrix} 3 x_1 - 5 x_2 \\ 2 x_1 + x_2 \\ 2 x_1 + 3 x_2 \end{bmatrix} \]
\[ T \left( \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \right) = \begin{bmatrix} x_1 + x_2^2 + x_3 \\ 2 x_2 + x_1 x_3 \\ 2 x_1 + 3 x_2 + x_3 \end{bmatrix} \]
3.4 Partial Information about a Linear Transformation
We are given that \(T: \mathbb{R}^4 \rightarrow \mathbb{R}^3\) is a linear transformation such that: \[ T\left(\begin{bmatrix} 3 \\ ~2~ \\ 1 \\ 2 \end{bmatrix} \right)=\begin{bmatrix} ~2~ \\ 3 \\ 6 \end{bmatrix} \qquad\hbox{and}\qquad T\left(\begin{bmatrix}~~2 \\ -1 \\ 0 \\ -1 \end{bmatrix} \right)=\begin{bmatrix} 2 \\ ~0~ \\ 1 \end{bmatrix}. \] Use this information to compute the value of \(T\) below? \[T\left(\begin{bmatrix} 5 \\ 8 \\ ~3~ \\ 8 \end{bmatrix} \right) = \hskip5in\] Hint: express the third input vector as a linear combination of the first two.
3.5 Square, Wide, and Tall Matrices
In each question below, choose all of the answers that apply, and give a succinct explanation of how you know that it is true.
[Wide] You are given a wide 5 x 8 matrix and you are studying the matrix equation \(A x = b\). These problems are said to be under constrained because there are fewer equations (constraints) than variables. \[ A=\begin{bmatrix} \ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot\ \\ \ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot\ \\ \ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot\ \\ \ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot\ \\ \ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot\ \\ \end{bmatrix} \]
- The vector that you are trying to get to is \(b \in \mathbb{R}^m\) and the solution vectors are \(x \in \mathbb{R}^n\). What are \(m\) and \(n\)?
- Decide which of the following statements are true and give a short and succinct justification of your answer using complete sentences and good punctuation.
- The columns of \(A\) must span \(\mathbb{R}^m\).
- The columns of \(A\) must be linearly dependent.
- There is a unique solution to \(A x = 0\).
- If the columns span \(\mathbb{R}^m\) then there is a unique solution to \(A x = b\) for all \(b\).
[Tall] You are given a wide 7 x 4 matrix and you are studying the matrix equation \(A x = b\). These problems are said to be over constrained because there are more equations (constraints) than variables. \[ A=\begin{bmatrix} \ \cdot & \cdot & \cdot & \cdot \ \\ \ \cdot & \cdot & \cdot & \cdot \ \\ \ \cdot & \cdot & \cdot & \cdot \ \\ \ \cdot & \cdot & \cdot & \cdot \ \\ \ \cdot & \cdot & \cdot & \cdot \ \\ \ \cdot & \cdot & \cdot & \cdot \ \\ \ \cdot & \cdot & \cdot & \cdot \ \\ \end{bmatrix} \]
- The vector that you are trying to get to is \(b \in \mathbb{R}^m\) and the solution vectors are \(x \in \mathbb{R}^n\). What are \(m\) and \(n\)?
- Decide which of the following statements are true and give a short and succinct justification of your answer using complete sentences and good punctuation.
- The columns of \(A\) might span \(\mathbb{R}^m\).
- The columns of \(A\) must be linearly dependent.
- There is a unique solution to \(A x = 0\).
[Square] You are given a square 6 x 6 matrix and you are studying the matrix equation \(A x = b\). \[ A=\begin{bmatrix} \ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot\ \\ \ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot\ \\ \ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot\ \\ \ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot\ \\ \ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot\ \\ \ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot\ \\ \end{bmatrix} \]
- The vector that you are trying to get to is \(b \in \mathbb{R}^m\) and the solution vectors are \(x \in \mathbb{R}^n\). What are \(m\) and \(n\)?
- Decide which of the following statements are true and give a short and succinct justification of your answer using complete sentences and good punctuation.
- The columns of \(A\) must span \(\mathbb{R}^m\).
- If the columns span \(\mathbb{R}^m\) then there is a unique solution to \(A x = b\) for all \(b\).
- If \({\bf 0}\) is the only solution to \(A x = {\bf 0}\), then there is a unique solution to \(A x = b\) for all \(b \in \mathbb{R}^m\).
- The columns of \(A\) span \(\mathbb{R}^m\) if and only if the columns of \(A\) are linearly independent.