Section 5 Problem Set 5

  • Due: Wednesday October 13 by 11:59pm CST.
  • You will get a chance to discuss some of these in class with your classmates.
  • 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.

5.1 Column and Null Space

Find a basis for the column space \(Col(A)\) and the null space \(Nul(A)\) of the matrix \(A\) below

(A = rbind(c(1, 2, 0, 2, 0, -1),c(1, 2, 1, 1, 0, -2),
           c(2, 4, -2, 6, 1, 2),c(1, 2,  0, 2, -1, -3 )))
##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    1    2    0    2    0   -1
## [2,]    1    2    1    1    0   -2
## [3,]    2    4   -2    6    1    2
## [4,]    1    2    0    2   -1   -3

5.2 Vectors Rescaled

If the function \(T: \mathbb{R}^n \to \mathbb{R}^n\) is a linear transformation, then show that the set below is a subspace of \(\mathbb{R}^n\) \[ E_{2021} = \{\ \vec{x} \in \mathbb{R}^n \mid T(\vec{x}) = 2021 \vec{x} \}. \] Is there anything special about 2021 in the definition? If it were replaced by another scalar, would it still be a subspace?

5.3 Fibonacci Vectors

The Fibonacci vectors \(F\) in \(\mathbb{R}^5\) are defined below: \[ F = \left\{ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix} ~\Bigg\vert~ \ x_3 = x_1 + x_2, x_4 = x_2 + x_3, x_5 = x_3 + x_4 \right\} \subseteq \mathbb{R}^5. \hskip5in \]

  1. Show that \(F\) is a subspace by showing that it is closed under addition and scalar multiplication

  2. Write \(F\) as the span of a set of vectors (hint: “bake” the subspace conditions into the vectors as we do in class for the zero-sum subspace).

  3. Find a basis for \(F\) from among your spanning vectors in the previous part. The previous part gives you spanning. Now you need to argue that the vectors are linearly independent.

  4. The dimension of a vector space is the number of vectors in a basis for the vector space. What is the dimension of \(F\)?

5.4 A Vector in Both Col(A) and Nul(A)

Give a \(3 \times 3\) matrix \(A\) for which the vector \(\mathsf{v} = \begin{bmatrix}3 \\ -2 \\ 5 \end{bmatrix}\) is in both \(\mathrm{Col}(A)\) and \(\mathrm{Nul}(A)\). Be sure to demonstrate that \(\mathsf{v} \in \mathrm{Col}(A)\) and \(\mathsf{v} \in \mathrm{Nul}(A)\).

5.5 Getting Into a Subspace

Let \(S \subseteq \mathbb{R}^n\) be a subspace and let \(\mathsf{v}, \mathsf{w} \in \mathbb{R}^n\). For each of the following statements, either give a specific example or explain why it cannot happen.

  1. If \(\mathsf{v}\) is not in \(S\) and \(c\) is a nonzero constant, can \(c\mathsf{v}\) be in \(S\)?

  2. If \(\mathsf{v}\) is not in \(S\) and \(\mathsf{w}\) is not in \(S\), can \(\mathsf{v} + \mathsf{w}\) be in \(S\)?

  3. If \(\mathsf{v}\) is in \(S\) and \(\mathsf{w}\) is not in \(S\) can \(\mathsf{v} + \mathsf{w}\) be in \(S\)?

5.6 Shrink to a basis

Let \(S\) be the subspace of \(\mathbb{R}^5\) spanned by the vectors below. Find a basis for \(S\) and find the dimension of \(S\) (the number of vectors in a basis).

\[ \mathsf{v}_1 = \begin{bmatrix} 1\\ 1\\ 1\\ 1\\ 3 \end{bmatrix}, \mathsf{v}_2 = \begin{bmatrix} 1\\ -1\\ 5\\ 1\\ 1 \end{bmatrix}, \mathsf{v}_3 = \begin{bmatrix} 1\\ 1\\ 1\\ 0\\ 2 \end{bmatrix}, \mathsf{v}_4 = \begin{bmatrix} 2\\ -1\\ 8\\ 1\\ 2 \end{bmatrix}, \mathsf{v}_5 = \begin{bmatrix} 1\\ 1\\ 1\\ 1\\ 3 \end{bmatrix}. \]

(A = cbind(c(1,1,1,1,3),c(1,-1,5,1,1),c(1,1,1,0,2),c(2,-1,8,1,2),c(1,1,1,1,3)))
##      [,1] [,2] [,3] [,4] [,5]
## [1,]    1    1    1    2    1
## [2,]    1   -1    1   -1    1
## [3,]    1    5    1    8    1
## [4,]    1    1    0    1    1
## [5,]    3    1    2    2    3

5.7 Extend to a basis

I am interested in the vectors below. I know that they do not span \(\mathbb{R}^5\), because there are not enough of them, but I want to extend this set to a basis of \(\mathbb{R}^5\) by adding some vectors to the set. \[ \begin{bmatrix} 5\\ 4\\ 3\\ 1\\ 2 \end{bmatrix}, \begin{bmatrix} 4\\ 4\\ 3\\ 1\\ 2 \end{bmatrix}, \begin{bmatrix} 1\\ 1\\ 1\\ 1\\ 1\end{bmatrix}. \] I searched online for ideas and one suggested that I make the matrix below and row reduce it.

(A = cbind(c(5,4,3,1,2),c(4,4,3,1,2),c(1,1,1,1,1),diag(5)))
##      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
## [1,]    5    4    1    1    0    0    0    0
## [2,]    4    4    1    0    1    0    0    0
## [3,]    3    3    1    0    0    1    0    0
## [4,]    1    1    1    0    0    0    1    0
## [5,]    2    2    1    0    0    0    0    1

  1. Row reduce this matrix.

  2. Use the result to come up with a basis for \(\mathbb{R}^5\) that includes my original 3 vectors

  3. Explain why this method works.