Due: October 23rd, 2017

Math 18 Assignment 3

Answers to these problems are now available here.

  1. Suppose that $T, S : \mathbb{R}^2 \to \mathbb{R}^3$ are maps given by \[T(x_1, x_2) = (x_1 + 2x_2 + 3, 2x_1 + 3x_2, x_1 - x_2) \qquad\text{ and }\qquad S(x_1, x_2) = (2x_1 - x_2, 2x_2 - x_1, 6x_1 -3x_2).\]
    1. Is $T$ linear? If it is, determine if it is one-to-one, onto, both, or neither.
    2. Is $S$ linear? If it is, determine if it is one-to-one, onto, both, or neither.
  2. Let $\vec{v_1}, \ldots, \vec{v_4},$ and $\vec{w}$ be the vectors defined as follows: \[ \vec{v}_1 = \begin{bmatrix}1\\0\\0\\1\\\end{bmatrix},\qquad \vec{v}_2 = \begin{bmatrix}0\\1\\0\\1\\\end{bmatrix},\qquad \vec{v}_3 = \begin{bmatrix}0\\0\\1\\1\\\end{bmatrix},\qquad \vec{v}_4 = \begin{bmatrix}0\\0\\0\\1\\\end{bmatrix},\qquad \text{ and }\qquad \vec{w} = \begin{bmatrix}1\\2\\1\\0\end{bmatrix}.\] Suppose that $T : \mathbb{R}^4 \to \mathbb{R}^3$ is a linear map such that: \[ \qquad T\left(\vec{v}_1\right) = \begin{bmatrix}2\\0\\0\\\end{bmatrix} \qquad T\left(\vec{v}_2\right) = \begin{bmatrix}1\\1\\1\\\end{bmatrix} \qquad T\left(\vec{v}_3\right) = \begin{bmatrix}-3\\-1\\4\\\end{bmatrix} \qquad T\left(\vec{v}_4\right) = \begin{bmatrix}0\\-2\\2\\\end{bmatrix} .\qquad\]
    1. Compute $T(\vec{w})$. (Hint: begin by expressing $\vec{w}$ as a linear combination of the four vectors given above.)
    2. Notice that $\vec{e}_1 = \vec{v}_1-\vec{v}_4$, $\vec{e}_2 = \vec{v}_2-\vec{v}_4$, $\vec{e}_3 = \vec{v}_3-\vec{v}_4$, and $\vec{e}_4 = \vec{v}_4$. Find the standard matrix for $T$. (Notice that if you multiply the vector $\vec{w}$ by the matrix which is your answer to this part, you should get the vector you found in the first part.)
  3. Let $v_1, \ldots, v_k \in \mathbb{R}^n$, and suppose that $w \in \mathrm{span}\{v_1, \ldots, v_k\}.$ Suppose further that $T : \mathbb{R}^n \to \mathbb{R}^m$ is linear. Show that $T(w) \in \mathrm{span}\{T(v_1), \ldots, T(v_k)\}$.
  4. Let $A$ and $B$ be the following matrices: \[A = \begin{bmatrix} 1&1&-2\\ -2&4&-2\\ -2&1&1\\ \end{bmatrix} \qquad\text{ and }\qquad B = \begin{bmatrix} 1&1&1\\ 1&1&1\\ 1&1&1\\ \end{bmatrix}.\]
    1. Compute $AB$ and $BA$.
    2. Let $T_{AB}$ and $T_{BA}$ be the linear maps from $\mathbb{R}^3$ to $\mathbb{R}^3$ given by multiplication by $AB$ and $BA$ respectively (i.e., $T_{AB}(\vec{x}) = AB\vec{x},$ and $T_{BA}(\vec{x}) = BA\vec{x}$). Does $\mathrm{image}(T_{AB}) = \mathrm{image}(T_{BA})$? (Hint: if you computed $AB$ and $BA$ correctly, answering this question should require very little computation.)
  5. Consider the following matrices: \[A = \begin{bmatrix} \frac{1}{2}&1&\frac{3}{2}\\2&4&-1\\0&4&-4 \end{bmatrix}, \qquad R_1 = \begin{bmatrix} 2&0&0\\0&1&0\\0&0&1 \end{bmatrix}, \qquad R_2 = \begin{bmatrix} 1&0&0\\-2&1&0\\0&0&1 \end{bmatrix}, \qquad R_3 = \begin{bmatrix} 1&0&0\\0&0&1\\0&1&0 \end{bmatrix},\] \[ R_5 = \begin{bmatrix} 1&0&0\\0&\frac{1}{4}&0\\0&0&-\frac{1}{7} \end{bmatrix}, \qquad R_6 = \begin{bmatrix} 1&0&0\\0&1&1\\0&0&1 \end{bmatrix}, \qquad\text{ and }\qquad R_7 = \begin{bmatrix} 1&-2&-3\\0&1&0\\0&0&1 \end{bmatrix}, \qquad \] Compute the following matrices: $$\begin{align*}R_1A,\\ R_2R_1A,\\ R_3R_2R_1A,\\ R_5R_3R_2R_1A,\\ R_6R_5R_3R_2R_1A,\\ R_7R_6R_5R_3R_2R_1A.\end{align*}$$ Notice that each matrix is the product of one more matrix with the previous one, so to compute $R_5R_3R_2R_1A$, for example, it is easiest to multiply $R_5\cdot(R_3R_2R_1A)$. It turns out that row reduction can always be performed by multiplying by a sequence of "row operation matrices", like those above.

    Correction: An earlier version of this problem accidentally repeated the matrix $\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}$ twice. Since the goal of this problem is to demonstrate row reduction as a sequence of matrix multiplications, its presence there defeated the point of the question. If you have already performed the computation with the original sequence of matrices, submitting it will also count as a correct solution to this problem.