Problem Set 7: Injective, Surjective, and Invertible

For each matrix $A$ below, decide whether $T_A$ is injective, surjective, and/or bijective. Justify your answers. You may use the Row Reducer to help you calculate.

1. $A = \begin{bmatrix} 1 & 1 \\ 2 & 3 \\ 3 & 4 \end{bmatrix}$
2. $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 1 & 1 \end{bmatrix}$
3. $A = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 5 \end{bmatrix}$
4. $A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}$

A tiny camera takes black-and-white photos that are a single row of four pixels, with brightness values $x_1, x_2, x_3, x_4$. The photographer's hands shake, so instead of the sharp image $x = (x_1, x_2, x_3, x_4)$ the camera records a blurred image $b = (b_1, b_2, b_3)$ in which each pixel is the average of two neighboring sharp pixels: \[ b_1 = \tfrac{1}{2}(x_1 + x_2),\qquad b_2 = \tfrac{1}{2}(x_2 + x_3),\qquad b_3 = \tfrac{1}{2}(x_3 + x_4). \] In other words, $b = Bx$ where \[ B = \frac{1}{2} \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}. \]

1. Explain why $T_B$ is not injective by finding a sharp image other than $(4, 4, 4, 4)$ that blurs to $b = (4, 4, 4)$.
2. Is every vector $b$ in $\RR^3$ the blurred image of some sharp image? Explain.
3. The camera is upgraded so that it also records the first pixel $x_1$ sharply. The recorded data is now $y = (x_1, b_1, b_2, b_3) = Cx$, where \[ C = \frac{1}{2} \begin{bmatrix} 2 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}. \] Explain why $C$ is invertible.
4. Find $C^{-1}$.
5. The upgraded camera records $y = (2, 3, 4, 5)$. Recover the sharp image.

For each statement below, decide whether it is true or false. If it is true, explain why. If it is false, give an example of a matrix $A$ for which it fails.

1. If $A$ is a $3 \times 5$ matrix, then $T_A$ is not injective.
2. If $A$ is a $3 \times 5$ matrix and $T_A$ is surjective, then the solution set of $Ax = 0$ is a plane through the origin.
3. If $A$ is a $2 \times 3$ matrix, then $T_A$ is surjective.
4. If $A$ is a $2 \times 2$ matrix with $A^{-1} = A$, then $A = I_2$ or $A = -I_2$.
5. If $A$ is a $4 \times 4$ matrix and there are two different vectors $x \neq y$ in $\RR^4$ with $Ax = Ay$, then $T_A$ is not surjective.
6. If the columns of $A$ are linearly independent, then $A$ is invertible.

Let \[ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 6 \end{bmatrix}. \] You may use the Row Reducer to help you calculate.

1. Find $A^{-1}$.
2. Use $A^{-1}$ to solve $Ax = b$ for \[ b = \begin{bmatrix} 2 \\ 3 \\ 5 \end{bmatrix} \qquad\text{and for}\qquad b = \begin{bmatrix} 3 \\ 6 \\ 10 \end{bmatrix}. \]
3. Explain why the matrix \[ B = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ 3 & 7 & 10 \end{bmatrix} \] is not invertible.

The face on the left sits on the parallelogram with corners at the origin, $u_1 = (2, 1)$, $u_2 = (1, 2)$, and $u_1 + u_2 = (3, 3)$. We want to find the matrix $A$ such that $T_A$ maps the face on the left to the face on the right. Let $w_1 = T_A(u_1)$ and $w_2 = T_A(u_2)$.

1. Define $U = \begin{bmatrix} u_1 & u_2 \end{bmatrix}$ and $W = \begin{bmatrix} w_1 & w_2 \end{bmatrix}$ be the $2 \times 2$ matrices with these vectors as columns. Explain why $AU = W$.
2. Use the equation $AU = W$ to calculate $A$.
   Hint: the $2\times 2$ inverse formula.