Corrected exercises: linear transformations and their matrices (MATH 133)

a) Take $u = (x_1, y_1)$, $v = (x_2, y_2)$ and a scalar $c$. Then $T(u + v) = (2(x_1 + x_2) - (y_1 + y_2),\ (x_1 + x_2) + 3(y_1 + y_2),\ -(x_1 + x_2))$, and regrouping each coordinate gives $(2x_1 - y_1, x_1 + 3y_1, -x_1) + (2x_2 - y_2, x_2 + 3y_2, -x_2) = T(u) + T(v)$. Likewise $T(cu) = (2cx_1 - cy_1,\ cx_1 + 3cy_1,\ -cx_1) = c\,T(u)$. Both identities hold for ALL inputs, so $T$ is linear. The proof works because every coordinate is a combination of $x$ and $y$ with no constant term, no product and no power: that is exactly what the chapter's theorem will turn into $T(x) = Ax$ with $A = \begin{pmatrix} 2 & -1 \\ 1 & 3 \\ -1 & 0 \end{pmatrix}$. Checking the identities on one or two particular vectors proves nothing: a proof of linearity must use letters.

b) The square has been slid away from the origin: its corner at $(0, 0)$ now sits at $(1, -2)$. A linear map can stretch, turn, flip or crush the square, but it can never move the origin, so the picture already says no. The smallest witness is the zero vector itself: $S(0, 0) = (1, -2) \neq (0, 0)$. If you want the additivity rule to fail visibly as well, $S(1, 0) + S(0, 1) = (2, -2) + (1, -1) = (3, -3)$ while $S(1, 1) = (2, -1)$. A translation is the textbook example of a map that is geometrically simple and algebraically NOT linear: the constant $+1$ and $-2$ are what break it.

c) $Q$ passes the origin test, and that proves nothing. Try the scaling rule with $c = 2$ on $e_1$: $Q(2, 0) = (4, 0)$, but $2\,Q(1, 0) = 2(1, 0) = (2, 0)$. The two differ, so $Q$ is not linear. The square is the culprit: doubling the input quadruples the first output. This is the trap the exercise is built around, and it costs the whole part on a midterm: $T(0) = 0$ is a NECESSARY condition, never a sufficient one. When the origin test passes you still owe either a proof with letters or a counterexample with numbers.

d) Using homogeneity with $c = 0$: $T(0) = T(0 \cdot 0) = 0 \cdot T(0) = 0$. (Additivity gives it too: $T(0) = T(0 + 0) = T(0) + T(0)$, then subtract $T(0)$.) So a map with $T(0) \neq 0$ is not linear, which decides b) in one line. It cannot decide c), because $Q(0) = 0$: the test only ever KILLS a candidate, it never saves one. On an exam it is the first thing to try, because it costs five seconds and settles every translation.

e) No. $f(0) = 2 \neq 0$, and additivity fails too: $f(1) + f(1) = 10$ while $f(2) = 8$. What high school calls linear, a straight-line graph, linear algebra calls AFFINE: a linear map plus a translation. The linear transformations $\mathbb{R} \to \mathbb{R}$ are exactly $f(x) = ax$, the lines through the origin, and their matrix is the $1 \times 1$ matrix $(a)$, whose single column is $f(1) = a$, the image of the one standard basis vector. Using the high-school meaning of the word in a proof is a vocabulary error that markers penalize every term.

a) Substitute each standard basis vector. $T(e_1) = T(1, 0, 0) = (1, 3)$, $T(e_2) = T(0, 1, 0) = (-2, 0)$, $T(e_3) = T(0, 0, 1) = (4, 1)$. Stand them up as COLUMNS: $A = \begin{pmatrix} 1 & -2 & 4 \\ 3 & 0 & 1 \end{pmatrix}$. The size is $2 \times 3$: as many rows as the dimension of the CODOMAIN, as many columns as the dimension of the DOMAIN, because there is one column per input basis vector. A $3 \times 2$ answer is the transposed matrix, the classic slip of writing the images as rows, and it cannot even be multiplied by a vector of $\mathbb{R}^3$. Notice that the rows of $A$ are the coefficient lists of the two output formulas: that is a useful cross-check, never the definition.

b) With the formula: $T(2, -1, 1) = (2 + 2 + 4,\ 6 + 0 + 1) = (8, 7)$. As a combination of columns: $Ax = 2\begin{pmatrix} 1 \\ 3 \end{pmatrix} - 1\begin{pmatrix} -2 \\ 0 \end{pmatrix} + 1\begin{pmatrix} 4 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 + 2 + 4 \\ 6 - 0 + 1 \end{pmatrix} = \begin{pmatrix} 8 \\ 7 \end{pmatrix}$. The second reading is the one to keep: $Ax$ is the combination of the columns of $A$ with the entries of $x$ as weights, so the image of $T$ is exactly the span of the columns, a fact exercise 10 will use.

c) Write $x = (x_1, \dots, x_n) = x_1 e_1 + x_2 e_2 + \cdots + x_n e_n$. Linearity lets $T$ pass through the sum and the scalars: $T(x) = x_1 T(e_1) + x_2 T(e_2) + \cdots + x_n T(e_n)$. The right-hand side is precisely the product of the matrix whose columns are $T(e_1), \dots, T(e_n)$ with the vector $x$. So $T(x) = Ax$ for every $x$. Two consequences worth a sentence on any exam: a linear map is COMPLETELY decided by $n$ vectors, the images of the basis; and two linear maps that agree on $e_1, \dots, e_n$ agree everywhere.

d) $B$ is $3 \times 2$, so it multiplies vectors of $\mathbb{R}^2$ and returns vectors of $\mathbb{R}^3$: $T_B : \mathbb{R}^2 \to \mathbb{R}^3$. $T_B(e_2)$ is the SECOND COLUMN, $(-1, 3, 1)$. The trap is to read the second row, $(0, 3)$, which is not even a vector of the codomain. Then $T_B(4, -3) = 4(2, 0, 1) - 3(-1, 3, 1) = (8 + 3,\ 0 - 9,\ 4 - 3) = (11, -9, 1)$. Reading a matrix as a list of images is the same gesture as building it, run backwards.

e) Linearity alone suffices: $U(4, -2) = U(4e_1 - 2e_2) = 4U(e_1) - 2U(e_2) = 4(2, 5) - 2(-1, 3) = (8 + 2,\ 20 - 6) = (10, 14)$. The standard matrix is $\begin{pmatrix} 2 & -1 \\ 5 & 3 \end{pmatrix}$, its columns being the two given images, and its rows give the formula $U(x, y) = (2x - y,\ 5x + 3y)$. Check: $U(4, -2) = (8 + 2, 20 - 6) = (10, 14)$, the same answer. Two data vectors decided the whole map, exactly as part c) promised.

a) $e_1$ is a unit vector along the $x$-axis; turned by $\theta$ it stays on the unit circle, at angle $\theta$, so $T(e_1) = (\cos\theta, \sin\theta)$. $e_2$ sits at angle $\pi/2$ and lands at angle $\theta + \pi/2$: $T(e_2) = (\cos(\theta + \pi/2), \sin(\theta + \pi/2)) = (-\sin\theta, \cos\theta)$. Columns: $R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$. The minus sign sits in the FIRST row because $T(e_2)$ points to the left of the $y$-axis for small positive $\theta$, as the figure shows. Why is a rotation linear at all? Turning the parallelogram built on $u$ and $v$ turns its diagonal $u + v$ with it, and turning a vector twice as long gives a vector twice as long: the two rules of the definition, read on the picture.

b) $\cos(2\pi/3) = -1/2$ and $\sin(2\pi/3) = \sqrt{3}/2$, so $R = \begin{pmatrix} -1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & -1/2 \end{pmatrix}$. Then $R(\sqrt{3}, 1) = (-\sqrt{3}/2 - \sqrt{3}/2,\ 3/2 - 1/2) = (-\sqrt{3}, 1)$. Geometric check: $(\sqrt{3}, 1)$ has length $2$ and angle $\pi/6$; after a turn of $2\pi/3$ it must have length $2$ and angle $5\pi/6$, which is $(2\cos(5\pi/6), 2\sin(5\pi/6)) = (-\sqrt{3}, 1)$. Both agree. A rotation must preserve lengths: if your image of a vector of length $2$ has another length, a sign is wrong.

c) A point at height $y$ moves by $3y$ horizontally: $(x, y) \mapsto (x + 3y, y)$. So $e_1$, at height $0$, does not move: $T(e_1) = (1, 0)$; $e_2$, at height $1$, moves to $(3, 1)$. Columns: $\begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}$. The trap is to put the $3$ below the diagonal, $\begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}$, which is the VERTICAL shear: it sends $e_1$ to $(1, 3)$ and moves points up in proportion to their abscissa. Asking where $e_1$ goes settles which one you wrote in two seconds.

d) The stretch sends $e_1$ to $(2, 0)$ and $e_2$ to $(0, 1/3)$: $\begin{pmatrix} 2 & 0 \\ 0 & 1/3 \end{pmatrix}$, and $(3, 6) \mapsto (6, 2)$. The dilation by $5$ sends every vector to $5$ times itself, so $e_1 \mapsto (5, 0)$, $e_2 \mapsto (0, 5)$ and the matrix is $5I = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix}$, not the $1 \times 1$ number $5$ and not a matrix full of fives. Diagonal matrices are exactly the maps that act on each axis separately, which is why their columns are multiples of $e_1$ and $e_2$.

e) A rotation matrix has columns of length $1$, perpendicular to each other, and determinant $+1$. For $\frac{1}{5}\begin{pmatrix} 3 & -4 \\ 4 & 3 \end{pmatrix}$: $(3^2 + 4^2)/25 = 1$ for each column, $3 \cdot (-4) + 4 \cdot 3 = 0$, and $\det = (9 + 16)/25 = 1$. It is the rotation by the angle $\theta$ with $\cos\theta = 3/5$ and $\sin\theta = 4/5$, about $0.93$ radian. For $\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$ the columns are perpendicular but of length $\sqrt{2}$ and $\det = 2$: not a rotation. Its first column $(1, -1)$ points at angle $-\pi/4$, so it is the rotation by $-\pi/4$ followed by the dilation by $\sqrt{2}$: $\sqrt{2}\,R_{-\pi/4} = \sqrt{2}\begin{pmatrix} \sqrt{2}/2 & \sqrt{2}/2 \\ -\sqrt{2}/2 & \sqrt{2}/2 \end{pmatrix}$, which is the given matrix.

a) $d \cdot d = 1 + 4 = 5$. For $e_1$: $e_1 \cdot d = 1$, so $P(e_1) = \frac{1}{5}(1, 2) = (1/5, 2/5)$, the green point of the figure. For $e_2$: $e_2 \cdot d = 2$, so $P(e_2) = \frac{2}{5}(1, 2) = (2/5, 4/5)$. Columns: $P = \frac{1}{5}\begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$. Check on a vector you know: $d$ itself lies on $L$ and must not move, and indeed $P(1, 2) = \frac{1}{5}(1 + 4,\ 2 + 8) = (1, 2)$. A vector perpendicular to $L$, such as $(2, -1)$, must be crushed to $0$: $\frac{1}{5}(2 - 2,\ 4 - 4) = (0, 0)$. The projection formula is from the dot product chapter; the new step here is only to apply it to $e_1$ and $e_2$ and stand the results up as columns.

b) On the figure $P(e_1)$ is the MIDPOINT of the segment from $e_1$ to $F(e_1)$, since the mirror image is on the perpendicular, at the same distance on the other side. The midpoint of $v$ and $F(v)$ is $\frac{1}{2}(v + F(v))$, so $P(v) = \frac{1}{2}(v + F(v))$, that is $F(v) = 2P(v) - v$. In matrices, $F = 2P - I = \frac{1}{5}\begin{pmatrix} 2 & 4 \\ 4 & 8 \end{pmatrix} - \frac{1}{5}\begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} = \frac{1}{5}\begin{pmatrix} -3 & 4 \\ 4 & 3 \end{pmatrix}$. So $F(e_1) = (-3/5, 4/5)$, the orange arrow: it has length $1$, as a reflection of a unit vector must.

c) With $d = (1, m)$ and $d \cdot d = 1 + m^2$: $P(e_1) = \frac{1}{1+m^2}(1, m)$ and $P(e_2) = \frac{m}{1+m^2}(1, m)$, so $P = \frac{1}{1+m^2}\begin{pmatrix} 1 & m \\ m & m^2 \end{pmatrix}$. Then $F = 2P - I$ has entries $\frac{2 - (1 + m^2)}{1+m^2} = \frac{1 - m^2}{1+m^2}$, $\frac{2m}{1+m^2}$ twice, and $\frac{2m^2 - (1 + m^2)}{1+m^2} = \frac{m^2 - 1}{1+m^2}$, which is the formula. For $m = 0$, the $x$-axis: $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$, which keeps $x$ and flips $y$. For $m = 1$, the line $y = x$: $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, which swaps the coordinates. Forgetting the factor $\frac{1}{1+m^2}$ is the classic loss: the columns then have length greater than $1$ and the map stretches the plane, which no mirror does.

d) $F(5, 0) = \frac{1}{5}(-15, 20) = (-3, 4)$. First check: a reflection preserves lengths, and $|(5, 0)| = 5 = \sqrt{9 + 16} = |(-3, 4)|$. Second check: the midpoint of $(5, 0)$ and $(-3, 4)$ is $(1, 2)$, which lies on $L$ since $2 = 2 \times 1$, and equals $P(5, 0) = \frac{1}{5}(5, 10) = (1, 2)$. Both facts come from the figure, not from the formula, which is what makes them real checks. On an exam, one such line catches a sign error that would otherwise cost the whole part.

e) Read the images directly. The reflection across the $y$-axis sends $e_1$ to $(-1, 0)$ and leaves $e_2$ in place: $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$. The projection onto the $y$-axis crushes $e_1$ to $0$ and keeps $e_2$: $\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$. The general formula agrees in the limit: as $m \to \infty$, $\frac{1 - m^2}{1 + m^2} \to -1$, $\frac{2m}{1 + m^2} \to 0$ and $\frac{m^2 - 1}{1 + m^2} \to 1$. The lesson is that a formula is a convenience; the images of $e_1$ and $e_2$ are the method, and they never fail on a vertical line.

a) The figure shows $v_1 + v_2 = (2, 0) = 2e_1$, so $e_1 = \frac{1}{2}(v_1 + v_2)$; similarly $v_1 - v_2 = (0, 2)$ gives $e_2 = \frac{1}{2}(v_1 - v_2)$. By linearity, $T(e_1) = \frac{1}{2}\big((3, 1) + (1, 5)\big) = (2, 3)$ and $T(e_2) = \frac{1}{2}\big((3, 1) - (1, 5)\big) = (1, -2)$. So $A = \begin{pmatrix} 2 & 1 \\ 3 & -2 \end{pmatrix}$. Check: $A(1, 1) = (3, 1)$ and $A(1, -1) = (1, 5)$. The trap of the exercise is to write $\begin{pmatrix} 3 & 1 \\ 1 & 5 \end{pmatrix}$, the given images used as columns: that matrix sends $e_1$ to $(3, 1)$, not $v_1$, and it fails the check at once. The columns are ALWAYS the images of $e_1$ and $e_2$, whatever vectors the statement happens to use.

b) $Av_1 = T(v_1)$ and $Av_2 = T(v_2)$ say, column by column, that $AV = W$. The columns of $V$ are independent, so $V$ is invertible and $A = WV^{-1}$. Here $V = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ has $\det V = -2$ and $V^{-1} = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$, so $WV^{-1} = \frac{1}{2}\begin{pmatrix} 3 & 1 \\ 1 & 5 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 4 & 2 \\ 6 & -4 \end{pmatrix}$, which is $A$. The order is forced: $V^{-1}$ on the RIGHT, because $V$ multiplies $A$ on the right. $V^{-1}W$ is a different matrix and a wrong answer.