Linear Transformations

A map $T$ is linear when $T(\mathbf u + \mathbf v) = T\mathbf u + T\mathbf v$ and $T(c\mathbf u) = cT\mathbf u$. Every $\mathbf x = x\mathbf e_1 + y\mathbf e_2$, so linearity forces $T\mathbf x = xT\mathbf e_1 + yT\mathbf e_2$ — the whole map is fixed by two vectors. For the rotation, $(3, 1) = 3\mathbf e_1 + \mathbf e_2$ goes to $3(0, 1) + (-1, 0) = (-1, 3)$, and indeed $(-1, 3)$ is $(3, 1)$ turned a quarter-turn.

Both are subspaces, and they measure opposite things: the kernel is what the map destroys, the image is what it can produce. If $T\mathbf u = T\mathbf w$, then $T(\mathbf u - \mathbf w) = \mathbf 0$, so $\mathbf u - \mathbf w$ lies in the kernel; a map is one-to-one exactly when its kernel is $\{\mathbf 0\}$. Every input dimension either survives into the image or is lost in the kernel, which is why the two dimensions add to the number of inputs.

Each claim is a determinant you can compute. A rotation matrix $\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$ has $\det = \cos^2\theta + \sin^2\theta = 1$; reflection across $y = x$ has $\det = 0 - 1 = -1$; projection onto the $x$-axis, $\begin{pmatrix}1&0\\0&0\end{pmatrix}$, has $\det = 0$. And $R_{90}^2 = \begin{pmatrix}0&-1\\1&0\end{pmatrix}^2 = -I$ with $\det(-I) = 1$: a half-turn keeps both area and orientation.

Split $\mathbf x$ into a piece along the line and a piece perpendicular to it. The along-piece is $\dfrac{\mathbf v\cdot\mathbf x}{\mathbf v\cdot\mathbf v}\mathbf v$, and $\mathbf v(\mathbf v^{\mathsf T}\mathbf x) = (\mathbf v\mathbf v^{\mathsf T})\mathbf x$ turns that into a matrix times $\mathbf x$. Reflection keeps the along-piece and negates the other: $F\mathbf x = P\mathbf x - (\mathbf x - P\mathbf x) = (2P - I)\mathbf x$. Two checks follow from the geometry: projecting twice changes nothing, so $P^2 = P$; reflecting twice returns home, so $F^2 = I$.

Apply it: $P\begin{pmatrix}3\\1\end{pmatrix} = \tfrac15\begin{pmatrix}3 + 2\\6 + 4\end{pmatrix} = \begin{pmatrix}1\\2\end{pmatrix}$. Two checks: $(1, 2)$ is on the line, and the leftover $(3, 1) - (1, 2) = (2, -1)$ is perpendicular to it, since $(2, -1)\cdot(1, 2) = 0$. The projection is the foot of the perpendicular from $(3, 1)$. Also $\det P = \tfrac{1}{25}(4 - 4) = 0$: projection flattens the plane onto a line, as it must.

Then $F\begin{pmatrix}3\\1\end{pmatrix} = \tfrac15\begin{pmatrix}-9 + 4\\12 + 3\end{pmatrix} = \begin{pmatrix}-1\\3\end{pmatrix}$. Check the geometry: the midpoint of $(3, 1)$ and $(-1, 3)$ is $(1, 2)$, the foot found above, and both points are $\sqrt{10}$ from the origin. The determinant is $\tfrac{1}{25}(-9 - 16) = -1$, the signature of a reflection. And $F^2 = \tfrac{1}{25}\begin{pmatrix}9 + 16&-12 + 12\\-12 + 12&16 + 9\end{pmatrix} = I$ ✓.

The tests say the map respects the two operations that define a vector space, which is what lets a matrix represent it. Geometrically, a linear map carries grid lines to grid lines: straight, parallel, and evenly spaced. $T(\mathbf 0) = \mathbf 0$ follows from scaling with $c = 0$: $T(\mathbf 0) = T(0\mathbf u) = 0\,T(\mathbf u) = \mathbf 0$. It is necessary but not sufficient — a map can fix the origin and still bend the grid.

So $T$ is not linear, and no matrix can represent it (a matrix sends $\mathbf 0$ to $\mathbf 0$). The additivity test fails too, for a concrete pair: $T((1, 0) + (1, 0)) = T(2, 0) = (3, 0)$, while $T(1, 0) + T(1, 0) = (2, 0) + (2, 0) = (4, 0)$. A translation is affine — linear plus a constant — which is why computer graphics adds an extra coordinate to turn translations into matrix products.

But $2T(\mathbf u) = 2(1, 1) = (2, 2) \ne (2, 4)$. Scaling fails, so $T$ is not linear. The culprit is the square: doubling $x$ quadruples $x^2$. In the figure, the horizontal grid lines $y = -1, 0, 1$ are bent into parabolas — a linear map could never do that.

Their coordinate vectors $(0, 0, 0)$, $(1, 0, 0)$, $(0, 2, 0)$ are the columns: $D = \begin{pmatrix}0&1&0\\0&0&2\\0&0&0\end{pmatrix}$. Test on $p = 3 + 2x + 5x^2$, coordinates $(3, 2, 5)$: $D\begin{pmatrix}3\\2\\5\end{pmatrix} = \begin{pmatrix}2\\10\\0\end{pmatrix}$, which is $2 + 10x$ — exactly $p'$ ✓. The kernel is the constants (dimension $1$) and the image is the polynomials of degree at most $1$ (dimension $2$): $1 + 2 = 3$ ✓. "Matrix" and "linear map" are the same idea even when the vectors are functions.

Why the two columns are enough: every vector is $x\mathbf e_1 + y\mathbf e_2$, so linearity forces $T(x, y) = x\,T(\mathbf e_1) + y\,T(\mathbf e_2)$ — which is exactly the matrix times $(x, y)$. Once you know where the two basis vectors land, nothing about the map is left to choose. Test it on the rotation with $\theta = 90^\circ$: the columns are $(0, 1)$ and $(-1, 0)$, so $(2, 1)$ goes to $2(0, 1) + 1(-1, 0) = (-1, 2)$ — the same length $\sqrt5$, and perpendicular to $(2, 1)$ because $2\cdot(-1) + 1\cdot 2 = 0$, just as a quarter turn should be.

Image: the span of the columns $(1,0), (1,1), (0,1)$; already the first two are independent, so the image is all of $\mathbb R^2$, dimension $2$. Rank–nullity audit: $3 = 2 + 1$ ✓. Interpretation: $T$ squashes three-dimensional space onto the plane, and the crushing happens exactly along the kernel line — every fiber $T^{-1}(\mathbf b)$ is a translate of that line, which is the geometric meaning of "general solution = particular + homogeneous."

Step 1 — matrices: reflection $F = \begin{pmatrix}0&1\\1&0\end{pmatrix}$, rotation $R = \begin{pmatrix}0&-1\\1&0\end{pmatrix}$.

Step 2 — compose in the right order: the reflection acts first, so the product is $RF$ (closest to the input acts first): $RF = \begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix} = \begin{pmatrix}-1&0\\0&1\end{pmatrix}$ — reflection across the $y$-axis.

Step 3 — the other order: $FR = \begin{pmatrix}1&0\\0&-1\end{pmatrix}$, reflection across the $x$-axis: different, as expected.

Step 4 — determinant ledger: $\det R\cdot\det F = (1)(-1) = -1 = \det(RF)$ ✓ — a rotation composed with a reflection is always another reflection.