Linear Transformations

Study Sheet

Linear Transformations

Kernel, image, rotations and reflections

Maps That Respect Addition

Tip
The matrix IS the transformation

A linear map is determined by where it sends the basis vectors — those images are the COLUMNS of its matrix. Rotation by 90∘90^\circ: (0−110)\begin{pmatrix}0&-1\\1&0\end{pmatrix}; reflection across y=xy = x: (0110)\begin{pmatrix}0&1\\1&0\end{pmatrix}; composition of maps is a matrix product.

Tip
Kernel and image

The KERNEL is everything sent to zero (the solution space of Ax=0A\mathbf x = \mathbf 0); the IMAGE is everything hit (the span of the columns). Their dimensions add to the number of input variables — rank–nullity, now with geometric meaning.

Example
Determinants classify the geometry

A reflection has determinant −1-1 (flips orientation, preserves area); rotations have determinant +1+1; a projection has determinant 00 (it flattens). Two quarter-turns compose to R902=−IR_{90}^2 = -I: every point goes to its antipode.

Side note
Significance: geometry becomes computation

Every rotation of a 3D game character, every camera projection, every robot-arm pose is a linear (or affine) transformation composed from simpler ones by matrix multiplication. Kernel and image answer engineering questions directly: the kernel of a measurement matrix is what your sensors CANNOT see.

Try it
Try it: build a matrix from geometry

Write the matrix that stretches xx by 33 and reflects yy. Work: e1↦(3,0)\mathbf e_1 \mapsto (3, 0) and e2↦(0,−1)\mathbf e_2 \mapsto (0, -1): columns give (300−1)\begin{pmatrix}3&0\\0&-1\end{pmatrix}. Determinant −3-3: areas scale by 33, orientation flips — both facts visible before multiplying anything.

Proofs & Why It Matters

Tip
Proof: the columns determine the map

Any x=x1e1+⋯+xnen\mathbf x = x_1\mathbf e_1 + \cdots + x_n\mathbf e_n, so by linearity T(x)=x1T(e1)+⋯+xnT(en)T(\mathbf x) = x_1T(\mathbf e_1) + \cdots + x_nT(\mathbf e_n) — a combination of the images of the basis vectors with weights given by the coordinates. Assembling T(ej)T(\mathbf e_j) as columns of a matrix AA makes this exactly AxA\mathbf x. Linearity leaves no freedom beyond the columns. ■\blacksquare

Tip
Proof: reflections have determinant −1

A reflection fixes a line and flips the perpendicular direction. In a basis adapted to those two directions its matrix is (100−1)\begin{pmatrix}1&0\\0&-1\end{pmatrix}, with determinant −1-1; changing basis conjugates the matrix, and det⁡(P−1MP)=det⁡M\det(P^{-1}MP) = \det M, so the determinant is −1-1 in EVERY basis. Areas are preserved (∣−1∣=1|-1| = 1) but orientation reverses. ■\blacksquare

Going Deeper: Explanations & Worked Problems

Concept
A gallery of maps, with their matrices derived

Derive, never memorize: a map's matrix has columns T(e1),T(e2)T(\mathbf e_1), T(\mathbf e_2). ROTATION by θ\theta: e1=(1,0)\mathbf e_1 = (1,0) lands at (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta); e2=(0,1)\mathbf e_2 = (0,1) at (−sin⁡θ,cos⁡θ)(-\sin\theta, \cos\theta); hence Rθ=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)R_\theta = \begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}, det⁡=cos⁡2+sin⁡2=1\det = \cos^2 + \sin^2 = 1. REFLECTION across y=xy = x: the basis vectors swap, so (0110)\begin{pmatrix}0&1\\1&0\end{pmatrix}, det⁡=−1\det = -1. PROJECTION onto the xx-axis: e1\mathbf e_1 stays, e2\mathbf e_2 dies: (1000)\begin{pmatrix}1&0\\0&0\end{pmatrix}, det⁡=0\det = 0 — information is destroyed, and indeed the kernel is the whole yy-axis. SHEAR: (1k01)\begin{pmatrix}1&k\\0&1\end{pmatrix} slides horizontals by their height, det⁡=1\det = 1 (areas survive shearing). The determinant column of this gallery — 1,−1,0,11, -1, 0, 1 — is a complete orientation-and-area story at a glance.

Example
Worked: kernel and image of one map, completely

Let T(x,y,z)=(x+y,;y+z)T(x, y, z) = (x + y,; y + z), matrix (110011)\begin{pmatrix}1&1&0\\0&1&1\end{pmatrix}. KERNEL: solve x+y=0x + y = 0, y+z=0y + z = 0: x=−yx = -y, z=−yz = -y, yy free — the line spanned by (−1,1,−1)(-1, 1, -1) (check: T(−1,1,−1)=(0,0)T(-1,1,-1) = (0, 0) ✓). Dimension 11.

IMAGE: the span of the columns (1,0),(1,1),(0,1)(1,0), (1,1), (0,1); already the first two are independent, so the image is ALL of R2\mathbb R^2, dimension 22. RANK–NULLITY audit: 3=2+13 = 2 + 1 ✓. Interpretation: TT squashes three-dimensional space onto the plane, and the crushing happens exactly along the kernel line — every fiber T−1(b)T^{-1}(\mathbf b) is a translate of that line, which is the geometric meaning of "general solution = particular + homogeneous."

Example
Worked: composing transformations by multiplying matrices

What single map is "reflect across y=xy = x, THEN rotate 90∘90^\circ counterclockwise"?

Step 1 — matrices: reflection F=(0110)F = \begin{pmatrix}0&1\\1&0\end{pmatrix}, rotation R=(0−110)R = \begin{pmatrix}0&-1\\1&0\end{pmatrix}.

Step 2 — compose in the right ORDER: the reflection acts first, so the product is RFRF (closest to the input acts first): RF=(0−110)(0110)=(−1001)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 yy-axis.

Step 3 — the other order: FR=(100−1)FR = \begin{pmatrix}1&0\\0&-1\end{pmatrix}, reflection across the xx-axis: DIFFERENT, as expected.

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

Formulas, Proofs & Tips

Tip
Rigid motions in the plane
(x,y)→(x+a, y+b),(x,y)→(x,−y),(x,y)→(−y,x) (90∘ CCW)(x,y)\to(x+a,\,y+b),\quad (x,y)\to(x,-y),\quad (x,y)\to(-y,x)\ (90^\circ\text{ CCW})

What it means. Translation, reflection in the xx-axis, and rotation about the origin.

Example. Reflecting (3,2)(3,2) over the xx-axis gives (3,−2)(3,-2).

Why it works. Rotating 90∘90^\circ counter-clockwise sends the direction "right" to "up" and "up" to "left", i.e. (1,0)→(0,1)(1,0)\to(0,1) and (0,1)→(−1,0)(0,1)\to(-1,0); applying that to (x,y)(x,y) gives (−y,x)(-y,x).

Tip. Rigid motions preserve lengths and angles, so the image is congruent. Only dilations change size.