algebra

Linear Operators and Matrices

Here is the truth that finally made matrices make sense to me: a matrix is not a box of numbers. It is a photograph of a machine, taken from one particular angle.

The machine is the linear operator. The angle is your choice of basis. Change the angle and the photo changes, but the machine doesn’t.

The core idea, in plain words

A linear map $T : V \to W$ between Vector Spaces is a function that respects adding and scaling:

T(u + v) = T(u) + T(v), \qquad T(c\,v) = c\,T(v).

When $V = W$ , so the machine feeds its output back into the same space, it’s called a linear operator.

Here’s the key fact. Because every vector is a combination of basis vectors, a linear map is completely determined by what it does to a basis. Choose a basis $e_1, \dots, e_n$ . Write $T(e_1)$ in coordinates and make it the first column; write $T(e_2)$ and make it the second column; and so on. That table is the matrix of $T$ .

Matrix multiplication isn’t an arbitrary rule handed down from on high. It is exactly what you get when you run one machine after another: the matrix of “ $S$ after $T$ ” is $ST$ . And since “rotate then flip” differs from “flip then rotate”, matrix multiplication is not commutative. Square matrices form a Ring, and the invertible ones form a Group.

An everyday analogy: currency exchange

Think of a money changer converting a basket of currencies. Double the basket, double what you get back; combine two baskets, get the sum of what each would give. That’s linear. The exchange-rate board is the matrix: each column says what one unit of a currency turns into. Now describe the same exchange using different currencies as your units and the board’s numbers all change, yet the actual trade is the same. That’s a change of basis.

A worked example: a reflection, seen two ways

Let $T$ reflect the plane across the line $y = x$ . It sends $(1, 0)$ to $(0, 1)$ and $(0, 1)$ to $(1, 0)$ . Those images become the columns:

A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.

Check: $A \begin{pmatrix} 3 \\ 5 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$ . The point swapped sides of the mirror. ✔

Now look at the same reflection from a cleverer angle. Use the basis $f_1 = (1, 1)$ , which lies on the mirror, and $f_2 = (1, -1)$ , which is perpendicular to it. The reflection leaves $f_1$ alone and flips $f_2$ : $T(f_1) = f_1$ , $T(f_2) = -f_2$ . In this basis the matrix is

D = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.

Same machine, much simpler photograph. The directions $f_1$ and $f_2$ are eigenvectors: $T$ only stretches them, by the eigenvalues $1$ and $-1$ . Finding such a basis is called diagonalising, and it’s the single most useful trick in linear algebra. The two photos are related by $D = P^{-1} A P$ , where $P$ has $f_1, f_2$ as its columns.

Notice the things that didn’t change: the determinant ( $-1$ both times, meaning area is kept but orientation flipped) and the trace ( $0$ both times). Quantities that survive a change of basis belong to the machine, not to the photograph.

A picture in words: kernel and image

Picture a machine that projects 3D space flat onto the floor, like the sun casting shadows at noon. Every point lands on the floor: that floor is the image of the machine. Every point straight above the origin, the whole vertical line, lands on the origin: that line is the kernel, everything the machine crushes to zero.

The rank–nullity theorem says the dimensions always balance:

\dim(\ker T) + \dim(\operatorname{im} T) = \dim V.

Here, $1 + 2 = 3$ . What’s crushed plus what survives is everything you started with.

graph LR
  T["Linear map T"] -->|choose a basis| A["Matrix A"]
  T -->|choose a better basis| D["Simpler matrix P⁻¹AP"]
  A -->|same determinant, trace, eigenvalues| D

Common misconceptions

“A matrix and a linear map are the same thing.” A matrix is the map in coordinates. Different bases, different matrices, same map.
" $AB = BA$ ." Almost never. Rotate 90° then reflect is not reflect then rotate 90°.
“Every matrix can be diagonalised.” No. The shear $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ has only one line of eigenvectors. Jordan normal form is the best you can do.
“A zero determinant means the matrix is zero.” It means the machine squashes space into a lower dimension and has a non-zero kernel, so it can’t be undone.

How it connects

Down to Vector Spaces, Rings (square matrices form one) and Groups (invertible matrices form one).
Sideways to Modules: an operator turns its vector space into a module over the polynomial ring, and that’s where Jordan form comes from.
Up to Inner Product Spaces, where special operators (symmetric, orthogonal, unitary) behave beautifully, and to Tensors and Exterior Algebras, where the determinant gets its true meaning.
Out to Quantum Mechanics: every measurable quantity is a linear operator, and the possible measurement results are its eigenvalues.

The truth revealed: don’t stare at the numbers. Find the angle from which the machine looks simple.