The Matrix of a Linear Map

The passage from linear map to matrix is the computational backbone of essentially all applied linear algebra. In numerical analysis, every algorithm — Gaussian elimination, QR decomposition, the singular value decomposition — operates on the matrix of a map relative to convenient bases, and choosing good bases (orthonormal ones, eigenbases) is often the entire strategy of the algorithm. In computer graphics, a 3D transformation (rotation, scaling, projection) is stored and applied to vertices as exactly the matrix constructed here, relative to the standard basis of $\mathbb{R}^3$ or $\mathbb{R}^4$ in homogeneous coordinates. In quantum mechanics, an observable is a linear (self-adjoint) operator on a Hilbert space, and physicists routinely write "the matrix of the operator in the energy basis" — a direct instance of this construction, with the choice of basis (energy eigenstates, position, momentum) chosen to make the matrix as simple as possible. In statistics and data science, a linear regression model or a principal component transformation is literally the matrix of a linear map between a feature space and a response space, computed relative to the standard coordinate basis of the data. In cryptography, linear codes and certain cipher components are described by matrices over finite fields, which are exactly matrices of linear maps between finite-dimensional vector spaces over $\mathbb{F}_2$ or larger finite fields. Finally, in the theoretical development of linear algebra itself, this construction is the hinge on which Matrix Multiplication as Composition and Change of Basis and the Change-of-Basis Matrix turn: composition of maps becomes matrix multiplication, and re-expressing a map in new coordinates becomes a similarity or equivalence transformation of its matrix.

Ask the tutor: How does the matrix of $T$ change if I replace the basis $\beta$ of $V$ with a new basis $\beta'$, keeping $\gamma$ fixed?

Think of a linear map $T$ as a machine that transforms vectors, and think of a basis as a fixed set of measuring sticks used to read off coordinates. Once measuring sticks are chosen for both the input space $V$ and the output space $W$, the only information you need to reconstruct the machine completely is: what happens to each measuring stick? This is precisely the content of Linear Maps Determined by Action on a Basis — a linear map is pinned down by its values on a basis, and those values, once expressed in coordinates via the output basis, are numbers you can write down. The key design choice is to store the image of the $j$-th input basis vector as the $j$-th column. This is not arbitrary: it is exactly what makes matrix-vector multiplication (a topic developed right after this one) recover $T$ automatically. If you write a general vector $v \in V$ in coordinates as $x = (x_1,\dots,x_n)$ relative to $\beta$, meaning $v = x_1v_1 + \cdots + x_nv_n$, then linearity gives

The coefficient of $w_i$ here, $\sum_j a_{ij}x_j$, is exactly the $i$-th entry of the matrix-vector product $\mathcal{M}(T)x$. So the columns-of-images convention is engineered precisely so that "apply $T$, then take $\gamma$-coordinates" equals "take $\beta$-coordinates, then multiply by the matrix." This identity, foreshadowed here, is the substance of the article on Matrix of Vector and Map-Vector Action. It is worth internalizing immediately that the matrix is a joint invariant of the map and the two bases — change either basis and, generically, every entry changes. This dependence is not a defect; it is the entire reason Change of Basis and the Change-of-Basis Matrix exists as a subsequent topic: understanding how the matrix changes under a change of basis is what lets you choose bases that make a given map look as simple as possible (diagonal, triangular, etc.).

Example 1 — a map between spaces of different dimension and structure, with a non-standard basis. Let $V = P_2(\mathbb{R})$, the polynomials of degree at most $2$, and let $W = \mathbb{R}^2$. Define $T : V \to W$ by

This is linear: for $p,q \in V$ and $\lambda \in \mathbb{R}$, $T(p+q) = ((p+q)(1),(p+q)'(0)) = (p(1)+q(1), p'(0)+q'(0)) = T(p)+T(q)$, and similarly $T(\lambda p) = \lambda T(p)$, using linearity of evaluation and differentiation. Choose the (non-monomial) ordered basis of $V$

One checks $\beta$ is a basis: three polynomials in a $3$-dimensional space are a basis iff independent, and if $a(1+x)+b(x-x^2)+c(2x^2)=0$ then comparing coefficients of $1,x,x^2$ gives $a=0$, $a+b=0$, $-b+2c=0$, forcing $a=b=c=0$. Choose the standard basis of $W = \mathbb{R}^2$: $\gamma = (e_1,e_2)$ with $e_1=(1,0)$, $e_2=(0,1)$. Compute the images of the basis vectors.

• $v_1 = 1+x$: $v_1(1) = 2$, $v_1'(x) = 1$ so $v_1'(0)=1$. Thus $T(v_1) = (2,1) = 2e_1 + 1e_2$.
• $v_2 = x - x^2$: $v_2(1) = 1-1=0$, $v_2'(x) = 1-2x$ so $v_2'(0)=1$. Thus $T(v_2) = (0,1) = 0e_1+1e_2$.
• $v_3 = 2x^2$: $v_3(1) = 2$, $v_3'(x)=4x$ so $v_3'(0)=0$. Thus $T(v_3) = (2,0) = 2e_1+0e_2$.

Reading off coordinates as columns,

Sanity check via the Corollary. Take $p = v_1 + 3v_2 - v_3 = (1+x) + 3(x-x^2) - 2x^2 = 1 + 4x - 5x^2$, so $[p]_\beta = (1,3,-1)^{\mathsf T}$. Directly, $p(1) = 1+4-5=0$ and $p'(x) = 4-10x$ so $p'(0)=4$; hence $T(p) = (0,4)$. Using the matrix:

This matches $T(p) = (0,4)$ exactly, confirming the matrix is correct.

Example 2 — an operator on $\mathbb{R}^3$, in the standard basis versus a different basis, showing the same map with genuinely different matrices. Let $T:\mathbb{R}^3 \to \mathbb{R}^3$ be the linear map defined on the standard basis $\varepsilon = (e_1,e_2,e_3)$ by

(This determines $T$ completely on all of $\mathbb{R}^3$ by linearity, per Linear Maps Determined by Action on a Basis: for general $(x,y,z) = xe_1+ye_2+ze_3$, $T(x,y,z) = x(3,1,0)+y(0,2,-1)+z(1,0,4) = (3x+z,\ x+2y,\ -y+4z)$.) Relative to $\beta = \gamma = \varepsilon$ (standard basis on both sides), the columns are just the images themselves:

Now use a different basis on the input side only:

keeping $\gamma = \varepsilon$ on the output side. ($\beta'$ is a basis: the determinant of the matrix with these columns is $1(1\cdot1 - 1\cdot0) - 0 + 1(1\cdot1-1\cdot0) = 1+1=2 \ne 0$, so the vectors are independent, hence a basis of the $3$-dimensional space.) Compute $T(u_j)$ using the explicit formula $T(x,y,z) = (3x+z,\,x+2y,\,-y+4z)$:

• $T(u_1) = T(1,1,0) = (3(1)+0,\ 1+2(1),\ -1+0) = (3,3,-1)$.
• $T(u_2) = T(0,1,1) = (3(0)+1,\ 0+2(1),\ -1+4(1)) = (1,2,3)$.
• $T(u_3) = T(1,0,1) = (3(1)+1,\ 1+2(0),\ -0+4(1)) = (4,1,4)$.

Since $\gamma = \varepsilon$ is the standard basis, coordinates relative to $\gamma$ of a vector $(a,b,c)$ are simply $(a,b,c)$ itself, so the columns are exactly these triples:

Notice this is a genuinely different array from $\mathcal{M}(T,\varepsilon)$ — same linear map $T$, different matrix, because the input basis changed while the output basis stayed standard. This is exactly the phenomenon that Change of Basis and the Change-of-Basis Matrix quantifies precisely, by relating $\mathcal{M}(T,\beta',\varepsilon)$ to $\mathcal{M}(T,\varepsilon,\varepsilon)$ via the change-of-basis matrix from $\beta'$ to $\varepsilon$.