Basis Images Determine a Linear Map

A proof that once you know where a linear map sends a basis, the whole map is fixed, which is the theorem underneath matrix representation.
Modified

April 26, 2026

Keywords

proof, basis, linear map, matrix representation

1 Claim

We want the theorem behind the slogan

a matrix is determined by what it does to the basis vectors.

Theorem 1 (Theorem) Let \(e_1,\dots,e_n\) be a basis of \(\mathbb{R}^n\), and let \(w_1,\dots,w_n \in \mathbb{R}^m\).

Then there exists a unique linear map \(T : \mathbb{R}^n \to \mathbb{R}^m\) such that

\[ T(e_j) = w_j \qquad \text{for } j=1,\dots,n. \]

2 Why This Proof Matters

This proof is the real reason matrices work.

It tells you:

  • why the columns of a matrix are enough to define the whole linear map
  • why linear maps are fully controlled by basis data
  • why “define the map on a basis and extend by linearity” is a legitimate construction

3 Dependencies

  • every vector in \(\mathbb{R}^n\) has a unique expansion in a basis
  • a linear map must preserve linear combinations

4 Strategy Before Details

There are two parts.

  1. Existence: define the map by expanding an arbitrary vector in the basis and replacing each basis vector by its target image.
  2. Uniqueness: any linear map with the same basis images must agree on every basis expansion.

5 Full Proof

Proof. Let \(x \in \mathbb{R}^n\) be arbitrary. Because \(e_1,\dots,e_n\) is a basis, there are unique scalars \(a_1,\dots,a_n\) such that

\[ x = a_1 e_1 + \cdots + a_n e_n. \]

Define

\[ T(x) = a_1 w_1 + \cdots + a_n w_n. \]

This is well defined because the coefficients \(a_1,\dots,a_n\) are unique.

Now check linearity. Let

\[ x = \sum_{j=1}^n a_j e_j, \qquad y = \sum_{j=1}^n b_j e_j. \]

Then

\[ x+y = \sum_{j=1}^n (a_j+b_j)e_j, \]

so

\[ T(x+y) = \sum_{j=1}^n (a_j+b_j)w_j = \sum_{j=1}^n a_j w_j + \sum_{j=1}^n b_j w_j = T(x)+T(y). \]

Similarly, for any scalar \(c\),

\[ cx = \sum_{j=1}^n (ca_j)e_j, \]

and therefore

\[ T(cx) = \sum_{j=1}^n (ca_j)w_j = c \sum_{j=1}^n a_j w_j = cT(x). \]

So \(T\) is linear.

Also, if we apply the definition to a basis vector \(e_i\), whose coefficient expansion has \(1\) in the \(i\)th position and \(0\) elsewhere, we get

\[ T(e_i) = w_i. \]

This proves existence.

For uniqueness, let \(S : \mathbb{R}^n \to \mathbb{R}^m\) be another linear map with \(S(e_j)=w_j\) for all \(j\). If

\[ x = a_1 e_1 + \cdots + a_n e_n, \]

then by linearity,

\[ S(x) = a_1 S(e_1) + \cdots + a_n S(e_n) = a_1 w_1 + \cdots + a_n w_n = T(x). \]

So \(S(x)=T(x)\) for every \(x\), hence \(S=T\).

6 Step Annotations

  1. The basis matters because it gives a unique coordinate expansion.
  2. Existence works by replacing coordinates in the domain basis with prescribed target vectors.
  3. Uniqueness works because every vector is built from the basis, and linearity forces the map to respect that construction.

7 Why The Assumptions Matter

  • If the list is not a basis, coordinates may fail to exist or fail to be unique, and the construction can break.
  • Linearity is essential. Nonlinear maps are not determined by basis images alone.

8 Common Failure Modes

  • forgetting to justify why the definition is well defined
  • assuming the result also holds for arbitrary generating sets
  • using linearity before proving the constructed map is linear

9 Reusable Pattern

This proof teaches the construction:

define on a basis -> extend by linearity -> use uniqueness of coordinates

That same pattern appears in linear operator design, Fourier-style expansions, and later basis changes.

10 Where This Shows Up Again

  • Matrices and Linear Maps
  • any argument that constructs a matrix from column images
  • operator definitions in control, numerics, and ML layers

11 Exercises

  1. Use the theorem to build the matrix of a linear map once the images of \(e_1,e_2,e_3\) are given.
  2. Explain why the theorem fails if the chosen list of vectors is linearly dependent.
  3. Adapt the proof from the standard basis to an arbitrary basis.

12 Sources and Further Reading

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