Change of Basis

The coordinates of one and the same vector depend on the choice of basis. A change of basis is the rule for recomputing coordinates from one basis to another; it often simplifies computations (for example, reducing an operator's matrix to diagonal form).

The Change-of-Basis Matrix

Suppose two bases are given in an $n$ -dimensional space:

the "old" one $e_{1}, e_{2}, \dots, e_{n}$ ,
the "new" one $e_{1}^{'}, e_{2}^{'}, \dots, e_{n}^{'}$ .

Expand each new basis vector in the old basis:

e_{j}^{'} = c_{1 j} e_{1} + c_{2 j} e_{2} + \dots + c_{nj} e_{n}, j = 1, \dots, n .

The coordinates of the new vectors, written by columns, form the change-of-basis matrix $C$ from the old basis to the new one:

C = c_{11} c_{21} ⋮ c_{n 1} c_{12} c_{22} ⋮ c_{n 2} \dots \dots ⋱ \dots c_{1 n} c_{2 n} ⋮ c_{nn}, (e_{1}^{'} e_{2}^{'} \dots e_{n}^{'}) = (e_{1} e_{2} \dots e_{n}) C .

Since the new basis vectors are linearly independent, $det C \neq = 0$ , that is, the change-of-basis matrix is invertible.

Transforming Coordinates

Suppose a vector $x$ has coordinates $X = (x_{1}, \dots, x_{n})^{T}$ in the old basis and $X^{'} = (x_{1}^{'}, \dots, x_{n}^{'})^{T}$ in the new one. Then

X = C X^{'}, X^{'} = C^{- 1} X .

Derivation. Expand $x$ in the new basis and substitute $e_{j}^{'} = \sum_{i} c_{ij} e_{i}$ :

x = j \sum x_{j}^{'} e_{j}^{'} = j \sum x_{j}^{'} i \sum c_{ij} e_{i} = i \sum (j \sum c_{ij} x_{j}^{'}) e_{i} .

On the other hand $x = \sum_{i} x_{i} e_{i}$ . By uniqueness of the expansion in a basis, the coordinates match: $x_{i} = \sum_{j} c_{ij} x_{j}^{'}$ , that is, $X = C X^{'}$ . Multiplying on the left by $C^{- 1}$ gives $X^{'} = C^{- 1} X$ . $■$

Note: the columns of $C$ are the coordinates of the new vectors in the old basis, yet it maps the new coordinates of a vector to the old ones ( $X = C X^{'}$ ).

Example

Let the old basis be $\overset{ˉ}{i}, \overset{ˉ}{j}, \overset{ˉ}{k}$ . The new basis is given by the vectors

\overset{a}{ˉ} = (5; 1; 3), \overset{ˉ}{b} = (0; 1; 2), \overset{c}{ˉ} = (- 1; 1; 1) .

Let's find the coordinates of the vector $\overset{x}{ˉ} = (8; - 4; 12)$ in the new basis.

The change-of-basis matrix (new vectors by columns) and its determinant:

C = 513012 - 1 11, det C = 5 (1 - 2) - 0 + (- 1) (2 - 3) = - 4 \neq = 0,

so $\overset{a}{ˉ}, \overset{ˉ}{b}, \overset{c}{ˉ}$ indeed form a basis. Computing the inverse matrix,

C^{- 1} = - \frac{1}{4} - 1 2 - 1 - 2 8 - 10 1 - 6 5,

we find the new coordinates:

X^{'} = C^{- 1} X = - \frac{1}{4} - 1 2 - 1 - 2 8 - 10 1 - 6 5 8 - 4 12 = - \frac{1}{4} 12 - 88 92 = - 3 22 - 23 .

Answer: $\overset{x}{ˉ} = - 3 \overset{a}{ˉ} + 22 \overset{ˉ}{b} - 23 \overset{c}{ˉ}$ (check: $- 3 \overset{a}{ˉ} + 22 \overset{ˉ}{b} - 23 \overset{c}{ˉ} = (8; - 4; 12)$ ).

Relation to the Matrix of a Linear Operator

The same change-of-basis matrix $C$ relates the matrices of one linear operator in two bases: if $A$ is the operator's matrix in the old basis, and $B$ in the new one, then

B = C^{- 1} A C

(see Linear Operator). If both bases are orthonormal, the change-of-basis matrix $C$ is orthogonal ( $C^{- 1} = C^{T}$ ).

The Change-of-Basis Matrix

Transforming Coordinates

Example

Relation to the Matrix of a Linear Operator

Related works