\[ \newcommand{R}{\mathbb{R}} \newcommand{defarrow}{\quad \stackrel{\text{def}}{\Longleftrightarrow} \quad} \]
Basis transformations
Change-of-basis matrix
Let \(e = \{e_1, \ldots, e_n\}\) and \(f = \{f_1, \ldots, f_n\}\) be two bases for a finite-dimensional real vector space \(X\). Pick any element \(x \in X\). Then \[ x = \sum_{j=1}^n x_j e_j \] has coordinates \((x_1, \ldots, x_n)_e\) in the basis \(e\). Since \(f\) is also a basis, we may express \[ e_j = \sum_{k=1}^n c_{k,j}\, f_k, \quad j = 1, \ldots, n, \qquad\text{ and }\qquad x = \sum_{j=1}^n x_j \sum_{k=1}^n c_{k,j}\, f_k = \sum_{k=1}^n \bigg(\underbrace{\sum_{j=1}^n c_{k,j}\, x_j}_{\text{coord. in } f}\bigg) f_k. \] Put differently, \[ \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix} = \begin{bmatrix} c_{1,1} & c_{1,2} & \ldots & c_{1,n} \\ c_{2,1} & c_{2,2} & \ldots & c_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ c_{n,1} & c_{n,2} & \ldots & c_{n,n} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}\] defines the coordinates \((y_1, \ldots, y_n)\) of \(x\) in the basis \(f\): \( x_f = C x_e. \) The \(n\times n\) scalar-valued matrix \(C \in M_{n \times n}(\R)\) is called a change-of-basis matrix.
- The change-of-basis matrix from \(e = \{(1,0,0),(0,1,0), (0,0,1)\}\) to \(f = \{(1,0,0), (1,1,0),(1,1,1)\}\) in \(\R^n\): \[ \begin{aligned} e_1 &= \sum_{k=1}^3 c_{k,1} f_k \quad\Longleftrightarrow\quad \begin{bmatrix} c_{1,1} \\ c_{2,1} \\ c_{3,1}\end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\0 \end{bmatrix} \\ e_2 &= \sum_{k=1}^3 c_{k,2} f_k \quad\Longleftrightarrow\quad \begin{bmatrix} c_{1,2} \\ c_{2,2} \\ c_{3,2}\end{bmatrix} = \begin{bmatrix} -1 \\ 1 \\0 \end{bmatrix} \\ e_3 &= \sum_{k=1}^3 c_{k,3} f_k \quad\Longleftrightarrow\quad \begin{bmatrix} c_{1,3} \\ c_{2,3} \\ c_{3,3}\end{bmatrix} = \begin{bmatrix} 0 \\ -1 \\1 \end{bmatrix} \end{aligned}. \tag{1} \] The change-of-basis matrix is \[ C = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix}. \] In particular,\[ \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix} \qquad\text{ yields that }\qquad (2,0,1)_e = (2-1,1)_f.\]
Change-of-basis matrix as an inverse
If we write (1) in column form, we get: \[ \begin{vmatrix} \begin{bmatrix} . \\ e_1 \\ . \end{bmatrix} \begin{bmatrix} . \\ e_2 \\ . \end{bmatrix} \begin{bmatrix} . \\ e_3 \\ . \end{bmatrix} \end{vmatrix} = \begin{vmatrix} \begin{bmatrix} . \\ f_1 \\ . \end{bmatrix} \begin{bmatrix} . \\ f_2 \\ . \end{bmatrix} \begin{bmatrix} . \\ f_3 \\ . \end{bmatrix} \end{vmatrix} \: \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix} \quad\Longleftrightarrow\quad \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix}. \] Thus \( I = [f]C\) and \( C = [f]^{-1}\), where \([f]\) is the matrix with the basis vectors \(f_1, \ldots, f_n\) as column vectors.
N.b. Matrices of this form—with zeros below the main diagonal—are called upper triangular. More precisely, \((a_{ij})_{ij}\) is upper triangular if \(a_{ij} = 0\) for \(i > j.\) Lower triangular matrices are defined in a similar fashion (\(a_{ij} = 0\) for \(j > i\)).
› The inverse of a basis matrix is its inverse change-of-basis matrix
Let \([f] = [f_1, \ldots, f_n] \in M_{n \times n}(\mathbb C)\) denote a matrix with column basis vectors \(f_1, \ldots, f_n \in \mathbb C^n\) expressed in the standard basis \(e\). Then \[ \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix}_e = \underbrace{\begin{vmatrix} \begin{bmatrix} . \\ f_1 \\ . \end{bmatrix} \ldots \begin{bmatrix} . \\ f_n \\ . \end{bmatrix} \end{vmatrix}}_{[f]} \: \begin{bmatrix} y_1 \\ \vdots \\ y_n \end{bmatrix}_f \quad \text{ expresses } (y_1, \ldots, y_n)_f \text{ in the basis } e,\] and \[ \begin{bmatrix} y_1 \\ \vdots \\ y_n \end{bmatrix}_f = \underbrace{\begin{vmatrix} \begin{bmatrix} . \\ f_1 \\ . \end{bmatrix} \ldots \begin{bmatrix} . \\ f_n \\ . \end{bmatrix} \end{vmatrix}^{-1}}_{[f]^{-1}} \: \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix}_e \quad \text{ expresses } (x_1, \ldots, x_n)_e \text{ in the basis } f.\]