Matrices in Different Bases

Definition Let $T \colon \mathbb{R}^n \to \mathbb{R}^m$ be linear, let $\mathcal{B} = (\mathbf{b}_1, \dots, \mathbf{b}_n)$ be an ordered basis of $\mathbb{R}^n$, and let $\mathcal{C}$ be an ordered basis of $\mathbb{R}^m$. The matrix of $T$ relative to $\mathcal{B}$ and $\mathcal{C}$ is the $m \times n$ matrix

whose $j$th column describes the image of the $j$th input basis vector in the output basis. It is the only matrix with

for every $\mathbf{x}$ in $\mathbb{R}^n$. This works because $T$ sends $c_1\mathbf{b}_1 + \cdots + c_n\mathbf{b}_n$ to $c_1T(\mathbf{b}_1) + \cdots + c_nT(\mathbf{b}_n)$, and taking $\mathcal{C}$-coordinates respects sums and scalar multiples. When both bases are the standard ones, it is the standard matrix of $T$.

Definition For a linear map $T \colon \mathbb{R}^n \to \mathbb{R}^m$ with standard matrix $A$, the input basis $\mathcal{B}$ is an ordered basis of the domain $\mathbb{R}^n$, with an invertible $n \times n$ basis matrix $B$, and the output basis $\mathcal{C}$ is an ordered basis of the codomain $\mathbb{R}^m$, with an invertible $m \times m$ basis matrix $C$. The two choices are independent of each other. For an input with $\mathcal{B}$-coordinates $\mathbf{c}$, the vector itself is $B\mathbf{c}$, its image is $AB\mathbf{c}$, and the $\mathcal{C}$-coordinates of that image are $C^{-1}AB\mathbf{c}$. Hence

The sizes match, since $C^{-1}$ is $m \times m$, $A$ is $m \times n$ and $B$ is $n \times n$, so the product is $m \times n$ like $A$. For the identity map of $\mathbb{R}^n$ the formula gives $C^{-1}B = P_{\mathcal{C} \leftarrow \mathcal{B}}$, the change-of-basis matrix.

Definition A linear operator is a linear map from a vector space to itself, such as $T \colon \mathbb{R}^n \to \mathbb{R}^n$. For an operator one normally uses the same ordered basis $\mathcal{B}$ for inputs and outputs and writes $[T]_{\mathcal{B}}$ for $[T]_{\mathcal{B} \leftarrow \mathcal{B}}$. With $C = B$ the formula becomes

an $n \times n$ matrix whose $j$th column is $[T(\mathbf{b}_j)]_{\mathcal{B}}$.

Definition Two $n \times n$ matrices $A$ and $A'$ are similar if there is an invertible $n \times n$ matrix $P$ with

The matrices of one linear operator in two ordered bases are always similar. If $[T]_{\mathcal{B}} = B^{-1}AB$ and $[T]_{\mathcal{D}} = D^{-1}AD$, then $[T]_{\mathcal{D}} = P^{-1}[T]_{\mathcal{B}}P$ with $P = B^{-1}D = P_{\mathcal{B} \leftarrow \mathcal{D}}$. Conversely, if $A$ is the standard matrix of $T$ and $P$ is any invertible $n \times n$ matrix, then $P^{-1}AP$ is the matrix of $T$ in the ordered basis formed by the columns of $P$. Similarity uses the same change of basis on the input side and the output side, so it is a relation for operators. Passing from $A$ to $C^{-1}AB$, with two unrelated bases for input and output, is more general and should not be called similarity. Similar matrices have the same rank and nullity, and later they turn out to share the determinant and the eigenvalues as well.

Definition For an operator $T$ on $\mathbb{R}^n$ with standard matrix $A$ and an ordered basis $\mathcal{B}$ with basis matrix $B$, the four descriptions of an input and its output fit in a square,

The diagram commutes when the top route and the route down, across and up give the same result for every starting column. A matrix $M$ on the top edge makes it commute exactly when $M = B^{-1}AB$, or equivalently $BM = AB$, so the matrix $[T]_{\mathcal{B}}$ is the one matrix that fits there. For a map between two spaces the same square works with $\mathcal{C}$-coordinates in the top right corner, $C^{-1}$ on the right edge and $[T]_{\mathcal{C} \leftarrow \mathcal{B}} = C^{-1}AB$ on the top edge. Checking one starting column is a useful test, but it proves agreement only for that column and its multiples. The matrix equation covers every input at once.

Definition A linear operator $T$ is a rule on vectors and does not depend on any choice of basis. A matrix such as $[T]_{\mathcal{B}}$ is one description of $T$, and it depends on the chosen basis. Changing the basis can change the entries, but it never changes the operator. In particular, the kernel and the image of $T$ are the same sets of vectors in every description, even though the coordinate lists that describe them depend on the basis. For a map between two spaces the same holds for $C^{-1}AB$. A column $\mathbf{c}$ lies in the null space of $C^{-1}AB$ exactly when $B\mathbf{c}$ lies in $\operatorname{Null} A$, and the column space of $C^{-1}AB$ is the set of all $C^{-1}\mathbf{y}$ with $\mathbf{y}$ in $\operatorname{Col} A$. Invertible matrices preserve dimension, so $C^{-1}AB$ has the same rank and the same nullity as $A$.