Change of basis and similarity

Every \mathbf{x} \in V has a unique expansion in the \mathcal{B}-basis, \mathbf{x} = x_1\mathbf{b}_1 + x_2\mathbf{b}_2 + \cdots + x_n\mathbf{b}_n, where the scalars x_1, \dots, x_n are exactly the entries of [\mathbf{x}]_{\mathcal B}. Take \mathcal{C}-coordinates of both sides. Coordinate-taking relative to a fixed basis is linear — the coordinate map sends sums to sums and scalar multiples to scalar multiples, since it just reads off the unique coefficients in a linear combination — so [\mathbf{x}]_{\mathcal C} = [x_1\mathbf{b}_1 + \cdots + x_n\mathbf{b}_n]_{\mathcal C} = x_1[\mathbf{b}_1]_{\mathcal C} + x_2[\mathbf{b}_2]_{\mathcal C} + \cdots + x_n[\mathbf{b}_n]_{\mathcal C}. The right side is precisely a linear combination of the columns of P_{\mathcal C \leftarrow \mathcal B} with weights x_1, \dots, x_n, which is the definition of the matrix-vector product: x_1[\mathbf{b}_1]_{\mathcal C} + \cdots + x_n[\mathbf{b}_n]_{\mathcal C} = \begin{bmatrix} [\mathbf{b}_1]_{\mathcal C} & \cdots & [\mathbf{b}_n]_{\mathcal C} \end{bmatrix}\begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix} = P_{\mathcal C \leftarrow \mathcal B}\,[\mathbf{x}]_{\mathcal B}. Combining the two displays gives [\mathbf{x}]_{\mathcal C} = P_{\mathcal C \leftarrow \mathcal B}\,[\mathbf{x}]_{\mathcal B}, as claimed.

For invertibility, run the same argument with the roles of \mathcal{B} and \mathcal{C} swapped: the matrix P_{\mathcal B \leftarrow \mathcal C} built from the \mathcal{B}-coordinates of the \mathcal{C}-basis vectors satisfies [\mathbf{x}]_{\mathcal B} = P_{\mathcal B \leftarrow \mathcal C}\,[\mathbf{x}]_{\mathcal C} for every \mathbf{x} \in V. Substituting the first formula into the second (and vice versa) gives [\mathbf{x}]_{\mathcal B} = P_{\mathcal B \leftarrow \mathcal C}P_{\mathcal C \leftarrow \mathcal B}\,[\mathbf{x}]_{\mathcal B}, \qquad [\mathbf{x}]_{\mathcal C} = P_{\mathcal C \leftarrow \mathcal B}P_{\mathcal B \leftarrow \mathcal C}\,[\mathbf{x}]_{\mathcal C} for every \mathbf{x} \in V. A matrix that fixes every coordinate vector in \mathbb{R}^n must be the identity matrix I_n (test it against the standard basis vectors, whose coordinate images run over all of \mathbb{R}^n), so both products equal I_n. Hence P_{\mathcal C \leftarrow \mathcal B} and P_{\mathcal B \leftarrow \mathcal C} are inverse to one another.

Characteristic polynomial (and hence eigenvalues). Write the characteristic polynomial of B using B = P^{-1}AP: \det(B - \lambda I) = \det\!\left(P^{-1}AP - \lambda I\right) = \det\!\left(P^{-1}AP - \lambda P^{-1}P\right) = \det\!\left(P^{-1}(A - \lambda I)P\right). The determinant is multiplicative, so this splits into three factors: \det(B - \lambda I) = \det(P^{-1})\,\det(A - \lambda I)\,\det(P) = \det(A - \lambda I), since \det(P^{-1}) = 1/\det(P) cancels against \det(P). So A and B share the same characteristic polynomial \det(A - \lambda I), and therefore the same roots — the same eigenvalues, each with the same algebraic multiplicity, since multiplicity is read off the same polynomial.

Determinant. Setting \lambda = 0 in the identity just proved gives \det(B) = \det(A) directly. (Equivalently, \det(B) = \det(P^{-1})\det(A)\det(P) = \det(A) by the same multiplicative argument.)

Trace. The characteristic polynomial of an n \times n matrix M has the form \det(M - \lambda I) = (-\lambda)^n + \operatorname{tr}(M)(-\lambda)^{n-1} + \cdots + \det(M), so \operatorname{tr}(M) is exactly the coefficient of (-\lambda)^{n-1}. Since A and B share the same characteristic polynomial, they share that coefficient, so \operatorname{tr}(A) = \operatorname{tr}(B).

Rank. Left-multiplication by the invertible matrix P^{-1} and right-multiplication by the invertible matrix P each preserve rank — multiplying by an invertible matrix on either side sends a basis of the column space to another basis of a column space of the same dimension, and never collapses two independent columns together, since invertible maps are one-to-one. So \operatorname{rank}(B) = \operatorname{rank}(P^{-1}AP) = \operatorname{rank}(AP) = \operatorname{rank}(A).

Why this matters. These quantities are exactly the ones that don't depend on which basis is used to write a linear transformation as a matrix — changing basis replaces A by P^{-1}AP for the change-of-basis matrix P, so \det, \operatorname{tr}, \operatorname{rank}, and the eigenvalues are properties of the underlying transformation itself, not artifacts of coordinates. The converse is false: matrices can share all of these invariants without being similar (a diagonalizable matrix is never similar to a non-diagonalizable one with the same eigenvalues), so this theorem gives necessary conditions for similarity, not sufficient ones.