Basis-Independence of the Determinant

The basis-independence of the determinant is what allows $\det(T)$ to be quoted as a genuine invariant of a linear operator in every downstream application. In differential geometry and multivariable calculus, the Jacobian determinant of a map represents the local volume-scaling factor of that map at a point, a quantity that must not depend on the coordinate chart used — precisely the same cancellation argument applies to Jacobian matrices under coordinate changes. In physics, the determinant of a linear transformation (e.g., a Lorentz transformation or a strain tensor) needs to be a physical, coordinate-independent quantity, and similarity-invariance is exactly why this is legitimate. In numerical linear algebra, algorithms that diagonalize or triangularize a matrix (via similarity transformations, as in the QR algorithm or Schur decomposition) rely on knowing the determinant — and more generally the characteristic polynomial — is unaffected by the similarity transformations performed at each step, so it can be tracked or recovered cheaply from the final, simpler form. In statistics, the determinant of a covariance matrix (related to the "generalized variance") is unchanged under a change of coordinate basis for the underlying random vector, a fact used when discussing multivariate normal densities in different coordinate systems. In cryptography and coding theory, when matrices act as linear codes or as parts of lattice-based schemes, similarity-invariant quantities like the determinant provide basis-independent classification data for equivalence classes of lattices or codes. Conceptually, this result is the first serious instance of a broader theme in linear algebra: separating properties of an abstract map from artifacts of a chosen coordinate system, a theme that continues through Multiplicativity of the Determinant and the Characteristic Polynomial and the Determinant.

Ask the tutor: Can you show step-by-step why $\det(P^{-1}) = 1/\det(P)$ specifically fails if $P$ is not invertible?

Picture a linear operator $T$ as a fixed geometric transformation of space — say, a shear-and-stretch map of the plane. A basis is a choice of coordinate grid used to write down $T$ as a matrix of numbers. Change the grid — rotate it, skew it, rescale its axes — and the numerical entries of the matrix change completely. Yet the underlying transformation $T$ has not moved at all; it still stretches the same parallelogram to the same scaled parallelogram, in the same absolute sense. The determinant measures the factor by which $T$ scales (signed) area or volume. Since $T$ itself hasn't changed, that scaling factor shouldn't change either, no matter what grid we used to compute it. The change-of-basis formula $B = P^{-1}AP$ encodes this: $P$ translates into the new coordinates, $A$ applies $T$ in the old coordinates, and $P^{-1}$ translates back. The determinant of this three-step composite is $\det(P^{-1})\det(A)\det(P)$, and because $\det(P^{-1}) = 1/\det(P)$, the $P$'s cancel algebraically, leaving exactly $\det(A)$.

The picture is meant to convey a ratio, not a raw area: whatever unit cell your basis draws, $T$ scales its (signed) volume by the same factor $6$. That factor is $\det(T)$, and it does not care which cell you started with.

The proof technique here is direct algebraic computation via multiplicativity: no induction, no cofactor expansion, no eigenvalue theory — just the multiplication rule $\det(XY)=\det(X)\det(Y)$ applied twice, plus the elementary consequence $\det(P^{-1})=\det(P)^{-1}$. This is deliberately minimal: the invariance of $\det$ under similarity is logically upstream of deeper facts (like the characteristic polynomial being similarity-invariant), so the proof must not lean on any of those later results.

Example 1: An explicit non-diagonal similarity in $\mathbb{R}^{2\times 2}$. Let

First check $P$ is invertible: $\det(P) = (1)(4) - (3)(1) = 1 \ne 0$. Its inverse is

Compute $AP$:

Now $B = P^{-1}AP$:

Now compute determinants directly, without invoking the theorem, to verify it:

Indeed $\det(B) = \det(A) = 6$, exactly as the theorem guarantees, even though $A$ and $B$ look nothing alike entrywise. Note also $\det(P^{-1})\det(P) = 1\cdot 1 = 1$, consistent with Step 1 of the proof (here $\det P = 1$, a special but non-forced coincidence, chosen to keep the arithmetic checkable by hand; the theorem places no requirement on $\det P$ at all).

Example 2: A genuine change-of-basis interpretation, with $\det P \ne \pm 1$. Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be the linear operator defined by

In the standard basis $\mathcal{E} = (e_1, e_2)$, the matrix is

Now switch to the basis $\mathcal{B}' = (u_1, u_2)$ where $u_1 = (2,1)$, $u_2 = (1,3)$. The change-of-basis matrix from $\mathcal{B}'$-coordinates to standard coordinates is

Compute $P^{-1} = \frac{1}{5}\begin{pmatrix} 3 & -1 \\ -1 & 2 \end{pmatrix}$. To find $[T]_{\mathcal{B}'} = P^{-1}[T]_\mathcal{E}P$, first compute $[T]_\mathcal{E}\,P$:

Then

Check the determinant of this new representation:

Again $\det([T]_{\mathcal{B}'}) = \det([T]_\mathcal{E}) = -5$, confirming that despite $\det(P) = 5 \ne 1$, the operator determinant $\det(T) = -5$ is completely independent of whether we used the standard basis or $\mathcal{B}'$.