Question about the proving that Characteristic Polynomial Independent From the Choice of Basis

characteristic polynomialdeterminantlinear algebra

Prove that definition of characteristic polynomial of a linear operator on a finite-dimensional vector space $V$ is independent of the choice of basis for $V$.

Proof: Let $A \colon V \rightarrow V $ be a linear operator. Let $e = \{e_1,\ldots , e_n\}$ and $f = \{f_1,\ldots , f_n\}$ be bases for $V$. Let $[A]_e$ and $[A]_{f}$ be the matrices of the linear operator $A$ in the given respective bases.
Then there exists an invertible matrix $C$ such that $[A]_f = C^{-1} [A]_e C$.

Let $P_A (\lambda)$ represent the characteristic polynomial. Then
$$P_{[A]_f }(\lambda) = \textrm{det } \left( {[A]_f \space – \lambda I}\right) = \textrm{det } \left( C^{-1} [A]_e C \space – \lambda I\right) = \textrm{det }\left( C^{-1} [A]_e C \space – \lambda C^{-1}C\right)$$

Now if we factor out $C$ then factor out $C^{-1}$ we get:

$$\textrm{det } \left( C^{-1} ([A]_e \space – \lambda I) C\right) = \textrm{det } C^{-1} \cdot P_{[A]_e }(\lambda) \cdot \textrm{det} \space C = \textrm{det } C^{-1} \cdot \textrm{det} \space C \cdot P_{[A]_e }(\lambda) \\= \textrm{det } C^{-1}C\cdot \chi_{[A]_e }(\lambda) = P_{[A]_e }(\lambda) $$

In other words, we have shown that $$P_{[A]_f }(\lambda) = P_{[A]_e }(\lambda) $$

My questions are
1). why does it claim that there exists an invertible matrix C?

2). What's the idea of this proof? Is it arguing by definition of characteristic polynomial?

3). $$\textrm{det } C^{-1}C\cdot \chi_{[A]_e }(\lambda)$$ In the end, what does that $\chi$ mean?

Thanks a lot in advance!!

Best Answer

  1. $C$ is the change-of-basis matrix from basis $e$ to basis $f$, which is always invertible. It is the matrix $[I]_{f}^e$.

  2. Yes; it's showing that if you calculate the characteristic polynomial of a linear transformation in terms of its coordinate matrix (which depends on the basis), then you get the same polynomial regardless of the basis you pick. It does so by using the facts that (i) two representations of the same linear transformation relative to two given bases are conjugate; and (ii) $\det(XY)=\det(X)\det(Y)$ for square matrices $X$ and $Y$ of the same size.

  3. $\chi_A$ stands for the characteristic polynomial of the matrix $A$. The argument seems to be switching between the notation using $P$ and the notation using $\chi$, which is not uncommon.

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