TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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For $A, B \in M_{n} (F)$ , the products $A B$ and $B A$ have the same characteristic polynomial

Statement

For $A, B \in M_{n} (F)$ ,

$χ_{A B} (x) = χ_{B A} (x) .$

Facts & Assumptions

Given: Matrices $A, B \in M_{n} (F)$ .

[L1]

The characteristic polynomial of $C$ is $\det (x I - C)$ , with value $1$ in size zero (For $A \in M_{n} (F)$ , the characteristic polynomial is $χ_{A} (x) = \det (x I_{n} - A)$ when $n \geq 1$ , with $χ_{A} (x) = 1$ for the unique $0 \times 0$ matrix).

[L2]

For a positive-sized square matrix over a commutative ring, the determinant is the signed sum over permutations of products selecting one entry in each row and column (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[L3]

Determinants are multiplicative for positive-sized square matrices over a commutative ring (For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ ).

[L4]

Block multiplication follows from associative and distributive matrix arithmetic (Matrix arithmetic over a commutative ring is associative, unital and distributive, and transpose reverses products).

Proof

technique · direct

1.1

If $n = 0$ , both characteristic polynomials are $1$ by [L1]. Assume henceforth that $n \geq 1$ , and work over $F [t]$ .

1.2

Put $N = (\begin{array}{cc} I & t A \\ - B & I \end{array})$ . Left multiplication by $(\begin{array}{cc} I & - t A \\ 0 & I \end{array})$ produces $(\begin{array}{cc} I + t A B & 0 \\ - B & I \end{array})$ , while left multiplication by $(\begin{array}{cc} I & 0 \\ B & I \end{array})$ produces $(\begin{array}{cc} I & t A \\ 0 & I + t B A \end{array})$ .

L4algebra

2.1

In the Leibniz sum [L2] for a block-triangular matrix, every nonzero term preserves the two index blocks, so its determinant is the product of the two diagonal-block determinants. Both multiplying matrices in step 1.2 consequently have determinant $1$ , and [L3] yields $\det (I + t A B) = \det (N) = \det (I + t B A)$ in $F [t]$ .

step 1.2L2L3algebra

3.1

Replacing $t$ by $- t$ , the coefficient of $t^{k}$ in $\det (I - t C)$ is the coefficient of $x^{n - k}$ in $\det (x I - C)$ , directly from the Leibniz formula. Equality in step 2.1 therefore gives equality of every coefficient of $χ_{A B}$ and $χ_{B A}$ .

step 2.1L1L2algebra

4.1

Together with the zero-sized case, $χ_{A B} = χ_{B A}$ for all $n$ .

step 1.1step 3.1∎

Depends on

Used by

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Sources

H. Pinkham, Linear Algebra, §12.3.5 (standard reference, not scraped)

For A,B∈Mn(F)A,B∈Mn​(F), the products ABAB and BABA have the same characteristic polynomial