TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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Similar matrices have the same characteristic polynomial

Statement

If $A, B \in M_{n} (F)$ are similar, then $χ_{A} (x) = χ_{B} (x)$ in $F [x]$ , including $n = 0$ .

Facts & Assumptions

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

[L1]

Similarity means that $B = P^{- 1} A P$ for some invertible $P \in M_{n} (F)$ (Similar matrices: $B = P^{- 1} A P$ for an invertible $P$ ).

[L2]

For positive size over a commutative ring, $\det (X Y) = \det (X) \det (Y)$ (For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ ).

[L3]

Field matrices embed entrywise into matrices over $F [x]$ , with the same matrix arithmetic and determinant (For a field, the ring-matrix operations, invertibility and similarity agree exactly with the established field-matrix interface).

[L4]

The characteristic polynomial is $\det (x I - A)$ in positive size and $1$ 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).

Proof

technique · direct

1.1L4

If $n = 0$ , [L4] gives $χ_{A} = 1 = χ_{B}$ .

1.2L1L3algebra

Suppose $n \geq 1$ and choose $P$ from [L1]. Over $F [x]$ , $x I - B = P^{- 1} (x I - A) P$ .

2.1step 1.2L2L3algebra

By [L2], $\det (x I - B) = \det (P^{- 1}) \det (x I - A) \det (P)$ . Since $P^{- 1} P = I$ , multiplicativity also gives $\det (P^{- 1}) \det (P) = 1$ .

3.1step 1.1step 2.1L4∎

Using [L4] in step 2.1 gives $χ_{B} (x) = χ_{A} (x)$ , and step 1.1 supplies the remaining size.

Depends on

Dependency tree · 2 levels

Sources

H. Pinkham, Linear Algebra, §12.1