Prove that the characteristic polynomial of a matrix is the product of its invariant factors.

[Editorial note: This proof (as suggested by D&F) depends on Theorem 21, which we will prove in the exercises after this one.]

Remember our three types of elementary row and column operations (ERCOs):

- Interchange two rows/columns
- Add a multiple of one row/column to another
- Multiply a row/column by a unit

Recall that the determinant function on is multilinear in the columns and the rows of its argument. (See the discussion on page 437 and 438 of D&F, specifically the definition of determinant, Theorem 24, and Corollary 25.) In particular, is not changed if we perform ERCOs of type (2), is negated if we perform type (1) ERCOs, and is multiplied by a unit under type (3) ERCOs. In particular, if is obtained from by a series of ERCOs, then and differ only by a unit factor.

Now let be a matrix. By Theorem 21 in D&F, is similar to a diagonal matrix whose (nonconstant) diagonal entries are precisely the invariant factors of and whose constant diagonal entries are all 1. That is, the determinant of is the product of the invariant factors of . On the other hand, the determinant of is the characteristic polynomial of , which is monic. So the characteristic polynomial of is a unit multiple of the product of its invariant factors; since both polynomials are monic, the ‘unit’ is 1, and in fact the characteristic polynomial *is* the product of the invariant factors.

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