## The characteristic polynomial of a matrix is the product of its invariant factors

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):

1. Interchange two rows/columns
2. Add a multiple of one row/column to another
3. Multiply a row/column by a unit

Recall that the determinant function on $\mathsf{Mat}_n(R)$ 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, $\mathsf{det}(A)$ 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 $A$ is obtained from $B$ by a series of ERCOs, then $\mathsf{det}(A)$ and $\mathsf{det}(B)$ differ only by a unit factor.

Now let $A$ be a matrix. By Theorem 21 in D&F, $xI-A$ is similar to a diagonal matrix $B$ whose (nonconstant) diagonal entries are precisely the invariant factors of $A$ and whose constant diagonal entries are all 1. That is, the determinant of $B$ is the product of the invariant factors of $A$. On the other hand, the determinant of $xI-A$ is the characteristic polynomial of $A$, which is monic. So the characteristic polynomial of $A$ 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.