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.

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