### Properties

##### 3. About polynomials of matrix

Polynomials can be applied to square matrices. For example,

p(t) = - 2 - t + 2t2 + t3p(A) = - 2I - A + 2A2 + A3.

Moreover, factorizations are preserved,

- 2 - t + 2t2 + t3 = (t - 1)(t + 1)(t + 2) → - 2I - A + 2A2 + A3 = (A - I)(A + I)(A + 2I).

The following summarizes what we know about eigenvalues and eigenvectors of p(A).

Let p(t) be a polynomial. Let A be a square matrix.

• v is an eigenvector of A with eigenvalue λv is an eigenvector of p(A) with eigenvalue p(λ).
• A is diagonalizable ⇔ p(A) is diagonalizable.
• Any eigenvalue of p(A) is of the form p(λ) for some eigenvalue λ of A.

Note that the third part is the "partial converse" of the first part. See the second example below and the subsequent discussion. Also note that the third part implicitly allows the use of complex numbers as eigenvalues.

Proof p(A) can be constructed from A and I by repeatedly making use of matrix product and linear combination. The first then follow from this result. Together with basis of eigenvectors interpretation, the second part also follows from the result.

To prove the third part, let μ be an eigenvalue of p(A). We may decompose the polynomial p(t) - μ

p(t) - μ = c(t - λ1)(t - λ2)...(t - λk), c ≠ 0

(this requires us to allow complex numbers) and get

p(A) - μI = c(A - λ1I)(A - λ2I)...(A - λkI).

Then by the property detAB = detA detB, we have

det(p(A) - μI) = cn det(A - λ1I)det(A - λ2I)...det(A - λkI).

Since μ is assumed to be an eigenvalue of p(A), we have det(p(A) - μI) = 0. Consequently, det(A - λiI) = 0 for some λi. Substituting t = λi into the decomposition of p(t) - μ, we find μ is of the form p(λi) for some eigenvalue λi of A.

Example Let A be a square matrix such that the eigenvalues of I + 2A + A2 are 0, 4, 16. Then any eigenvalue λ of A must be the solution of one of the following equations.

1 + 2λ + λ2 = 0 (solution: λ = -1)
1 + 2λ + λ2 = 4 (solution: λ = 1, -3)
1 + 2λ + λ2 = 16 (solution: λ = 3, -5)

Thus λ must be one of the following five numbers: -5, -3, -1, 1, 3.

Moreover, the eigenvalues of A must include at least one from each of the three groups above. For example, the eigenvalues of A cannot be -3, -1, 1 only, because either 3 or -5 must also be an eigenvalue of A.

Example The matrices

 A = [ 0 0 ], B = [ 0 1 ] 0 0 0 0

satisfy A2 = B2 = O. Note that although eigenvalues (= 0) of A and B can be determined from the eigenvalues (= 0) of O, the eigenvectors cannot be determined.

In general, all we can say is that the inclusion of the eigenspace nul(A - λI) ⊂ nul(p(A) - p(λ)I) (equivalent to the first property above). The equality does not necessarily hold.

Example In an earlier example, we found the matrix

 A = [ 13 -4 ] -4 7

has a basis {(1, 2), (-2, 1)} of eigenvectors, with eigenvalues 5, 15. For the characteristic polynomial p(t) = 75 - 20t + t2 = (5 - t)(15 - t), we know that {(1, 2), (-2, 1)} is a basis of eigenvectors for p(A) = 75I - 20A + A2, with p(5) = 0, p(15) = 0 as eigenvalues. Then it is easy to conclude that p(A) = O.

In another earlier example, we found the matrix

 B = [ 1 3 -3 ] -3 7 -3 -6 6 -2

has a basis {(1, 1, 0), (1, 0, -1), (1, 1, 2)} of eigenvectors, with eigenvalues 4, 4, -2. For the polynomial p(t) = (4 - t)(- 2 - t) = - 8 - 2t + t2, we know that {(1, 1, 0), (1, 0, -1), (1, 1, 2)} is a basis of eigenvectors for p(B) = - 8I - 2B + B2, with p(4) = 0, p(4) = 0, p(-2) = 0 as eigenvalues. Then it is easy to conclude that p(B) = O.

Note that the equality - 8I - 2B + B2 = O is the same as B(-2I + B) = 8I. Therefore

 B-1 = (1/8)(-2I + B) = (1/8)[ -1 3 -3 ]. -3 5 -3 -6 6 -4