math111_logo Properties

2. Polynomials of matrix

Polynomials can be applied to square matrices. For example,

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

Moreover, the product of polynomials can also be applied to matrix. For example,

- 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).

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 The first part follows from this result, because p(A) is a combination of A and I.

The second part follows from this first part, and the use of basis of eigenvectors, similar to the proof for this result on similar matrices.

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

p(t) - μ = λ0(t - λ1)(t - λ2)...(t - λk), λ0 ≠ 0

(this requires us to allow complex numbers) and get

p(A) - μI = λ0(A - λ1I)(A - λ2I)...(A - λkI)

Then

0 = det(p(A) - μI) = λ0n det(A - λ1I)det(A - λ2I)...det(A - λkI),

where the first equality is due to the assumption that μ be an eigenvalue of p(A), and the second equality comes from the property detAB = detA detB. Thus we have det(A - λiI) = 0 for some λi, which means some λi is an eigenvalue of A. Substituting t = λi in the decomposition of p(t) - μ, we find μ = p(λi), which shows the eigenvalue μ of p(A) is p(some eigenvalue of A).

Example Let A be a square matrix such that the eigenvalues of I + 2A + A2 are 0, 4, 16. Then for 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 Let

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

Then A2 = B2 = O. The example shows that although we can determine eigenvalues (= 0) of A and B from the eigenvalues (= 0) of O, the eigenvectors cannot be determined.

In general, all we can say is that the eigenspace nul(A - λI) ⊂ nul(p(A) - p(λ)I). 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

The polynomials 75 - 20t + t2 and - 8 - 2t + t2 of the example above are called minimal polynomials of the matrices. For a diagonalizable matrix A with λ1, λ2, , ..., λk as all the eigenvalues (no repitition), the minomial polynomial is p(t) = (λ1 - t)(λ2 - t)...(λk - t). In particular, we see that p(t) divides the characteristic polynomial.

The minimial polynomials for non-diagonalizable matrices are more complicated. However, it still divides the characteristic polynomial. More detailed discussion on the minimal polynomial can be found here.


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