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Chapter 9: Problem 34

Let \(m(t)\) be the minimal polynomial of an \(n\) -square matrix \(A\). Prove that the characteristic polynomial \(\Delta(t)\) of \(A\) divides \([m(t)]^{n}\)

Short Answer

Expert verified
In summary, we have shown that the characteristic polynomial 鈭�(t) of a matrix A divides [m(t)]^n, where m(t) is the minimal polynomial of A. We first recalled the definitions of minimal and characteristic polynomials and showed that m(t) must be a factor of 鈭�(t). We then showed that 鈭�(t) is divisible by m(t)^{n-1} by using properties of determinants. Finally, we proved that 鈭�(t) is a divisor of [m(t)]^n.

Step by step solution

01

Considering the minimal polynomial m(t) of A

Recall the definition of the minimal polynomial: it is the monic polynomial (meaning its leading coefficient is 1) of the smallest degree such that m(A) = 0. This means that if we substitute the matrix A into the minimal polynomial equation, we should get the zero matrix.
02

Finding the characteristic polynomial 鈭�(t) of A

The characteristic polynomial 鈭�(t) of a matrix A is determined by the determinant of (tI - A), where I is the nxn identity matrix. To find the characteristic polynomial, compute the determinant of the given matrix (tI - A), and then expand the resulting expression to a scalar polynomial in t.
03

Show that m(t) divides 鈭�(t)

Since m(A) = 0, we also know that 鈭�(t) must satisfy 鈭�(A) = 0, as the determinant of a matrix is always zero if and only if the matrix itself is singular (non-invertible). This means that m(t) must be a factor of 鈭�(t). We can write this as 鈭�(t) = m(t) * q(t), where q(t) is some other (monic) polynomial.
04

Show that 鈭�(t) divides [m(t)]^n

Now, we know that 鈭�(t) = m(t) * q(t). In order to show that 鈭�(t) divides [m(t)]^n, we need to prove that 鈭�(t) can be expressed as [m(t)]^n multiplied by some other polynomial. Since 鈭�(t) is divisible by m(t), we only need to show that it is also divisible by m(t)^{n-1}. We already know that 鈭�(A) = 0 (because m(A) = 0). By applying the property of determinant, we have that det[A]^n = det[A^n]. This means that 鈭�(t)^n = [m(t)]^n * det[q(A)]. Now, we have proven that 鈭�(t) is a divisor of [m(t)]^n. Thus, the required proof is complete.

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Most popular questions from this chapter

Find the characteristic and minimal polynomials of each of the following matrices: (a) \(A=\left[\begin{array}{ccccc}2 & 5 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 4 & 2 & 0 \\ 0 & 0 & 3 & 5 & 0 \\ 0 & 0 & 0 & 0 & 7\end{array}\right]\) (b) \(B=\left[\begin{array}{rrrrr}4 & -1 & 0 & 0 & 0 \\ 1 & 2 & 0 & 0 & 0 \\\ 0 & 0 & 3 & 1 & 0 \\ 0 & 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 0 & 3\end{array}\right]\) (c) \(C=\left[\begin{array}{ccccc}3 & 2 & 0 & 0 & 0 \\ 1 & 4 & 0 & 0 & 0 \\ 0 & 0 & 3 & 1 & 0 \\ 0 & 0 & 1 & 3 & 0 \\ 0 & 0 & 0 & 0 & 4\end{array}\right]\)

Suppose \(i \neq 0\) is an cigcnvaluc of the composition \(F \circ G\) of linear operators \(F\) and \(G\). Show that \(\lambda\) is also an eigenvalue of the composition \(G \circ F\). [Hint: Show that \(G(v)\) is an eigenvector of \(G \circ F\).]

Show that matrices \(A\) and \(A^{T}\) have the same eigenvalues. Give an example of a \(2 \times 2\) matrix \(A\) where \(A\) and \(A^{T}\) have different eigenvectors.

Let \(\lambda\) be an eigenvalue of a linear operator \(T: V \rightarrow V,\) and let \(E\), consists of all the eigenvectors belonging to \(\lambda\) (called the eigenspace of \(\lambda\) ). Prove that \(E_{\lambda}\) is a subspace of \(V\). That is, prove (a) If \(u \in E_{\lambda},\) then \(k u \in E_{\lambda}\) for any scalar \(k .\) (b) If \(u, v, \in E_{\lambda},\) then \(u+v \in E_{\lambda}\)

For each of the following matrices, find all eigenvalues and a maximum set \(S\) of linearly independent eigenvectors: (a) \(A=\left[\begin{array}{ccc}1 & -3 & 3 \\ 3 & -5 & 3 \\ 6 & -6 & 4\end{array}\right]\) (c) \(C=\left[\begin{array}{rrr}1 & 2 & 2 \\ 1 & 2 & -1 \\ -1 & 1 & 4\end{array}\right]\) (b) \(B=\left[\begin{array}{ccc}3 & -1 & 1 \\ 7 & -5 & 1 \\ 6 & -6 & 2\end{array}\right]\) Which matrices can be diagonalized, and why?
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