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Chapter 9: Problem 65

Show that a matrix \(A\) and its transpose \(A^{T}\) have the same minimal polynomial.

Short Answer

Expert verified
To show that matrix $A$ and its transpose $A^T$ have the same minimal polynomial, we use the facts that they have the same eigenvalues and the same characteristic polynomial. Since $A$ and $A^T$ share the same set of eigenvalues, their characteristic polynomials are the same. Thus, for any polynomial $P(\lambda)$ that annihilates $A$, $P(\lambda)$ must also annihilate $A^T$. As both minimal polynomials $m_A(\lambda)$ and $m_{A^T}(\lambda)$ annihilate both $A$ and $A^T$, we can conclude that they are the same, i.e., $m_A(\lambda) = m_{A^T}(\lambda)$.

Step by step solution

01

Eigenvalues of A and A^T

Consider an eigenvalue 位 of A with corresponding eigenvector x, i.e., Ax = 位x. To show that 位 is also an eigenvalue of A^T, we need to find an eigenvector for A^T corresponding to 位. Taking the transpose of the above equation, we get (Ax)^T = (位x)^T x^T A^T = 位 x^T. Now, notice that if x is an eigenvector of A, then x^T is a left-eigenvector of A^T. Thus, 位 is indeed an eigenvalue of both A and A^T. As a result, A and A^T share the same set of eigenvalues.
02

Characteristic polynomial of A and A^T

The characteristic polynomial p(位) of a matrix A is given by the determinant of (A - 位I), where I is the identity matrix of the same size as A. And the characteristic polynomial of A^T is given by the determinant of (A^T - 位I). We need to show that these polynomials are the same. Using the property of determinants, we have: \(\text{det}(A^T - \lambda I) = \text{det}((A-\lambda I)^T)\) Since the determinant of a matrix and its transpose are equal, we get: \(\text{det}(A^T - \lambda I) = \text{det}(A - \lambda I)\). Thus, the characteristic polynomials of A and A^T are the same.
03

Minimal polynomial of A and A^T

Recall that the minimal polynomial m_A(位) of a matrix A is the monic polynomial of least degree such that m_A(A) = 0. Since A and A^T have the same characteristic polynomial, they also have the same algebraic multiplicities for their eigenvalues. Hence, for any polynomial P(位) that annihilates A, P(A) = 0, P(位) must also annihilate A^T, satisfying P(A^T) = 0. Now, consider the minimal polynomial m_A(位) of A. Since m_A(位) divides any polynomial P(位) that annihilates A, m_A(位) must also divide any polynomial Q(位) that annihilates A^T. Therefore, m_A(位) annihilates A^T as well. Similarly, if m_{A^T}(位) is the minimal polynomial of A^T, then m_{A^T}(位) annihilates A too. Since m_A(位) and m_{A^T}(位) are monic polynomials, and they annihilate both A and A^T, we can conclude that both minimal polynomials are the same, i.e., m_A(位) = m_{A^T}(位). Thus, a matrix A and its transpose A^T have the same minimal polynomial.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Eigenvalues and Their Role
Eigenvalues are scalars associated with a matrix that provide significant insights into the matrix's characteristics. If you have a matrix \(A\) and an eigenvalue \(\lambda\), there exists a non-zero vector \(x\) such that the matrix transformation \(Ax\) is proportional to \(x\), specifically \(Ax = \lambda x\). This relationship illustrates that \(\lambda\) is an eigenvalue of \(A\) with \(x\) as its eigenvector.
Eigenvalues are vital in understanding the properties of a matrix, including stability in systems and vibrational modes in physical structures. When dealing with the transpose of a matrix, \(A^T\), we find that the eigenvalues remain unchanged. This is because taking the transpose does not alter the scalar values that allow the transformation to remain scalar multiples of their original vectors.
  • Eigenvalues help determine the span of data in fields like machine learning and physics.
  • They are used to simplify matrix functions in various computations.
Understanding eigenvalues is crucial for diving deeper into the structure and behavior of matrices in different mathematical and applied contexts.
Matrix Transpose Basics
The transpose of a matrix, denoted as \(A^T\), is achieved by flipping the matrix over its diagonal. This means that the row and column indices are swapped. For example, if the element in the first row and second column of the matrix \(A\) is \(a_{12}\), then in \(A^T\), this element would be in the second row and first column.
The transpose operation has a variety of interesting properties:
  • The transpose of the transpose brings you back to the original matrix, i.e., \((A^T)^T = A\).
  • The determinant of a matrix is the same as the determinant of its transpose, \(\text{det}(A^T) = \text{det}(A)\).
  • Transposing a product of matrices \((AB)^T\) results in the reverse order of the product of their transposes \(B^T A^T\).
These properties make \(A^T\) an essential tool in linear algebra, simplifying many operations, including solving equations and finding characteristic polynomials.
Characteristic Polynomial Explained
The characteristic polynomial of a matrix \(A\) is a fundamental concept in understanding the matrix's properties. It is defined as the polynomial obtained from the determinant of \(A - \lambda I\), where \(I\) is the identity matrix and \(\lambda\) is a scalar. This polynomial is represented as \(p(\lambda) = \text{det}(A - \lambda I)\).
The roots of this polynomial correspond to the eigenvalues of the matrix. Since the transpose of a matrix has the same determinant properties, the characteristic polynomial of \(A\) and its transpose \(A^T\) are the same. This equality stems from the fact that determinants remain unchanged under transposition:
  • The characteristic polynomial provides a method to calculate eigenvalues.
  • This concept is crucial for studies involving linear transformations and system stability.
In essence, the characteristic polynomial not only helps in finding eigenvalues but also in understanding how transformations influence the matrix's original structure.
Algebraic Multiplicity Unveiled
Algebraic multiplicity refers to the number of times an eigenvalue appears as a root in the characteristic polynomial of a matrix. For a matrix \(A\) with a characteristic polynomial \(p(\lambda)\), if \(\lambda\) is a root with algebraic multiplicity \(m\), \(\lambda\) would appear exactly \(m\) times in the factorization of the polynomial.
It gives us insight into the linear dependency among eigenvectors associated with that eigenvalue. In essence, if the algebraic multiplicity of an eigenvalue is greater than one, it indicates that the matrix has multiple linearly dependent eigenvectors corresponding to that eigenvalue.
  • The matrix \(A\) and its transpose \(A^T\) have the same algebraic multiplicities for their eigenvalues.
  • This means they share the same minimal polynomial, as both the matrices are annihilated by the same polynomials.
Understanding algebraic multiplicity is fundamental to grasping the deeper algebraic structure of matrices and is essential when studying minimal polynomials in linear algebra.

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Most popular questions from this chapter

Let \(A=\left[\begin{array}{rr}1 & -2 \\ 4 & 5\end{array}\right] .\) Find \(f(A),\) where (a) \(f(t)=t^{2}-3 t+7\) (b) \(f(t)=t^{2}-6 t+13\)

Let \(A=\left[\begin{array}{rr}7 & 3 \\ 3 & -1\end{array}\right]\). Find an orthogonal matrix \(P\) such that \(D=P^{-1} A P\) is diagonal.First find the characteristic polynomial \(\Delta(t)\) of \(A\). We have $$ \Delta(t)=t^{2}-\operatorname{tr}(A) t+|A|=t^{2}-6 t-16=(t-8)(t+2) $$ Thus, the eigenvalues of \(A\) are \(\lambda=8\) and \(\lambda=-2\). We next find corresponding eigenvectors. Subtract \(\lambda=8\) down the diagonal of \(A\) to obtain the matrix $$ M=\left[\begin{array}{rr} -1 & 3 \\ 3 & -9 \end{array}\right], \quad \text { corresponding to } \quad \begin{array}{r} -x+3 y=0 \\ 3 x-9 y=0 \end{array} \text { or } x-3 y=0 $$ A nonzero solution is \(u_{1}=(3,1)\) Subtract \(\lambda=-2\) (or add 2 ) down the diagonal of \(A\) to obtain the matrix $$ M=\left[\begin{array}{ll} 9 & 3 \\ 3 & 1 \end{array}\right] \text {, corresponding to } \quad \begin{aligned} &9 x+3 y=0 \\ &3 x+y=0 \end{aligned} \quad \text { or } \quad 3 x+y=0 $$ A nonzero solution is \(u_{2}=(1,-3)\). As expected, because \(A\) is symmetric, the eigenvectors \(u_{1}\) and \(u_{2}\) are orthogonal. Normalize \(u_{1}\) and \(u_{2}\) to obtain, respectively, the unit vectors $$ \hat{u}_{1}=(3 / \sqrt{10}, 1 / \sqrt{10}) \quad \text { and } \quad \hat{u}_{2}=(1 / \sqrt{10},-3 / \sqrt{10}) $$ Finally, let \(P\) be the matrix whose columns are the unit vectors \(\hat{u}_{1}\) and \(\hat{u}_{2}\), respectively. Then $$ P=\left[\begin{array}{cc} 3 / \sqrt{10} & 1 / \sqrt{10} \\ 1 / \sqrt{10} & -3 / \sqrt{10} \end{array}\right] \quad \text { and } \quad D=P^{-1} A P=\left[\begin{array}{rr} 8 & 0 \\ 0 & -2 \end{array}\right] $$ As expected, the diagonal entries in \(D\) are the eigenvalues of \(A\).

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

Let \(A=\left[\begin{array}{ll}3 & -4 \\ 2 & -6\end{array}\right]\) (a) Find all eigenvalues and corresponding eigenvectors. (b) Find matrices \(P\) and \(D\) such that \(P\) is nonsingular and \(D=P^{-1} A P\) is diagonal. (a) First find the characteristic polynomial \(\Delta(t)\) of \(A\) : $$ \Delta(t)=t^{2}-\operatorname{tr}(A) t+|A|=t^{2}+3 t-10=(t-2)(t+5) $$ The roots \(\lambda=2\) and \(\lambda=-5\) of \(\Delta(t)\) are the eigenvalues of \(A\). We find corresponding eigenvectors. (i) Subtract \(\lambda=2\) down the diagonal of \(A\) to obtain the matrix \(M=A-2 I\), where the corresponding homogeneous system \(M X=0\) yields the eigenvectors corresponding to \(\lambda=2 .\) We have $$ M=\left[\begin{array}{ll} 1 & -4 \\ 2 & -8 \end{array}\right], \quad \text { corresponding to } \quad \begin{array}{r} x-4 y=0 \\ 2 x-8 y=0 \end{array} \text { or } x-4 y=0 $$ The system has only one free variable, and \(v_{1}=(4,1)\) is a nonzero solution. Thus, \(v_{1}=(4,1)\) is an eigenvector belonging to (and spanning the eigenspace of) \(\lambda=2\). (ii) Subtract \(\lambda=-5\) (or, equivalently, add 5) down the diagonal of \(A\) to obtain $$ M=\left[\begin{array}{ll} 8 & -4 \\ 2 & -1 \end{array}\right], \quad \text { corresponding to } \quad \begin{aligned} &8 x-4 y=0 \\ &2 x-y=0 \end{aligned} \text { or } 2 x-y=0 $$ The system has only one free variable, and \(v_{2}=(1,2)\) is a nonzero solution. Thus, \(v_{2}=(1,2)\) is an eigenvector belonging to \(\lambda=5\) (b) Let \(P\) be the matrix whose columns are \(v_{1}\) and \(v_{2}\). Then $$ P=\left[\begin{array}{ll} 4 & 1 \\ 1 & 2 \end{array}\right] \quad \text { and } \quad D=P^{-1} A P=\left[\begin{array}{rr} 2 & 0 \\ 0 & -5 \end{array}\right] $$ Note that \(D\) is the diagonal matrix whose diagonal entries are the eigenvalues of \(A\) corresponding to the eigenvectors appearing in \(P\).

For each of the following matrices, find all eigenvalues and corresponding linearly independent eigenvectors: (a) \(A=\left[\begin{array}{ll}2 & -3 \\ 2 & -5\end{array}\right]\) (b) \(\quad B=\left[\begin{array}{rr}2 & 4 \\ -1 & 6\end{array}\right]\) , (c) \(C=\left[\begin{array}{ll}1 & -4 \\ 3 & -7\end{array}\right]\) When possible, find the nonsingular matrix \(P\) that diagonalizes the matrix.
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