/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 Cayley-Hamilton Theorem In Exerc... [FREE SOLUTION] | 91Ó°ÊÓ

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Cayley-Hamilton Theorem In Exercises \(49-52\) demonstrate the Cayley-Hamilton Theorem for the matrix \(A\). The Cayley-Hamilton Theorem states that a matrix satisfies its characteristic equation. For example, the characteristic equation of \(A=\left[\begin{array}{rr}1 & -3 \\ 2 & 5\end{array}\right]\) is \(\lambda^{2}-6 \lambda+11=0,\) and by the theorem you have \(A^{2}-6 A+11 I_{2}=O\). $$A=\left[\begin{array}{rr}5 & 0 \\\\-7 & 3\end{array}\right]$$

Short Answer

Expert verified
The Cayley-Hamilton Theorem implies that a matrix \(A\) satisfies its characteristic equation. In this case, the matrix \(A=\left[\begin{array}{rr}5 & 0 \\-7 & 3\end{array}\right]\) has the characteristic equation \( \lambda^2 - 8\lambda + 15 = 0 \). When \(A\) is plugged into this equation, it results in a zero matrix, thereby verifying the theorem.

Step by step solution

01

Determine the Characteristic Equation

The characteristic equation of a matrix \(A\) can be found by taking the determinant of the matrix subtracted from a 2x2 identity matrix multiplied by an arbitrary variable (usually \( \lambda \)). The matrix \(A=\left[\begin{array}{rr}5 & 0 \\-7 & 3\end{array}\right]\) would have a characteristic equation of \( \text{det}(\lambda I-A) = 0 \). This would result in \( \lambda^2 - 8\lambda + 15 = 0 \).
02

Verify the Cayley-Hamilton Theorem

The Cayley-Hamilton Theorem states that the matrix \(A\) satisfies its characteristic equation. Therefore, if you plug in the matrix \(A\) into its characteristic equation ( \(A^{2}-8 A+15 I=0\)), you should get a zero matrix. By doing so, \[ \left[\begin{array}{rr}5 & 0 \\-7 & 3\end{array}\right]^2 -8*\left[\begin{array}{rr}5 & 0 \\-7 & 3\end{array}\right]+ 15*I=0 \] you'll find that indeed equals to the zero matrix, verifying the Cayley-Hamilton theorem.

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Key Concepts

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

Characteristic Equation
The characteristic equation of a matrix is a special equation that plays a crucial role in understanding the matrix's properties. It is typically represented as a polynomial equation where the matrix variable is replaced by the variable, usually denoted as \( \lambda \).

To find the characteristic equation of any square matrix \( A \), you would calculate \( \det(\lambda I - A) = 0 \), where \( \det \) stands for determinant, and \( I \) is the identity matrix of the same dimension as \( A \). The resulting polynomial's roots are the eigenvalues of the matrix \( A \). The given matrix \( A = \left[\begin{array}{rr}5 & 0 \-7 & 3\end{array}\right] \) thus has a characteristic equation \( \lambda^2 - 8\lambda + 15 = 0 \).
Matrix Satisfies Its Characteristic Equation
According to the Cayley-Hamilton Theorem, every square matrix \( A \) will satisfy its own characteristic equation. This means if you replace each \( \lambda \) in the characteristic polynomial with the matrix \( A \), you should end up with the zero matrix \( O \).

For our example matrix \( A \), the characteristic equation obtained is \( \lambda^2 - 8\lambda + 15 = 0 \). When applying Cayley-Hamilton, you replace \( \lambda \) with \( A \) to get \( A^2 - 8A + 15I = O \), where \( I \) is the identity matrix. Computing each term \( A^2 \), \( 8A \), and \( 15I \), and then combining them should result in the zero matrix, confirming that \( A \) indeed satisfies its characteristic equation.
Determinant of a Matrix
The determinant is a scalar attribute of a square matrix that provides important information about the matrix, such as whether it is invertible, and characteristics of its transformation in geometry. The determinant for a 2x2 matrix \( A = \left[\begin{array}{cc}a & b \c & d\end{array}\right] \) is calculated as \( ad - bc \).

In our example, the matrix \( A = \left[\begin{array}{rr}5 & 0 \-7 & 3\end{array}\right] \), the determinant is \( (5)(3) - (0)(-7) = 15 \), which is also the constant term in the characteristic equation. Det(A) is essential in finding the characteristic equation, where its calculation is part of the process to determine the polynomial.
Identity Matrix
An identity matrix \( I \) is a square matrix in which all the elements of the principal diagonal are ones and all other elements are zeros. For instance, a 2x2 identity matrix looks like \( I_2 = \left[\begin{array}{cc}1 & 0 \0 & 1\end{array}\right] \). It acts as the multiplicative identity in matrix algebra, meaning any matrix \( A \) when multiplied by the identity matrix of the same dimension results in \( A \) itself.

In the context of the Cayley-Hamilton Theorem, \( I \) is used with a scalar multiplier from the characteristic equation, such as \( 15I \) in our example, to maintain the matrix dimensions consistent when forming the theorem's expression to be satisfied.

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Most popular questions from this chapter

Use the Principal Axes Theorem to perform a rotation of axes to eliminate the \(x y\) -term in the quadratic equation. Identify the resulting rotated conic and give its equation in the new coordinate system. $$2 x^{2}+4 x y+2 y^{2}+6 \sqrt{2} x+2 \sqrt{2} y+4=0$$

Prove that \(A\) and \(A^{T}\) have the same eigenvalues. Are the eigenspaces the same?

In Exercises \(33-38,\) show that any two eigenvectors of the symmetric matrix corresponding to distinct eigenvalues are orthogonal. $$\left[\begin{array}{rr} -1 & -2 \\ -2 & 2 \end{array}\right]$$

(a) Explain how to model population growth using an age transition matrix and an age distribution vector, and how to find a stable age distribution vector. (b) Explain how to use a matrix equation to solve a system of first-order linear differential equations. (c) Explain how to use the Principal Axes Theorem to perform a rotation of axes for a conic and a quadric surface. (d) Explain how to solve a constrained optimization problem.

Prove that if \(A^{2}=O,\) then 0 is the only eigenvalue of \(A\) Getting Started: You need to show that if there exists a nonzero vector \(\mathbf{x}\) and a real number \(\lambda\) such that \(A \mathbf{x}=\lambda \mathbf{x},\) then if \(A^{2}=O, \lambda\) must be zero. (i) \(A^{2}=A \cdot A,\) so you can write \(A^{2} \mathbf{x}\) as \(A(A \mathbf{x})\) (ii) Use the fact that \(A \mathbf{x}=\lambda \mathbf{x}\) and the properties of matrix multiplication to show that \(A^{2} \mathbf{x}=\lambda^{2} \mathbf{x}\) (iii) \(A^{2}\) is a zero matrix, so you can conclude that \(\lambda\) must be zero.

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