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Chapter 7: Problem 33

Find all eigenvalues and corresponding eigenvectors of $$A=\left[\begin{array}{ccc} 1+i & 0 & 0 \\\2-2 i & 1-3 i & 0 \\\2 i & 0 & 1\end{array}\right]$$. Note that the eigenvectors do not occur in complex conjugate pairs. Does this contradict Theorem \(7.1 .8 ?\) Explain.

Short Answer

Expert verified
The eigenvalues of the matrix A are 位鈧� = 2 + i, 位鈧� = 2 - i, and 位鈧� = 1. Corresponding eigenvectors are: v鈧� = \(c_1\left[\begin{array}{c} 0 \\ 1 \\ -2i \end{array}\right]\), v鈧� = \(c_2\left[\begin{array}{c} 0 \\ 1 \\ 2 \end{array}\right]\), and v鈧� = \(c_3\left[\begin{array}{c} 2 \\ 4i \\ 1 \end{array}\right]\), where c鈧�, c鈧�, and c鈧� are scalar constants. The presence of eigenvectors that do not occur in complex conjugate pairs in this example does NOT contradict Theorem 7.1.8 since the matrix A does not have all real entries.

Step by step solution

01

Find the characteristic equation of the given matrix

The characteristic equation of a matrix A is given by the determinant of (A - 位I), where 位 is an eigenvalue and I is the identity matrix. The matrix (A - 位I) is given by: \(A - 位I = \left[\begin{array}{ccc} 1+i-位 & 0 & 0 \\ 2-2i & 1-3i-位 & 0 \\ 2i & 0 & 1-位 \end{array}\right]\) The determinant of this matrix will give us the characteristic equation.
02

Calculate the determinant of (A - 位I)

The determinant of the (A - 位I) matrix is: \(|A - 位I| = (1+i-位)((1-3i-位)(1-位))\) Now, we can simplify the determinant expression: \((1-位+i)(1-位-3i+3i位) = (1-位+i)(1-位+(-3+i位))\)
03

Solve the characteristic equation for eigenvalues

Now we need to find the eigenvalues by setting the determinant of (A - 位I) equal to zero: \((1-位+i)(1-位+(-3+i位)) = 0\) Upon expanding the equation, we get: \((1-位+i)(1-位-3+i\lambda) = (1-位-3\lambda + i\lambda -i\lambda^2 + 3i -3i\lambda -i^2\lambda^2) = 0\) Simplifying the equation: \(1-4\lambda + \lambda^2-2i\lambda = 0\) Solving the quadratic equation for 位, we get the eigenvalues: 位鈧� = 2 + i, 位鈧� = 2 - i, 位鈧� = 1
04

Find the eigenvectors corresponding to each eigenvalue

Now we will find the eigenvectors corresponding to each eigenvalue: For 位鈧� = 2 + i: \((A - (2+i)I)v_1 = 0\) Substitute the eigenvalue into the (A - 位I) matrix: \(\left[\begin{array}{ccc} (1+i)-(2+i) & 0 & 0 \\ 2-2i & (1-3i)-(2+i) & 0 \\ 2i & 0 & 1-(2+i) \end{array}\right]v_1 = 0\) Simplify the matrix and solve for v鈧�: \(\left[\begin{array}{ccc} -1 & 0 & 0 \\ 2-2i & -1-2i & 0 \\ 2i & 0 & -1-i \end{array}\right]v_1 = 0\) v鈧� = \(c_1\left[\begin{array}{c} 0 \\ 1 \\ -2i \end{array}\right]\), where c鈧� is a scalar. Now, for 位鈧� = 2 - i: \((A - (2-i)I)v_2 = 0\) Do the same procedure as above: \(\left[\begin{array}{ccc} (1+i)-(2-i) & 0 & 0 \\ 2-2i & (1-3i)-(2-i) & 0 \\ 2i & 0 & 1-(2-i) \end{array}\right]v_2 = 0\) Simplify the matrix and solve for v鈧�: \(\left[\begin{array}{ccc} 1 & 0 & 0 \\ 2-2i & -1+2i & 0 \\ 2i & 0 & -1+i \end{array}\right]v_2 = 0\) v鈧� = \(c_2\left[\begin{array}{c} 0 \\ 1 \\ 2 \end{array}\right]\), where c鈧� is a scalar. Finally, for 位鈧� = 1: \((A - (1)I)v_3 = 0\) Do the same procedure as before: \(\left[\begin{array}{ccc} (1+i)-(1) & 0 & 0 \\ 2-2i & (1-3i)-(1) & 0 \\ 2i & 0 & 1-(1) \end{array}\right]v_3 = 0\) Simplify the matrix and solve for v鈧�: \(\left[\begin{array}{ccc} i & 0 & 0 \\ 2-2i & -2i & 0 \\ 2i & 0 & 0 \end{array}\right]v_3 = 0\) v鈧� = \(c_3\left[\begin{array}{c} 2 \\ 4i \\ 1 \end{array}\right]\), where c鈧� is a scalar.
05

Answer the question about Theorem 7.1.8

Theorem 7.1.8 states that if a matrix A has real entries, then any complex eigenvalue must have a corresponding eigenvector that is a complex conjugate of the first eigenvector. In this case, the matrix A does not have all real entries, and so the theorem does not apply. As a result, we can have complex eigenvalues with eigenvectors that are not complex conjugate pairs. So, the presence of eigenvectors that do not occur in complex conjugate pairs in this example does NOT contradict Theorem 7.1.8.

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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 plays a pivotal role in the study of linear algebra, particularly in the analysis of eigenvalues and eigenvectors. It is derived from a square matrix and provides a polynomial that we solve to find the eigenvalues of the matrix. To formulate the characteristic equation, one must compute the determinant of the matrix subtracted by the product of the eigenvalue and the identity matrix, denoted as \( A - \lambda I \). This process converts our matrix problem into an algebraic one, allowing us to find eigenvalues as the roots of the characteristic equation.

As seen in the example, after setting the determinant of \( A - \lambda I \) to zero, we obtain a polynomial that, when simplified, gives us the desired equation. It is crucial to mention that the characteristic equation can produce real, complex, or a combination of both types of eigenvalues, depending on the properties of the matrix.
Complex Eigenvalues
When dealing with matrices with complex numbers, as in our example, it's not unusual to encounter complex eigenvalues. These are values for \(\lambda\) that include both a real part and an imaginary part, represented typically as \(\lambda = a + bi\), where \(i\) is the square root of -1. Complex eigenvalues often arise in systems that exhibit oscillatory behavior, such as in the case of differential equations modeling wave motion.

Despite the presence of complex numbers, the process of finding these eigenvalues remains the same: solve the characteristic equation. Complex eigenvalues do not always come in conjugate pairs when the matrix itself has complex entries, as reflection across the real axis is not a guaranteed symmetry. This is an important distinction, as it affects the eigenvectors and the potential real-world interpretations of the solution.
Determinant of a Matrix
The determinant is a scalar attribute of a square matrix that provides critical information about the matrix. For instance, a matrix is invertible if and only if its determinant is non-zero. The calculation of the determinant is based on a recursive process if the matrix is larger than 2x2, often involving minor matrices and cofactors, or via direct formula for 2x2 matrices.

In the context of eigenvalues, as demonstrated, the determinant of \( A - \lambda I \) is used to ascertain the characteristic equation. The value of the determinant helps us understand the geometric properties of the matrix, such as scaling factors in linear transformations, and is therefore key to identifying the eigenvalues.
Linear Algebra
Linear algebra is the branch of mathematics that deals with vectors, vector spaces, linear mappings, and the systems of linear equations. It is fundamental in various areas of mathematics and its applications, such as computer science, engineering, physics, and more. In our context, linear algebra provides the framework for understanding how transformations are applied to vectors and how systems can change over time.

Eigenvalues and eigenvectors are cornerstone concepts of linear algebra. They reveal much about the nature of a linear transformation represented by a matrix: eigenvalues tell us the factor by which a transformation scales space, while eigenvectors indicate the direction in which this scaling occurs. This understanding is not just theoretical 鈥� it has practical implications in studies of stability, oscillations, and even in sophisticated algorithms in machine learning.

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Most popular questions from this chapter

The characteristic polynomial \(p(\lambda)\) for a square matrix \(A\) is given. Write down a set \(S\) of matrices such that every square matrix with characteristic polynomial \(p(\lambda)\) is guaranteed to be similar to exactly one of the matrices in the set \(S\). \(p(\lambda)=(-2-\lambda)^{2}(6-\lambda)^{5}\).

We call a matrix \(B\) a square root of \(A\) if \(B^{2}=A\) (a) Show that if \(D=\operatorname{diag}\left(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}\right),\) then thematrix $$\sqrt{D}=\operatorname{diag}(\sqrt{\lambda_{1}}, \sqrt{\lambda_{2}}, \ldots, \sqrt{\lambda_{n}})$$ is a square root of \(D .\) (b) Show that if \(A\) is a nondefective matrix with \(S^{-1} A S=D\) for some invertible matrix \(S\) and diagonal matrix \(D,\) then \(S \sqrt{D} S^{-1}\) is a square root of \(A\). (c) Find a square root for the matrix $$A=\left[\begin{array}{rr}6 & -2 \\\\-3 & 7\end{array}\right]$$

Determine the general solution to the system \(\mathbf{x}^{\prime}=A \mathbf{x}\) for the given matrix \(A\). \(A=\left[\begin{array}{rr}-3 & -2 \\ 2 & 1\end{array}\right]\).

Determine all eigenvalues and corresponding eigenvectors of the given matrix. $$\left[\begin{array}{rrr}1 & 0 & 0 \\\0 & 0 & 1 \\\0 & -1 & 0\end{array}\right]$$.

Let \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) be eigenvectors of \(A\) corresponding to the distinct eigenvalues \(\lambda_{1}\) and \(\lambda_{2},\) respectively. Prove that \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) are linearly independent. [Hint: Model your proof on the general case considered in Theorem 7.2.5.]
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