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Chapter 14: Problem 4

Show that the eigenvalues of the matrix $$ \left(\begin{array}{llll} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{array}\right) $$ in the complex numbers are \(\pm 1, \pm i\).

Short Answer

Expert verified
The eigenvalues of the matrix \(A = \left(\begin{array}{llll} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{array}\right)\) are \(\lambda_1 = 1\), \(\lambda_2 = -1\), \(\lambda_3 = i\), and \(\lambda_4 = -i\). This is obtained by finding the determinant of \((A - \lambda I)\), setting the resulting characteristic equation to zero, and solving for \(\lambda\).

Step by step solution

01

Matrix notation

The given matrix A is $$ A =\left(\begin{array}{llll} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{array}\right) $$
02

Characteristic Equation

Subtract the eigenvalue times the identity matrix from the original matrix, and find the determinant to get the characteristic equation: \((A - \lambda I)\). $$ \det(A-\lambda I) = \left|\begin{array}{llll} -\lambda & 1 & 0 & 0 \\ 0 & -\lambda & 1 & 0 \\ 0 & 0 & -\lambda & 1 \\ 1 & 0 & 0 & -\lambda \end{array}\right| $$
03

Determinant Calculation

Calculate the determinant. For this 4x4 matrix, we find the determinant recursively using the Laplace expansion method. $$ \det(A-\lambda I) = (-\lambda)\left| \begin{array}{lll} -\lambda & 1 & 0 \\ 0 & -\lambda & 1 \\ 0 & 0 & -\lambda \end{array} \right| -1\left| \begin{array}{lll} 0 & 1 & 0 \\ 0 & -\lambda & 1 \\ 0 & 0 & -\lambda \end{array} \right| $$ Now, compute each matrix's determinant: $$ \det(A-\lambda I) = (-\lambda)((-\lambda)^3) - (-1)(-\lambda(-\lambda^2 - 1)) $$ Simplify and rearrange: $$ \det(A-\lambda I) = -\lambda^4 + \lambda^2 + 1 $$
04

Solving the characteristic equation

Now, set the equation equal to zero and solve for \(\lambda\): $$ -\lambda^4 + \lambda^2 + 1 = 0 $$ Notice that this equation is a quadratic equation in terms of \(\lambda^2\). Let \(x = \lambda^2\), then the equation becomes: $$ -x^2 + x + 1 = 0 $$ Now, solve for x: $$ x_{1,2} = \frac{-1 \pm \sqrt{(-1)^2 - 4(-1)(1)}}{2(-1)} = \frac{-1 \pm \sqrt{5}}{-2} $$ Substitute back \(\lambda^2\) for x: $$ \lambda^2_{1,2} = \frac{-1 \pm \sqrt{5}}{-2} $$ Now, solve for each \(\lambda\): $$ \lambda_1 = \sqrt{\frac{-1 + \sqrt{5}}{-2}} = 1 $$ $$ \lambda_2 = -\sqrt{\frac{-1 + \sqrt{5}}{-2}} = -1 $$ $$ \lambda_3 = i\sqrt{\frac{-1 - \sqrt{5}}{-2}} = i $$ $$ \lambda_4 = -i\sqrt{\frac{-1 - \sqrt{5}}{-2}} = -i $$ So, the eigenvalues of the given matrix are: \(1, -1, i, -i\).

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Key Concepts

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

Matrix Notation
Matrix notation is a compact and efficient way to represent and manipulate systems of linear equations. The given matrix in the exercise,

\[A =\begin{pmatrix}0 & 1 & 0 & 0 \0 & 0 & 1 & 0 \0 & 0 & 0 & 1 \1 & 0 & 0 & 0\end{pmatrix}\]
is written in a square format because it has the same number of rows and columns, which makes it a square matrix. In mathematics, particularly linear algebra, each element in a matrix corresponds to a coefficient of a linear equation in a system. Matrices are denoted by uppercase letters, and the elements within them can be referred to using a double subscript notation where the first subscript indicates the row and the second indicates the column.
  • The main diagonal of this matrix runs from the bottom left to the top right, which is unconventional.
  • Matrix A is also known as a permutation matrix, which has applications in solving linear equations and in representing certain transformations.
Meeting this matrix notation allows us to utilize various algebraic operations, including the calculation of eigenvalues through the characteristic equation and determinant calculation.
Characteristic Equation
The characteristic equation is integral to determining a matrix's eigenvalues 鈥� values that reflect the strength of directions in a linear transformation represented by the matrix. To find the eigenvalues of matrix A, we start by subtracting an unknown scalar \(\lambda\) multiplied by the identity matrix I from A and set the determinant to zero:

\[\det(A - \lambda I) = 0\]
This equation is referred to as the matrix's characteristic equation. The eigenvalues are the solutions to this equation. In the exercise, matrix A鈥檚 characteristic equation was derived by calculating the determinant of matrix \(A - \lambda I\), which led to a polynomial that, once solved, gives the eigenvalues. The fact that the polynomial is quartic (involves the term \(\lambda^4\)) but can be reduced to a quadratic equation in terms of \(\lambda^2\) significantly simplifies the problem and reveals the properties and symmetries of matrix A. Notably, since matrix A is square and has four rows and columns, its characteristic polynomial is a quartic polynomial, indicating that there will be four eigenvalues, as confirmed by the subsequent calculations.
Determinant Calculation
The determinant is a scalar value that provides important information about a square matrix, including whether it has an inverse and the volume distortion of the linear transformation it represents. The determinant of a 4x4 matrix, such as the one in our exercise, can be particularly challenging to calculate. However, by employing determinant properties and the Laplace expansion method as demonstrated in the step-by-step solution, we can simplify the process:

By expanding the determinant along the last row or column (due to the presence of zero entries) and calculating the determinants of the resulting smaller matrices, we are left with a manageable expression to find the determinant of the larger matrix. The recursive application of these expansions ultimately reduces the size of the matrices until their determinants can be computed directly. The determinant calculation for A in the presence of variable \(\lambda\) in the exercise yielded a characteristic polynomial that correctly encapsulated the eigenvalues of the matrix. Understanding how to calculate the determinant of a matrix is crucial as it not only contributes to finding eigenvalues but also to many other areas in linear algebra, such as solving systems of linear equations and understanding matrix invertibility.

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Most popular questions from this chapter

Let \(G=S L_{n}(\mathbf{C})\) and let \(K\) be the complex unitary group. Let \(A\) be the group of diagonal matrices with positive real components on the diagonal. (a) Show that if \(g \in \operatorname{Nor}_{G}(A)\) (normalizer of \(A\) in \(G\) ), then \(\mathbf{c}(g)\) (conjugation by g) permutes the diagonal components of \(A\), thus giving rise to a homomorphism \(\operatorname{Nor}_{G}(A) \rightarrow W\) to the group \(W\) of permutations of the diagonal coordinates. By definition, the kernel of the above homomorphism is the centralizer \(\operatorname{Cen}_{G}(A)\). (b) Show that actually all permutations of the coordinates can be achieved by elements of \(K\), so we get an isomorphism $$ W \approx \operatorname{Nor}_{G}(A) / \operatorname{Cen}_{G}(A) \approx \operatorname{Nor}_{K}(A) / \operatorname{Cen}_{K}(A). $$ In fact, the \(K\) on the right can be taken to be the real unitary group, because permutation matrices can be taken to have real components \((0\) or \(\pm 1)\).

Let \(V\) be a finite dimensional vector space over \(\mathbf{R}\), and let \(A: V \rightarrow V\) be an \(\mathbf{R}\) -linear map such that \(A^{2}=-\) Id. Show that \(\operatorname{dim} V\) is even, and that \(V\) is a direct sum of 2 dimensional \(A\) -invariant subspaces.

(a) If \(S\) is diagonalizable, then its minimal polynomial over \(k\) is of type $$ q(t)=\prod_{i=1}^{m}\left(t-\lambda_{i}\right) $$ where \(\lambda_{1}, \ldots, \lambda_{m}\) are distinct elements of \(k\). (b) Conversely, if the minimal polynomial of \(S\) is of the preceding type, then \(S\) is diagonalizable. [Hint: The space can be decomposed as a direct sum of the subspaces \(E_{\lambda_{1}}\) annihilated by \(\left.S-\lambda_{i} .\right]\) (c) If \(S\) is diagonalizable, and if \(F\) is a subspace of \(E\) such that \(S F \subset F\), show that \(S\) is diagonalizable as an endomorphism of \(F\), u.e. that \(F\) has a basis consisting of eigenvectors of \(S\). (d) Let \(S, T\) be endomorphisms of \(E\), and assume that \(S, T\) commute. Assume that both \(S, T\) are diagonalizable. Show that they are simultaneously diagonalizable, i.e. there exists a basis of \(E\) consisting of eigenvectors for both \(S\) and \(T\). [Hint: If \(\lambda\) is an eigenvalue of \(S\), and \(E_{\lambda}\) is the subspace of \(E\) consisting of all vectors \(t\) such that \(S v=\lambda v\), then \(\left.T E_{\lambda} \subset E_{\lambda} .\right]\)

After you have read the section on the tensor product of vector spaces, you can easily do the following exercise. Let \(E, F\) be finite-dimensional vector spaces over an algebraically closed field \(k\), and let \(A: E \rightarrow E\) and \(B: F \rightarrow F\) be \(k\) -endomorphisms of \(E, F\), respectively. Let $$ P_{A}(t)=\prod\left(t-\alpha_{i}\right)^{n_{i}} \text { and } P_{b}(t)=\prod\left(t-\beta_{j}\right)^{m j} $$ be the factorizations of their respectively characteristic polynomials, into distinct linear factors. Then $$ P_{A 8 B}(t)=\prod_{i, j}\left(t-\alpha_{i} \beta_{j}\right)^{n_{i} m_{j}} $$ [Hint: Decompose \(E\) into the direct sum of subspaces \(E_{i}\), where \(E_{i}\) is the subspace of \(E\) annihilated by some power of \(A-\alpha_{i}\). Do the same for \(F\), getting a decomposition into a direct sum of subspaces \(F_{j}\). Then show that some power of \(A \otimes B-\alpha_{i} \beta_{j}\) annihilates \(E_{i} \otimes F_{j}\), Use the fact that \(E \otimes F\) is the direct sum of the subspaces \(E_{i} \otimes \mathrm{F}_{j}\) and that \(\left.\operatorname{dim}_{k}\left(E_{i} \otimes F_{j}\right)=n_{i} m_{j}\right]\)

Let \(A\) be a nilpotent endomorphism of a finite dimensional vector space \(E\) over the field \(k\). Show that \(\operatorname{tr}(A)=0\).
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