Question

'The product of the eigen values of \(\left[\begin{array}{rrr}1 & 0 & 0 \\ 0 & 3 & -1 \\ 0 & -1 & 3\end{array}\right]\) iss

Short answer

The product of the eigenvalues is 8.

Step-by-step solution

Understand the Problem

We need to find the product of the eigenvalues of the given matrix. The matrix is \[\begin{bmatrix} 1 & 0 & 0 \ 0 & 3 & -1 \ 0 & -1 & 3 \end{bmatrix}.\] Eigenvalues are scalars that satisfy the characteristic equation \(\det(A - \lambda I) = 0\), where \(\lambda\) are the eigenvalues and \(I\) is the identity matrix.

Identify the Special Matrix Type

Recognize that the given matrix is a block diagonal matrix. This is because it is divided into blocks such as the \(1 \times 1\) matrix [1] and the \(2 \times 2\) submatrix \(\begin{bmatrix} 3 & -1 \ -1 & 3 \end{bmatrix}\). Block diagonal matrices have eigenvalues that can be determined from the blocks separately.

Calculate Eigenvalues for Each Block

For the \(1 \times 1\) matrix [1], the eigenvalue is simply 1. For the \(2 \times 2\) matrix \(\begin{bmatrix} 3 & -1 \ -1 & 3 \end{bmatrix}\), find its eigenvalues using its characteristic polynomial. The characteristic polynomial is found by calculating \(\det(\begin{bmatrix} 3-\lambda & -1 \ -1 & 3-\lambda \end{bmatrix}) = 0\).

Compute Characteristic Polynomial for 2x2 Submatrix

Calculate the determinant of the matrix \(\begin{bmatrix} 3-\lambda & -1 \ -1 & 3-\lambda \end{bmatrix}\).\[(3-\lambda)(3-\lambda) - (-1)(-1) = \lambda^2 - 6\lambda + 8 = 0.\]

Solve the Characteristic Equation

Find the roots of the quadratic equation \(\lambda^2 - 6\lambda + 8 = 0\). Use the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1, b=-6, c=8\). Solving gives \(\lambda = \frac{6 \pm \sqrt{36 - 32}}{2} = \frac{6 \pm 2}{2}\).Thus, the eigenvalues are \(4\) and \(2\).

Calculate the Product of All Eigenvalues

Now, aggregate the eigenvalues from the blocks: 1 (from the first block), 4, and 2 (from the second block). The product of the eigenvalues is given by their multiplication: \[1 \times 4 \times 2 = 8.\]

Conclude with the Calculation Result

The product of the eigenvalues of the matrix \[\begin{bmatrix} 1 & 0 & 0 \ 0 & 3 & -1 \ 0 & -1 & 3 \end{bmatrix}\] is 8.

Key concepts

Block Diagonal Matrix

Block diagonal matrices are a special form of matrices with distinct submatrices along their main diagonal. These submatrices or "blocks" are separated by zero entries. For example, considering the matrix: \[\begin{bmatrix} 1 & 0 & 0 \ 0 & 3 & -1 \ 0 & -1 & 3 \end{bmatrix}\]We can identify two clear blocks: a \(1 \times 1\) block \([1]\) and a \(2 \times 2\) block \[\begin{bmatrix} 3 & -1 \ -1 & 3 \end{bmatrix}\]This formation simplifies calculations, particularly with eigenvalues. Each block's eigenvalues can be considered separately. Thus, you'd only need to find eigenvalues of individual blocks and then combine them.
For students, understanding this concept will ease the process of calculating the eigenvalues for larger matrices that can be broken down into smaller blocks.

Characteristic Polynomial

To find the eigenvalues of a matrix, you must compute the characteristic polynomial. This polynomial is derived from the matrix's determinant equation:\(\det(A - \lambda I) = 0\)where \(A\) is your matrix and \(I\) is the identity matrix of the same size as \(A\). For the \(2 \times 2\) submatrix \[\begin{bmatrix} 3 & -1 \ -1 & 3 \end{bmatrix}\]we calculate its characteristic polynomial by \(\det\begin{vmatrix} 3-\lambda & -1 \ -1 & 3-\lambda \end{vmatrix}\)The determinant results in \(\lambda^2 - 6\lambda + 8 = 0\).This polynomial is crucial because solving it will yield the eigenvalues, providing specific scalars that have important implications regarding the matrix's properties.

Quadratic Equation

In the context of matrices, the quadratic equation arises when determining the eigenvalues from the characteristic polynomial. Given our characteristic polynomial \(\lambda^2 - 6\lambda + 8 = 0\),we can solve for \(\lambda\) using the quadratic formula:\(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\),where \(a=1\), \(b=-6\), and \(c=8\). Here, solving gives us:

• \(4\)
• \(2\)

These eigenvalues are solutions to the quadratic equation derived from the characteristic polynomial. It is important for students to note that recognizing patterns like quadratic equations greatly facilitates finding eigenvalues when dealing with polynomial characteristic equations.

Matrix Determinant

The determinant of a matrix is a special number that can be calculated from its elements. It is essential in various matrix computations, such as solving systems of equations and finding eigenvalues. For a block diagonal matrix, like the one used in the exercise:\[\begin{bmatrix} 1 & 0 & 0 \ 0 & 3 & -1 \ 0 & -1 & 3 \end{bmatrix}\]you calculate the determinant for each block separately.
For a \(2 \times 2\) matrix \[\begin{bmatrix} 3 & -1 \ -1 & 3 \end{bmatrix}\]calculate \(\det = (3-\lambda)(3-\lambda) - (-1)(-1) = \lambda^2 - 6\lambda + 8\).Knowing how to calculate the determinant is key to forming the characteristic polynomial. Once the characteristic polynomial is formed, you can further deduce the eigenvalues from it, understanding their importance in analyzing and predicting matrix behavior.