Question

Find the eigenvalues of the given matrix. For each eigenvalue, give an eigenvector. $$\left[\begin{array}{cc}0 & 1 \\ 25 & 0\end{array}\right]$$

Short answer

Eigenvalues: 5 and -5. Eigenvectors: (1, 5) and (1, -5).

Step-by-step solution

Write the Characteristic Equation

To find the eigenvalues of a matrix, we need to solve the characteristic equation, which is given by \( \det(A - \lambda I) = 0 \). For the matrix \( A = \left[\begin{array}{cc}0 & 1 \ 25 & 0\end{array}\right] \), we substitute \( A \) into the equation: \[ \det\left(\begin{array}{cc}-\lambda & 1 \ 25 & -\lambda\end{array}\right) = 0.\]

Calculate the Determinant

Now we calculate the determinant of the matrix \( \left(\begin{array}{cc}-\lambda & 1 \ 25 & -\lambda\end{array}\right) \). The determinant is given by \( (-\lambda)(-\lambda) - (1)(25) = \lambda^2 - 25 \).

Solve the Characteristic Equation

The characteristic equation is \( \lambda^2 - 25 = 0 \). Solve for \( \lambda \) by setting the equation to zero: \[ \lambda^2 = 25. \] Taking the square root of both sides gives \( \lambda = 5 \) and \( \lambda = -5 \). So, the eigenvalues are \( 5 \) and \( -5 \).

Find Eigenvector for \( \lambda = 5 \)

For each eigenvalue, find the eigenvector by solving \( (A - \lambda I)\mathbf{v} = \mathbf{0} \). For \( \lambda = 5 \): \[ \left(\begin{array}{cc}-5 & 1 \ 25 & -5\end{array}\right)\left(\begin{array}{c}x \ y\end{array}\right) = \left(\begin{array}{c}0 \ 0\end{array}\right). \]This gives the system of equations:\(-5x + y = 0\) and \(25x - 5y = 0\).Simplify to get \( y = 5x \). Let \( x = 1 \), then \( y = 5 \). Eigenvector: \( \mathbf{v} = \left(\begin{array}{c}1 \ 5\end{array}\right) \).

Find Eigenvector for \( \lambda = -5 \)

For \( \lambda = -5 \): \[ \left(\begin{array}{cc}5 & 1 \ 25 & 5\end{array}\right)\left(\begin{array}{c}x \ y\end{array}\right) = \left(\begin{array}{c}0 \ 0\end{array}\right). \]This gives the system of equations:\(5x + y = 0\) and \(25x + 5y = 0\).Simplify to get \( y = -5x \). Let \( x = 1 \), then \( y = -5 \). Eigenvector: \( \mathbf{v} = \left(\begin{array}{c}1 \ -5\end{array}\right) \).

Key concepts

Understanding Eigenvectors

Eigenvectors are a critical concept in linear algebra. They are vectors that, when transformed by a matrix, result in a scaled version of themselves, specifically scaled by the eigenvalue. Eigenvectors offer insights into the properties of the matrix and help simplify complex transformations.
Imagine a matrix operating on a vector. The direction remains unchanged, though the magnitude might vary. This invariant property signifies an eigenvector. When dealing with square matrices, eigenvectors are paired with their respective eigenvalues. These pairs help in diagonalizing matrices, solving differential equations, and in many applications across physics and engineering.

• An eigenvector associated with eigenvalue \(\lambda\) satisfies \(A\mathbf{v} = \lambda \mathbf{v}\).
• This means the matrix \(A\) acts on vector \(\mathbf{v}\) and scales it by \(\lambda\).

Characteristic Equation Explained

To find eigenvalues of a matrix, the characteristic equation is your starting point. It is derived from the equation \(\det(A - \lambda I) = 0\), where \(A\) is a given matrix and \(I\) is the identity matrix of the same size. Here, \(\lambda\) represents eigenvalues.
The equation comes from setting up a new matrix, \(A - \lambda I\), which adjusts the original matrix by subtracting \(\lambda\) from each principal diagonal element. The aim is to find values of \(\lambda\) that make this matrix singular, i.e., its determinant is zero, reflecting that it has no inverse.

• \(\lambda\) represents the eigenvalues for which the matrix becomes singular.
• Solving \(\det(A - \lambda I) = 0\) yields a polynomial equation.

Polynomials obtained this way provide potential eigenvalues as their roots. Each root corresponds to one eigenvalue, critical for finding associated eigenvectors.

Matrices in Eigenvalue Problems

Matrices are structured arrays of numbers that represent linear transformations. When tackling eigenvalue problems, understanding how matrices operate is vital.
Basic operations include addition, subtraction, multiplication, and finding determinants. These operations are foundational for manipulating matrices to derive eigenvectors and eigenvalues. The matrix in our exercise is a 2x2 matrix represented as \(\begin{array}{cc} 0 & 1 \ 25 & 0 \end{array}\). This specific form hints at particular symmetries or properties, such as an interchange of elements, which will influence the solution.

• For 2x2 matrices like this one, finding the determinant plays a key role in solving their characteristic equation.
• These matrices are manageable, making hand-calculation and determinant solving straightforward processes.

Determinant Calculation Made Simple

The determinant of a matrix is a critical value used to analyze the matrix's properties. It helps determine if a matrix is invertible, its volume scaling factor in transformations, and in our case, plays a role in finding eigenvalues.
For a 2x2 matrix such as \(\begin{array}{cc} a & b \ c & d \end{array}\), the determinant \(|A|\) is given by \(ad - bc\). This determinant calculation is crucial to solving the characteristic equation.In our scenario, calculating the determinant of \(A - \lambda I\) leads us directly to the eigenvalues, as this determinant holds the key to whether a solution exists:

• A zero determinant means the matrix \(A - \lambda I\) is singular, indicating \(\lambda\) is an eigenvalue.
• For the problem matrix, we calculated it as \(\lambda^2 - 25\), highlighting the square nature making solution simple.