Finding Eigenvalues and Eigenvectors | Generated by AI
Finding the eigenvalues and eigenvectors of a matrix is a fundamental concept in linear algebra. It involves solving the characteristic equation derived from the matrix.
🔢 Step 1: Set up the Characteristic Equation
For an \(n \times n\) square matrix \(\mathbf{A}\), the eigenvalues \((\lambda)\) and the corresponding non-zero eigenvectors \((\mathbf{v})\) satisfy the equation:
\(\)\mathbf{A}\mathbf{v} = \lambda\mathbf{v}\(\)
This equation can be rewritten as:
\(\)\mathbf{A}\mathbf{v} - \lambda\mathbf{v} = \mathbf{0}\(\) \(\)(\mathbf{A} - \lambda\mathbf{I})\mathbf{v} = \mathbf{0}\(\)
where \(\mathbf{I}\) is the \(n \times n\) identity matrix and \(\mathbf{0}\) is the zero vector.
For a non-zero solution \(\mathbf{v}\) to exist, the matrix \((\mathbf{A} - \lambda\mathbf{I})\) must be singular (not invertible). This means its determinant must be equal to zero:
\(\)\text{det}(\mathbf{A} - \lambda\mathbf{I}) = 0\(\)
This equation is called the characteristic equation.
💡 Step 2: Find the Eigenvalues (\(\lambda\))
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Form the matrix \((\mathbf{A} - \lambda\mathbf{I})\): Subtract \(\lambda\) from each element on the main diagonal of \(\mathbf{A}\).
- For a \(2 \times 2\) matrix \(\mathbf{A} = \begin{pmatrix} a & b \ c & d \end{pmatrix}\), the matrix is: \(\)\mathbf{A} - \lambda\mathbf{I} = \begin{pmatrix} a-\lambda & b \ c & d-\lambda \end{pmatrix}\(\)
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Calculate the determinant: Set \(\text{det}(\mathbf{A} - \lambda\mathbf{I}) = 0\).
- For the \(2 \times 2\) case, the characteristic equation is: \(\)(a-\lambda)(d-\lambda) - bc = 0\(\)
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Solve the resulting polynomial: This equation will be a polynomial in \(\lambda\) (called the characteristic polynomial). The roots of this polynomial are the eigenvalues of the matrix \(\mathbf{A}\).
🔎 Step 3: Find the Eigenvectors (\(\mathbf{v}\))
For each eigenvalue (\(\lambda_i\)) found in Step 2, you must solve the following system of linear equations to find its corresponding eigenvector \((\mathbf{v}_i)\):
\(\)(\mathbf{A} - \lambda_i\mathbf{I})\mathbf{v}_i = \mathbf{0}\(\)
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Substitute the eigenvalue: Substitute the specific eigenvalue \(\lambda_i\) back into the matrix equation \((\mathbf{A} - \lambda\mathbf{I})\mathbf{v} = \mathbf{0}\).
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Solve the system: Use techniques like Gaussian elimination to solve the resulting homogeneous system of equations. Since the matrix \((\mathbf{A} - \lambda_i\mathbf{I})\) is singular, you will always get an infinite number of solutions (a non-zero solution space).
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Express the eigenvector: The solutions define the eigenspace corresponding to \(\lambda_i\). You express the eigenvector \(\mathbf{v}_i\) in terms of one or more free variables, and usually choose a simple, non-zero vector to represent the direction of the eigenspace.
Example for a \(2 \times 2\) Matrix
Let \(\mathbf{A} = \begin{pmatrix} 2 & 1 \ 1 & 2 \end{pmatrix}\).
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Characteristic Equation: \(\)\text{det}\begin{pmatrix} 2-\lambda & 1 \ 1 & 2-\lambda \end{pmatrix} = 0\(\) \(\)(2-\lambda)(2-\lambda) - 1(1) = 0\(\) \(\)\lambda^2 - 4\lambda + 4 - 1 = 0\(\) \(\)\lambda^2 - 4\lambda + 3 = 0\(\)
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Eigenvalues: \(\)(\lambda - 3)(\lambda - 1) = 0\(\) The eigenvalues are \(\mathbf{\lambda_1 = 3}\) and \(\mathbf{\lambda_2 = 1}\).
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Eigenvectors:
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For \(\lambda_1 = 3\): \(\)\begin{pmatrix} 2-3 & 1 \ 1 & 2-3 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \implies \begin{pmatrix} -1 & 1 \ 1 & -1 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}\(\) The first row gives \(-v_1 + v_2 = 0\), or \(v_1 = v_2\). A simple, non-zero eigenvector is \(\mathbf{v_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix}}\).
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For \(\lambda_2 = 1\): \(\)\begin{pmatrix} 2-1 & 1 \ 1 & 2-1 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \implies \begin{pmatrix} 1 & 1 \ 1 & 1 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}\(\) The first row gives \(v_1 + v_2 = 0\), or \(v_1 = -v_2\). A simple, non-zero eigenvector is \(\mathbf{v_2 = \begin{pmatrix} 1 \ -1 \end{pmatrix}}\).
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Would you like to try finding the eigenvalues and eigenvectors for a different matrix?Finding the eigenvalues and eigenvectors of a matrix is a three-step process: setting up the characteristic equation, solving it for the eigenvalues, and then solving a system of linear equations for the corresponding eigenvectors.
1️⃣ Step 1: Set up the Characteristic Equation
For an \(n \times n\) matrix \(\mathbf{A}\), the relationship between an eigenvalue (\(\lambda\)) and its corresponding non-zero eigenvector (\(\mathbf{v}\)) is:
\(\)\mathbf{A}\mathbf{v} = \lambda\mathbf{v}\(\)
To find the eigenvalues, you must solve the characteristic equation, which is derived from the requirement that the system of equations \((\mathbf{A} - \lambda\mathbf{I})\mathbf{v} = \mathbf{0}\) must have non-zero solutions (i.e., the matrix is singular):
\(\)\text{det}(\mathbf{A} - \lambda\mathbf{I}) = 0\(\)
Here, \(\mathbf{I}\) is the \(n \times n\) identity matrix. To form \((\mathbf{A} - \lambda\mathbf{I})\), you simply subtract \(\lambda\) from every element on the main diagonal of \(\mathbf{A}\).
2️⃣ Step 2: Find the Eigenvalues (\(\lambda\))
- Form the Matrix: Calculate \(\mathbf{A} - \lambda\mathbf{I}\).
- Calculate the Determinant: Compute the determinant of the resulting matrix.
- Solve the Polynomial: Set the determinant to zero and solve the resulting polynomial equation in \(\lambda\). This polynomial is the characteristic polynomial, and its roots are the eigenvalues.
For a \(2 \times 2\) matrix \(\mathbf{A} = \begin{pmatrix} a & b \ c & d \end{pmatrix}\), the characteristic equation is: \(\)\text{det}\begin{pmatrix} a-\lambda & b \ c & d-\lambda \end{pmatrix} = (a-\lambda)(d-\lambda) - bc = 0\(\)
3️⃣ Step 3: Find the Eigenvectors (\(\mathbf{v}\))
For each eigenvalue (\(\lambda_i\)) found in Step 2, you must solve the following homogeneous system of equations for the non-zero vector \(\mathbf{v}_i\):
\(\)(\mathbf{A} - \lambda_i\mathbf{I})\mathbf{v}_i = \mathbf{0}\(\)
- Substitute \(\lambda_i\): Replace \(\lambda\) in the matrix \((\mathbf{A} - \lambda\mathbf{I})\) with the specific eigenvalue \(\lambda_i\).
- Solve the System: Use methods like Gaussian elimination to solve the system. Since the matrix is guaranteed to be singular, you will find a line or plane of solutions (an eigenspace).
- Express the Vector: Choose a simple, non-zero vector that satisfies the equations to represent the eigenvector \(\mathbf{v}_i\).
For example, if the solution leads to \(v_1 = 2v_2\), you can choose \(v_2 = 1\) and \(v_1 = 2\), making the eigenvector \(\mathbf{v} = \begin{pmatrix} 2 \ 1 \end{pmatrix}\).
Would you like to practice with a specific matrix?