Real eigenvalues and eigenvectors of 3x3 matrices, example 2
In these examples, the eigenvalues of matrices will turn out to be real values. In other words, the eigenvalues and eigenvectors are in \(\mathbb{R}^n\).
Suppose we have the following matrix:
\[\mathbf{A} = \begin{pmatrix} 8 & 0 & -5 \\ 9 & 3 & -6 \\ 10 & 0 & -7 \end{pmatrix}.\]
The objective is to find the eigenvalues and the corresponding eigenvectors.
Characteristic equation
Firstly, formulate the characteristic equation and solve it. The solutions are the eigenvalues of matrix \(\mathbf{A}\).
If \(\mathbf{I}\) is the \(3 \times 3\) identity matrix and \(\lambda\) represents the scalar eigenvalue, the characteristic equation is:
\[\det(\mathbf{A} - \lambda \mathbf{I}) = 0.\]
Written in determinant form:
\[\begin{vmatrix} 8-\lambda & 0 & -5 \\ 9 & 3-\lambda & -6 \\ 10 & 0 & -7-\lambda \end{vmatrix} = 0.\]
Choose the row or column easiest to use to compute the determinant. In this case, we expand along the second column because it contains two zeros and only one non-zero entry, \((3-\lambda)\):
\[-0 \begin{vmatrix} 9 & -6 \\ 10 & -7-\lambda \end{vmatrix} + (3-\lambda) \begin{vmatrix} 8-\lambda & -5 \\ 10 & -7-\lambda \end{vmatrix} - 0 \begin{vmatrix} 8-\lambda & -5 \\ 9 & -6 \end{vmatrix} = 0,\]
\[(3-\lambda) \begin{vmatrix} 8-\lambda & -5 \\ 10 & -7-\lambda \end{vmatrix} = 0.\]
Simplifying the \(2 \times 2\) determinant:
\[\begin{aligned} (3-\lambda)\left[(8-\lambda)(-7-\lambda) - (-5)(10)\right] &= 0, \\ (3-\lambda)\left[-56 - 8\lambda + 7\lambda + \lambda^2 + 50\right] &= 0, \\ (3-\lambda)(\lambda^2 - \lambda - 6) &= 0. \end{aligned}\]
Factoring the quadratic term gives a manageable form:
\[(3-\lambda)(\lambda + 2)(\lambda - 3) = 0,\]
or equivalently:
\[-(\lambda + 2)(\lambda - 3)^2 = 0.\]
The solutions for \(\lambda\) are immediately apparent:
\[\lambda = -2 \quad \vee \quad \lambda = 3 \quad (\text{multiplicity } 2).\]
Verifying the eigenvalues
We can verify these values by checking the matrix trace and determinant:
- Trace check: The sum of the eigenvalues must equal the trace of \(\mathbf{A}\) (the sum of the main diagonal elements): \[\text{tr}(\mathbf{A}) = 8 + 3 - 7 = 4 \quad \Longleftrightarrow \quad (-2) + 3 + 3 = 4. \quad \checkmark\]
- Determinant check: The product of the eigenvalues must equal \(\det(\mathbf{A})\): \[\det(\mathbf{A}) = 3\left[(8)(-7) - (-5)(10)\right] = 3(-56 + 50) = -18 \quad \Longleftrightarrow \quad (-2) \times 3 \times 3 = -18. \quad \checkmark\]
Eigenvector equations
To determine the eigenvectors, let:
\[\mathbf{v} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix},\]
satisfying:
\[(\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0}.\]
In matrix form:
\[\begin{pmatrix} 8-\lambda & 0 & -5 \\ 9 & 3-\lambda & -6 \\ 10 & 0 & -7-\lambda \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix},\]
which gives the system of linear equations:
\[\begin{cases} (8-\lambda)x_1 - 5x_3 = 0, \\ 9x_1 + (3-\lambda)x_2 - 6x_3 = 0, \\ 10x_1 + (-7-\lambda)x_3 = 0. \end{cases}\]
Finding the eigenvectors
Eigenvalue \(\lambda = -2\)
Substituting \(\lambda = -2\) into the eigenvector equations:
\[\begin{cases} (8 - (-2))x_1 - 5x_3 = 0 \implies 10x_1 - 5x_3 = 0, \\ 9x_1 + (3 - (-2))x_2 - 6x_3 = 0 \implies 9x_1 + 5x_2 - 6x_3 = 0, \\ 10x_1 + (-7 - (-2))x_3 = 0 \implies 10x_1 - 5x_3 = 0. \end{cases}\]
The first and third equations both reduce to \(2x_1 = x_3\). Substituting \(x_3 = 2x_1\) into the second equation:
\[9x_1 + 5x_2 - 6(2x_1) = 0 \implies -3x_1 + 5x_2 = 0 \implies 3x_1 = 5x_2.\]
Setting the smallest positive integer values, if \(x_1 = 5\), then \(x_2 = 3\) and \(x_3 = 10\).
Thus, an eigenvector corresponding to \(\lambda = -2\) is:
\[\mathbf{v}_1 = \begin{pmatrix} 5 \\ 3 \\ 10 \end{pmatrix}.\]
We verify this via matrix-vector multiplication \(\mathbf{A}\mathbf{v} = \lambda \mathbf{v}\):
\[\begin{pmatrix} 8 & 0 & -5 \\ 9 & 3 & -6 \\ 10 & 0 & -7 \end{pmatrix} \begin{pmatrix} 5 \\ 3 \\ 10 \end{pmatrix} = \begin{pmatrix} 40 + 0 - 50 \\ 45 + 9 - 60 \\ 50 + 0 - 70 \end{pmatrix} = \begin{pmatrix} -10 \\ -6 \\ -20 \end{pmatrix} = -2 \begin{pmatrix} 5 \\ 3 \\ 10 \end{pmatrix}. \quad \checkmark\]
Eigenvalue \(\lambda = 3\)
Substituting \(\lambda = 3\) into the eigenvector equations:
\[\begin{cases} (8 - 3)x_1 - 5x_3 = 0 \implies 5x_1 - 5x_3 = 0, \\ 9x_1 + (3 - 3)x_2 - 6x_3 = 0 \implies 9x_1 - 6x_3 = 0, \\ 10x_1 + (-7 - 3)x_3 = 0 \implies 10x_1 - 10x_3 = 0. \end{cases}\]
The first and third equations require \(x_1 = x_3\), while the second equation requires \(9x_1 = 6x_3\) (or \(3x_1 = 2x_3\)). The only simultaneous solution is \(x_1 = 0\) and \(x_3 = 0\).
Because the term \((3-\lambda)x_2\) becomes \(0 \cdot x_2 = 0\), \(x_2\) is completely unconstrained and acts as a free variable. Setting \(x_2 = 1\) yields the eigenvector:
\[\mathbf{v}_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}.\]
Verification:
\[\begin{pmatrix} 8 & 0 & -5 \\ 9 & 3 & -6 \\ 10 & 0 & -7 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 3 \\ 0 \end{pmatrix} = 3 \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}. \quad \checkmark\]
References and downloads
- Hero banner: Handwritten mathematics notes by KJ Runia.