Eigenvalues
- Eigenvalues are denoted as λ (Lambda).
- The following equation is used to calculate the eigenvalues of a matrix.
\[\Large
det(A – \lambda I) = 0
\]
\[\Large
|A – \lambda I| = 0
\]
- The above equation is also called the characteristic equation.
Calculating eigenvalues
Now let’s calculate the eigenvalues of the following 2 × 2 matrix step-by-step.
\[\Large
A =
\begin{bmatrix}
4 & 1 \\
3 & 2
\end{bmatrix}
\]
Step 1: Substitute values
- Substitute the matrix to A.
- Substitute an identity matrix (Of the same order) to the I.
\[\Large
|A – \lambda I| = 0
\]
\[\Large A – \lambda I = \begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix} – \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] \[\Large = \begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix} – \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} \] \[\Large = \begin{bmatrix} 4 – \lambda & 1 \\ 3 & 2 – \lambda \end{bmatrix} \]
\[\Large A – \lambda I = \begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix} – \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] \[\Large = \begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix} – \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} \] \[\Large = \begin{bmatrix} 4 – \lambda & 1 \\ 3 & 2 – \lambda \end{bmatrix} \]
Step 2: Find the determinant of the above resulting expression
To learn about how to find the determinant of a 2 × 2 matrix, go through Determinant of a 2 × 2 matrix.
\[\Large
|A – \lambda I| =
\left\vert
\begin{array}{ccc}
4 – \lambda & 1 \\
3 & 2 – \lambda
\end{array}
\right\vert
= 0
\]
\[\Large \color{purple}{Determinant,} \] \[\Large (4-\lambda)(2-\lambda)-3=0 \] \[\Large 8-4\lambda-2\lambda+\lambda^{2}-3=0 \] \[\Large \lambda^{2}-6\lambda+5 = 0 \]
\[\Large \color{purple}{Determinant,} \] \[\Large (4-\lambda)(2-\lambda)-3=0 \] \[\Large 8-4\lambda-2\lambda+\lambda^{2}-3=0 \] \[\Large \lambda^{2}-6\lambda+5 = 0 \]
- Now factorize the above quadratic equation.
\[\Large
\lambda^{2}-5\lambda-\lambda+5=0
\]
\[\Large
\lambda(\lambda-5)-1(\lambda-5)=0
\]
\[\Large
(\lambda-5)(\lambda-1)=0
\]
Step 3: Find the eigenvalues
- The solutions (or roots) of the above quadratic equations will we eigenvalues.
\[\Large
\textcolor{purple}{if,~~~}
(\lambda-5)(\lambda-1)=0
\]
\[\Large
\lambda=5 ~~or~~ \lambda=1
\]
\[\Large ∴~~\lambda_{1}=5 ~~and~~ \lambda_{2}=1 \]
\[\Large ∴~~\lambda_{1}=5 ~~and~~ \lambda_{2}=1 \]
- The above two values are the eigenvalues of the matrix ‘A’.