Given a square matrix $A$, an eigenvalue $\lambda$ and an eigenvector $v$ satisfy the equation:
\[A v = \lambda v\]This can be rearranged to:
\[(A - \lambda I) v = 0\]where $I$ is the identity matrix. For non-trivial solutions (where $v$ is not the zero vector), the determinant of $(A - \lambda I)$ must be zero:
\[\det(A - \lambda I) = 0\]Solving this characteristic equation will yield the eigenvalues. Substituting each eigenvalue back into the equation $(A - \lambda I) v = 0$ allows for solving the corresponding eigenvectors.
Example: Calculating Eigenvalues
For the matrix:
\[A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}\]The characteristic equation is:
\[\det \left( \begin{bmatrix} 2 - \lambda & 1 \\ 1 & 2 - \lambda \end{bmatrix} \right) = 0\]which simplifies to:
\[(2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = 0\]The solutions are $\lambda = 1$ and $\lambda = 3$. Substituting these back into the eigenvector equation provides the corresponding eigenvectors.
Example: Calculating Eigenvectors
Given the matrix $A$:
\[A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}\]We found the eigenvalues $\lambda = 1$ and $\lambda = 3$. Now, we’ll calculate the eigenvectors corresponding to each eigenvalue.
Eigenvector for $\lambda = 1$
Substitute $\lambda = 1$ into the equation $(A - \lambda I)v = 0$:
\[\begin{bmatrix} 2 - 1 & 1 \\ 1 & 2 - 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\]This simplifies to:
\[\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\]Which further reduces to $x + y = 0$. Solving for $y$, we have $y = -x$. Thus, the eigenvectors are any scalar multiple of:
\[\begin{bmatrix} 1 \\ -1 \end{bmatrix}\]Eigenvector for $\lambda = 3$
Substitute $\lambda = 3$ into the equation $(A - \lambda I)v = 0$:
\[\begin{bmatrix} 2 - 3 & 1 \\ 1 & 2 - 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\]This simplifies to:
\[\begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\]Which reduces to $-x + y = 0$. Solving for $y$, we have $y = x$. Thus, the eigenvectors are any scalar multiple of:
\[\begin{bmatrix} 1 \\ 1 \end{bmatrix}\]For the matrix $A$, the eigenvectors corresponding to $\lambda = 1$ are any scalar multiples of $\begin{bmatrix} 1 \ -1 \end{bmatrix}$, and for $\lambda = 3$, they are any scalar multiples of $\begin{bmatrix} 1 \ 1 \end{bmatrix}$. These vectors indicate the directions in which the matrix $A$ stretches the space by the factors $1$ and $3$, respectively.
Consider the matrix $B$:
\[B = \begin{bmatrix} 3 & 2 & 4 \\ 2 & 0 & 2 \\ 4 & 2 & 3 \end{bmatrix}\]Step 1: Find the Eigenvalues
First, we need to determine the eigenvalues by solving the characteristic equation derived from $\det(B - \lambda I) = 0$.
\[\det \left( \begin{bmatrix} 3 - \lambda & 2 & 4 \\ 2 & -\lambda & 2 \\ 4 & 2 & 3 - \lambda \end{bmatrix} \right) = 0\]Expanding this determinant:
\[(3-\lambda)\left((-\lambda)(3-\lambda) - 4\right) - 2\left(2(3-\lambda) - 8\right) + 4\left(4 - 2\lambda\right) = 0\]Simplifying, we find:
\[-\lambda^3 + 6\lambda^2 + 8\lambda - 48 = 0\]The eigenvalues are found to be $\lambda_1 = 8$, $\lambda_2 = 4$, and $\lambda_3 = -4$.
Step 2: Find the Eigenvectors
For $\lambda_1 = 8$:
Substitute $\lambda = 8$ back into the equation $(B - 8I)v = 0$:
\[\begin{bmatrix} -5 & 2 & 4 \\ 2 & -8 & 2 \\ 4 & 2 & -5 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\]Solving this system (e.g., using row reduction), we find the eigenvector:
\[\begin{bmatrix} x \\ y \\ z \end{bmatrix} = t \begin{bmatrix} 2 \\ 1 \\ -2 \end{bmatrix}, \quad t \in \mathbb{R}\]For $\lambda_2 = 4$:
Substitute $\lambda = 4$ into $(B - 4I)v = 0$:
\[\begin{bmatrix} -1 & 2 & 4 \\ 2 & -4 & 2 \\ 4 & 2 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\]Solving this, we find the eigenvector:
\[\begin{bmatrix} x \\ y \\ z \end{bmatrix} = s \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix}, \quad s \in \mathbb{R}\]For $\lambda_3 = -4$:
Substitute $\lambda = -4$ into $(B - (-4)I)v = 0$:
\[\begin{bmatrix} 7 & 2 & 4 \\ 2 & 4 & 2 \\ 4 & 2 & 7 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\]Solving this, we find the eigenvector:
\[\begin{bmatrix} x \\ y \\ z \end{bmatrix} = u \begin{bmatrix} 2 \\ -5 \\ 2 \end{bmatrix}, \quad u \in \mathbb{R}\]Consider a matrix $A$ defined as:
\[A = \begin{bmatrix} 1 & -2 \\ 1 & 3 \end{bmatrix}\]Finding Complex Eigenvalues
First, we determine the eigenvalues by setting up and solving the characteristic equation derived from $\text{det}(A - \lambda I) = 0$.
The characteristic polynomial is calculated as: \((1-\lambda)(3-\lambda) + 2 = \lambda^2 - 4\lambda + 5 = 0\)
Solving this using the quadratic formula yields: \(\lambda = \frac{4 \pm \sqrt{16 - 20}}{2} = \frac{4 \pm \sqrt{-4}}{2} = 2 \pm i\)
Finding the Corresponding Eigenvectors
For $\lambda = 2 + i$: Substitute $\lambda = 2 + i$ into $(A - \lambda I)v = 0$:
\[\begin{bmatrix} -1 - i & -2 \\ 1 & 1 - i \end{bmatrix}\]Through row reduction and simplification, we find that an eigenvector corresponding to $\lambda = 2 + i$ is:
\[v_1 = \begin{bmatrix} -1 + i \\ 1 \end{bmatrix}\]For $\lambda = 2 - i$: Similarly, substituting $\lambda = 2 - i$ yields another eigenvector:
\[v_2 = \begin{bmatrix} -1 - i \\ 1 \end{bmatrix}\]