Question

Find the eigenvalues and eigenvectors of the following matrices.$\left(\begin{array}{rrr}2& 0& 2\\ 0& 2& 0\\ 2& 0& -1\end{array}\right)$

Short answer

The eigenvalues are 2 (with multiplicity 2) and -3. Corresponding eigenvectors are $\left(\begin{array}{r}100\end{array}\right),\left(\begin{array}{r}010\end{array}\right)$ and $\left(\begin{array}{r}-205\end{array}\right)$.

Step-by-step solution

Understand the matrix

Identify the given matrix and its elements. The matrix provided is the 3x3 matrix $$A=\left(\begin{array}{rrrrrrr}2& 0& 20& 2& 02& 0& -1\end{array}\right)$$

Set up the characteristic equation

The characteristic equation is found using the determinant of $A-\lambda I$, where $\lambda$ is an eigenvalue and $I$ is the identity matrix. Set up the equation:$$\text{det}(A-\lambda I)=0$$$$A-\lambda I=\left(\begin{array}{rrrrrrr}2-\lambda & 0& 20& 2-\lambda & 02& 0& -1-\lambda \end{array}\right)$$

Calculate the determinant

Find the determinant of matrix $A-\lambda I$. The determinant is expanded as follows:$$\text{det}(A-\lambda I)=(2-\lambda )[(2-\lambda )(-1-\lambda )]-2[0-2(2-\lambda )]$$Simplify the determinant calculation:$$(2-\lambda )({\lambda }^2-\lambda -2)+4(2-\lambda )=(2-\lambda )({\lambda }^2-\lambda -6)$$

Solve for $\lambda$

The simplified determinant equation is:$$(2-\lambda {)}^2(\lambda +3)=0$$Thus, the eigenvalues are given by:$${\lambda }_1=2,{\lambda }_2=2,{\lambda }_3=-3$$

Find the eigenvectors for each eigenvalue

For each eigenvalue, solve $(A-\lambda I)\mathbf{v}=0$ to find the corresponding eigenvector $\mathbf{v}$. For ${\lambda }_1=2$:$$(A-2I)\mathbf{v}=\left(\begin{array}{rrrrrrr}0& 0& 20& 0& 02& 0& -3\end{array}\right)\mathbf{v}=0$$Solving this system, we get the eigenvector ${\mathbf{v}}_{\mathbf{1}}=\left(\begin{array}{r}100\end{array}\right)$.For ${\lambda }_2=2$:$$(A-2I)\mathbf{v}=\left(\begin{array}{rrrrrrr}0& 0& 20& 0& 02& 0& -3\end{array}\right)\mathbf{v}=0$$Solving this system, we get the eigenvector ${\mathbf{v}}_{\mathbf{2}}=\left(\begin{array}{r}010\end{array}\right)$.For ${\lambda }_3=-3$:$$(A+3I)\mathbf{v}=\left(\begin{array}{rrrrrrr}5& 0& 20& 5& 02& 0& 2\end{array}\right)\mathbf{v}=0$$Solving this system, we get the eigenvector ${\mathbf{v}}_{\mathbf{3}}=\left(\begin{array}{r}-205\end{array}\right)$.

Key concepts

Characteristic Equation

To find the eigenvalues of a matrix, we start by setting up the characteristic equation. This equation is derived from the matrix minus the eigenvalue times the identity matrix. In mathematical terms, this is represented as $A-\lambda I$, where \A\ is our given matrix, \lambda\ is the eigenvalue, and \I\ is the identity matrix.

The characteristic equation is found by computing the determinant of $A-\lambda I$ and setting it equal to zero. This yields an equation in \lambda\, which we can solve to find the eigenvalues. For example, if we have a 3x3 matrix $\left(\begin{array}{rrrrrrr}2& 0& 20& 2& 02& 0& -1\end{array}\right)$, we set up the equation:

$$\text{det}(A-\lambda I)=0$$

and proceed with finding the determinant.

Determinant

The determinant is a scalar value that can be computed from a square matrix and provides important information about the matrix, including solutions to linear systems, and eigenvalues. To find the determinant of a matrix $A-\lambda I$, for our example matrix, we get:

\A - \lambda I = \left($$\begin{array}{rrrrrrr}2-\lambda & 0& 20& 2-\lambda & 02& 0& -1-\lambda \end{array}$$\right)\

The determinant is calculated by expanding this matrix. For our matrix, it simplifies to:

$$(2-\lambda )[(2-\lambda )(-1-\lambda )]-2[0-2(2-\lambda )]$$

which upon further simplification gives:

$$(2-\lambda {)}^2(\lambda +3)=0$$

• Each term in the equation is solved to find the eigenvalues: ${\lambda }_1=2$, ${\lambda }_2=2$, and ${\lambda }_3=-3$.

Eigenvector Calculation

After finding the eigenvalues, the next step is to find the corresponding eigenvectors. For each eigenvalue, we solve the equation $A-\lambda I$\textbf{v} = 0\) to find the vector \textbf{v}\ that satisfies this.

For our matrix and eigenvalue ${\lambda }_1=2$, we have:
\A - 2I = \left($$\begin{array}{rrrrrrr}0& 0& 20& 0& 02& 0& -3\end{array}$$\right)\

Solving the equation $\left(\begin{array}{rrrrrrr}0& 0& 20& 0& 02& 0& -3\end{array}\right)\mathbf{v}=0$ gives us the eigenvector ${\mathbf{v}}_{\mathbf{1}}=\left(\begin{array}{r}100\end{array}\right)$.

We perform similar steps for ${\lambda }_2=2$ and get the eigenvector ${\mathbf{v}}_{\mathbf{2}}=\left(\begin{array}{r}010\end{array}\right)$.

For ${\lambda }_3=-3$, the matrix changes to:
$A+3I=\left(\begin{array}{rrrrrrr}5& 0& 20& 5& 02& 0& 2\end{array}\right)Solving\text{(}\left(\begin{array}{rrrrrrr}5& 0& 20& 5& 02& 0& 2\end{array}\right)\mathbf{v}=0$ results in the eigenvector ${\mathbf{v}}_{\mathbf{3}}=\left(\begin{array}{r}-205\end{array}\right)$. These eigenvectors help describe the directions in which the linear transformation acts by only scaling.