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The eigenvalue $λ$ and the eigenvector $φ$ of the matrix ${\bf A}$ satisfy the following equation.

$${\bf A}φ=λφ$$

Let’s see how to find this eigenvalue $λ$ and eigenvector $φ$.

Practice

You often see the problem of finding the eigenvalues and eigenvectors of the matrix ${\bf A}$ “, then if this is the case, the first thing you need is eigenvalues.

We’ll find out the eigenvalues and the eigenvectors of the following matrix.

$$\left(\begin{array}{cc} 3&4\\ 2&5\\ \end{array}\right)$$

How to obtain eigenvalues

First of all, we will obtain eigenvalues of a matrix. The first step is to assemble the equation $|{\bf A}-λ {\bf E}|=0$ with the eigenvalue as $λ$. ${\bf E}$ is an identity matrix. An identity matrix is a square matrix which the matrix elements whose row and column numbers are the same are \ (1 \) and other elements are $0$.

$${\bf E}=\left(\begin{array}{cccc} 1&0&\cdots&0\\0&1&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\0&0&0&1\end{array}\right)$$

Then the determinant $|{\bf A}-λ{\bf E}|$ is as follows.

\begin{eqnarray}\left|\begin{array}{cc} 3&4\\ 2&5\\  \end{array}\right|-λ\left|\begin{array}{cc} 1&0\\ 0&1\\  \end{array}\right|&=&\left|\begin{array}{cc} 3-λ&4\\ 2&5-λ\\  \end{array}\right|\\&=&(3-λ)(5-λ)-8\\&=&λ^{2}-8λ+7\end{eqnarray}

We will find the eigenvalues $λ$ which the determinant \ (| {\ bf A} – λ {\ bf E} | \) becomes 0.

$$λ^{2}-8λ+7=0$$

It is just a quadratic equation. If factoring the left-hand side, $λ$ will be the next two.

$$(λ-1)(λ-7)=0$$
$$λ=1,7$$

Thus, eigenvalues were obtained.

How to obtain eigenvectors

Since eigenvalues have been determined, eigenvectors of each eigenvalue can be obtained now.

The Eigenvector of the Eigenvalue $λ=1$

First consider the equation $({\bf A}-λ{\bf E})φ={\bf 0}$. In this practice, since the matrix ${\bf A}$ is a 2 × 2 matrix, the number of elements of the eigenvector is also 2. Substitute $λ = 1$ to obtain the following expression.

$$\left(\begin{array}{cc} 3-1&4\\ 2&5-1\\ \end{array}\right)\left(\begin{array}{c} x\\ y\\ \end{array}\right)=\left(\begin{array}{c} 0\\ 0\\ \end{array}\right)$$

Calculate the left side to obtain the following equation.

\begin{cases} 2x+4y=0 \\ 2x+4y=0 \end{cases}

In this example, two equations were obtained. Normally, at this stage, one equation is a real multiple of the other equation (it can be said that they are substantially the same function). If that does not happen, there is a possibility that the eigenvalues are wrong, so review the calculation once more.

When one of the above equations is transformed,

$$x=-2y$$

From this equation, the ratio of the x component and the y component of the eigenvector \ (φ \) is as follows.

$$x:y=-2:1$$

Given the above, the eigenvector \ (φ \) in \ (λ = 1 \) can be expressed as follows using the constant \ (C \).

Here, normalize the eigenvector $φ$. Normalization is to decide the constant \ (C \) so that the magnitude of the eigenvector is 1.

The magnitude of the eigenvector $|φ|$ in the case of not normalizing is

$$|φ|=\sqrt{(-2)^2+1^2}=\sqrt{5}$$

using the Pythagorean theorem.

Therefore, if $C=1/\sqrt{5}$, the magnitude of the eigenvector $|φ|$ is 1. Finally the standardized eigenvector has been found.

$$φ=\frac{1}{\sqrt{5}}\left(\begin{array}{c} -2\\ 1\\ \end{array}\right)$$

By the way, even if we decide the ratio of the x and y components of the eigenvector $φ$ is $2:-1$, the correct eigenvector is found. In that case, the eigenvector $φ$ is as follows.

$$φ=\frac{1}{\sqrt{5}}\left(\begin{array}{c} 2\\ -1\\ \end{array}\right)$$

In order to confirm both are correct answers, you only have to confirm that it meets ${\ bf A} φ=λφ$.

The Eigenvector of the Eigenvalue $λ=7$

In exactly the same way as above, eigenvectors in $λ=7$ are also obtained.

$$({\bf A}-7･{\bf E})φ={\bf 0}$$
\begin{cases} -4x+4y=0 \\ 2x-2y=0 \end{cases}

From these equations, the ratio of the x component and the y component of the eigenvector $φ$ is

$$x:y=1:1.$$

When eigenvector \ (φ \) is represented by a constant \ (C \),

$$φ=C\left(\begin{array}{c} 1\\ 1\\ \end{array}\right).$$

After that, normalize $|φ|$ and find the eigenvector $φ$.

$$φ=\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1\\ 1\\ \end{array}\right)$$

Thus, eigenvectors corresponding to each eigenvalue were obtained.