How to Obtain Eigenvalues and Eigenvectors

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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.

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