How to Obtain Eigenvalues and Eigenvectors


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 \(φ\).


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.


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


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,


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


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


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


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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