**
What Eigenvalues and Eigenvectors Ask:
"When a vector x undergoes a linear transformation A, what vector remains parallel to the original vector x but only changes in magnitude?"
"Then, how much did the magnitude change?"
**

# What Does it Mean to Perform Matrix Operations on Vectors?

A matrix is an operation of linear transformation. The term ‘linear’ may sound complicated, so let’s just call it a transformation. What does it transform? A matrix transforms a vector into another vector^{1}.

Figure 1. A matrix is an operator that transforms a vector.

As we can see from Figure 1, when a vector is transformed using a matrix, the resulting vector ($A\vec{x}$) can change in both magnitude and direction compared to the original vector ($\vec{x}$). Let’s use the applet below to explore the results of linear transformations using arbitrary vectors and matrices.

# Meaning of Eigenvalues and Eigenvectors

As shown in Figure 1 and the applet above, when a vector undergoes matrix operations, it can result in a different vector than the original. However, for certain vectors and matrices, the linear transformation may only change the magnitude of the vector without changing its direction^{2}. For example, in the above applet, if we input the vector

and the matrix

\[\left[\begin{array}{c} 2 & 1\\ 1 & 2\end{array}\right] \notag\]and check the result, we can see that the output vector is parallel to the input vector but with a different magnitude.

Figure 2. Some vectors and matrices output vectors that are parallel to the input vector but with a different magnitude.

In other words, the result of linearly transforming the input vector $\vec{x}$ with matrix $A$ can be expressed as a scalar multiple of the input vector.

\[A\vec{x} = \lambda \vec{x}\]

Figure 3. Some vectors and matrices output vectors that are parallel to the input vector but with a different magnitude.

# Definition of Eigenvalues and Eigenvectors

DEFINITION 1. Definition of Eigenvalues and Eigenvectors |
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$$A\vec{x}=\lambda \vec{x}$$ The solution vector $\vec{x}$ is the corresponding eigenvector for eigenvalue $\lambda$. |

In this case, equation (2) can be rewritten by the properties of matrices as follows.

\[(A-\lambda I)\vec{x}=0\]Here, $I$ is the identity matrix.

There are two conditions for equation (3) to hold, which are when the expression inside the parentheses becomes 0, and when $\vec{x} = 0$. If only the first condition is met, we can find an appropriate $\lambda$ and a non-zero $\vec{x}$. However, if the second condition is satisfied, we obtain a ‘trivial solution’ with $\lambda$ and $\vec{x} =0$.

Therefore, in order to avoid obtaining the ‘trivial solution’ of $\vec{x}=0$ from the expression inside the parentheses of equation (3), the necessary and sufficient condition for having nontrivial solutions is

\[det(A-\lambda I)=0\]# Calculating Eigenvectors and Eigenvalues through an Example

Let’s consider the following matrix $A$:

\[A = \left[\begin{array}{c} 2 & 1\\ 1 & 2\end{array}\right]\]We will now find the eigenvalues and eigenvectors for this matrix.

According to the definition of eigenvalues and eigenvectors, the eigenvalue $\lambda$ and eigenvector $\vec{x}$ satisfy the following equation:

\[A\vec{x} = \lambda\vec{x}\]Therefore, by the properties of matrices, we have $(A-\lambda I)\vec{x}=0$. Moreover, in order for $\vec{x}$ to have nontrivial solutions, the following condition must be satisfied.

\[det(A-\lambda I) = 0\]Hence,

\[det(A-\lambda I) = det \left( \left[\begin{array}{c} 2 - \lambda& 1\\ 1 & 2 -\lambda\end{array}\right] \right) = 0\] \[\Rightarrow (2-\lambda)^2-1\] \[= (4-4\lambda + \lambda ^2)-1\] \[=\lambda ^2 -4\lambda + 3 = 0\]Therefore, $\lambda_1=1$ and $\lambda_2=3$.

In other words, the eigenvalues of the linear transformation $A$ are 1 and 3. This means that when a vector that does not change in direction but only scales in size is transformed, its size will be multiplied by 1 and 3, respectively. Now, let’s find the eigenvectors.

Again, Equation (2) must satisfy both eigenvalues $\lambda_1=1$ and $\lambda_2=3$. Therefore, for the case of $\lambda_1=1$,

\[A\vec{x} = \lambda_1 \vec{x}\] \[\Rightarrow \left[\begin{array}{c} 2 & 1\\ 1 & 2\end{array}\right] \left[\begin{array}{c} x_1\\ x_2\end{array}\right] = 1\left[\begin{array}{c} x_1\\ x_2\end{array}\right]\]must be satisfied, which leads to the following system of linear equations:

\[2x_1+x_2=x_1\] \[x_1+2x_2=x_2\]Therefore, the eigenvector for $\lambda_1=1$ is

\[\left[\begin{array}{c} x_1\\ x_2\end{array}\right]= \left[\begin{array}{c} 1\\ -1\end{array}\right]\]Similarly, using the same method, the eigenvector for $\lambda_2=3$ is

\[\left[\begin{array}{c} x_1\\ x_2\end{array}\right]= \left[\begin{array}{c} 1\\ 1\end{array}\right]\]Geometrically, this means that the vector $\vec{x} = [1,1]$ remains unchanged in direction but scales by a factor of 3 under the linear transformation $A$. Similarly, the vector $\vec{x} = [1, -1]$ remains unchanged in direction but scales by a factor of 1 under the linear transformation $A$.