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title: Eigenvalues and Eigenvectors of a Matrix

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# Eigenvalues and Eigenvectors of a Matrix

Eigenvaues and Eigenvectors give us an interesting interpertation about the relationship between certain matrices and specific vectors. These two concepts bring together nicely the idea of the product of a matrix $A$ with a vector $\vec x$ $\textit{transforming}$ $\vec x$ into the vector  $A \vec x$. But first, let us formally define what we mean by by Eigenvalues and Eigenvectors of a Matrix.

Let $A$ be an $n \times n$ matrix. Then an $\textbf{Eigenvector}$ of $A$ is a non-zero vector $\vec x$ where there exists a scalar $\lambda$ such that 

$$A\vec x = \lambda \vec x .$$


Stated another way, vector $\vec x$ is an $\textbf{Eigenvector}$ of $A$ provided that the vector $\vec x$ is not the zero vector, and vector $A \vec x$ is a $\textit{scalar multiple}$ of $\vec x$. It should be noted that the condition that $A$ be a $\textit{square matrix}$ is to ensure that $A \vec x$ is the same dimension as vector $\vec x$. That is, both $\vec x$ and vector $A \vec x$ will both be in $\mathbb{R}^n$ if and only if $A$ is an $n \times n$ matrix. 


We call the scalar $\lambda$ that makes the matrix equation true the corresponding $\textbf{Eigenvalue}$ of the $\textit{Eigenvector}$ $\vec x$.  

What is interesting to note about $\textit{Eigenvectors}$ is the fact the product of these vectors with a partiuclar matrix produces a vector with the same orientaton as the original vector. The resulting vector $A \vec x$ is a $\textit{dilation}$ or $\textit{contraction}$ of the initial vector $\vec x$ when $\vec x$ is an Eigenvector of Matrix $A$. This geometric intepertatioin is shown below.





![](https://i.imgur.com/ZHgqzsr.png)











$\underline{Example 1}$:

Let 

$$A = \begin{bmatrix} \ 8 & 0 \\ 0 & 8 \\ \end{bmatrix}.$$

Determine whether the vector $\vec x =  \begin{bmatrix} \ 3 \\ -7 \end{bmatrix}$ is an eigenvector of $A$. If so, determine the value of $\lambda$.


To determine whether or not $\vec x$ is an eignevector of $A$, we need to see if there exists a scalar $\lambda$ such that 


$$A \vec x = \lambda \vec x.$$   

Hence, we must evaluate the vector $A \vec x$. Doing so, we get



$$A \vec x = \begin{bmatrix} \ 8 & 0 \\ 0 & 8 \\ \end{bmatrix} \begin{bmatrix} \ 3 \\ -7 \end{bmatrix}. $$

$$A \vec x = \begin{bmatrix} \ 24 \\ -56 \end{bmatrix}$$



$$A \vec x = 8 \begin{bmatrix} \ 3 \\ -7 \end{bmatrix}$$


Since $A \vec x = 8 \vec x$, we see that $\vec x$ is an indeed an eigevector of matrix $A$ and that $\lambda = 8$. 


It should be noted, that for the matrix $A$ above, $\textit{every}$ vector in $\mathbb{R}^2$ is actually an eigenvector of $A$. To verify this, consider the product of an aribitary vector $\vec y$ in $\mathbb{R}^2$ with $A$. We will let 

$$ \vec y = \begin{bmatrix} \ y_1 \\ y_2 \end{bmatrix}.$$


It follows that 



$$A \vec y = \begin{bmatrix} \ 8 & 0 \\ 0 & 8 \\ \end{bmatrix} \begin{bmatrix} \ y_1 \\ y_2 \end{bmatrix}. $$ 



$$ A \vec y = \begin{bmatrix} \ 8y_1  \\  8y_2 \\ \end{bmatrix}. $$


$$ A \vec y = 8 \begin{bmatrix} \ y_1  \\ y_2 \\ \end{bmatrix}. $$








Observe that the equation $A \vec y = 8 \vec y$ implies that any vector $\vec y$ in $\mathbb{R}^2$ is an eigenvector of $A$. 



## Text Question


Let $B = \begin{bmatrix} \ 1 & 2 \\ 3 & 4 \\ \end{bmatrix}.$

Determine whether the vector $\vec x = \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}$ is an Eigenvector of $B$.



$$\textit{Citations}$$

Bourne, M. (n.d.). 7. Eigenvalues and Eigenvectors. Retrieved from https://www.intmath.com/matrices-determinants/7-eigenvalues-eigenvectors.php.


Kishor. (n.d.). Eigenvalues and Eigenvectors. Retrieved November 8, 2019, from https://math.mit.edu/~gs/linearalgebra/ila0601.pdf.


Lay, D. C. (2006). Linear Algebra and its Applications. Reading: Pearson AddisonWesley Publishing Company


Margalit, D., & Rabinoff, J. (n.d.). Interactive Linear Algebra [1553]. Retrieved from https://textbooks.math.gatech.edu/ila/1553/eigenvectors.html.



