# 14.1 Eigenvectors and eigenvalues

 Page 1 / 1
This module defines eigenvalues and eigenvectors and explains a method of finding them given a matrix. These ideas are presented, along with many examples, in hopes of leading up to an understanding of the Fourier Series.

In this section, our linear systems will be n×n matrices of complex numbers. For a little background into some of theconcepts that this module is based on, refer to the basics of linear algebra .

## Eigenvectors and eigenvalues

Let $A$ be an n×n matrix, where $A$ is a linear operator on vectors in $\mathbb{C}^{n}$ .

$Ax=b$
where $x$ and $b$ are n×1 vectors ( [link] ).

eigenvector
An eigenvector of $A$ is a vector $v\in \mathbb{C}^{n}$ such that
$Av=\lambda v$
where $\lambda$ is called the corresponding eigenvalue . $A$ only changes the length of $v$ , not its direction.

## Graphical model

If $v$ is an eigenvector of $A$ , then only its length changes. See [link] and notice how our vector's length is simply scaled by our variable, $\lambda$ , called the eigenvalue :

When dealing with a matrix $A$ , eigenvectors are the simplest possible vectors to operate on.

## Examples

From inspection and understanding of eigenvectors, find the two eigenvectors, ${v}_{1}$ and ${v}_{2}$ , of $A=\begin{pmatrix}3 & 0\\ 0 & -1\\ \end{pmatrix}()$ Also, what are the corresponding eigenvalues, ${\lambda }_{1}$ and ${\lambda }_{2}$ ? Do not worry if you are having problems seeing these values from the information given so far,we will look at more rigorous ways to find these values soon.

The eigenvectors you found should be: ${v}_{1}=\left(\begin{array}{c}1\\ 0\end{array}\right)$ ${v}_{2}=\left(\begin{array}{c}0\\ 1\end{array}\right)$ And the corresponding eigenvalues are ${\lambda }_{1}=3$ ${\lambda }_{2}=-1$

Show that these two vectors, ${v}_{1}=\left(\begin{array}{c}1\\ 1\end{array}\right)$ ${v}_{2}=\left(\begin{array}{c}1\\ -1\end{array}\right)$ are eigenvectors of $A$ , where $A=\begin{pmatrix}3 & -1\\ -1 & 3\\ \end{pmatrix}$ . Also, find the corresponding eigenvalues.

In order to prove that these two vectors are eigenvectors, we will show that these statements meetthe requirements stated in the definition . $A{v}_{1}=\begin{pmatrix}3 & -1\\ -1 & 3\\ \end{pmatrix}()\left(\begin{array}{c}1\\ 1\end{array}\right)()=\left(\begin{array}{c}2\\ 2\end{array}\right)()$ $A{v}_{2}=\begin{pmatrix}3 & -1\\ -1 & 3\\ \end{pmatrix}()\left(\begin{array}{c}1\\ -1\end{array}\right)()=\left(\begin{array}{c}4\\ -4\end{array}\right)()$ These results show us that $A$ only scales the two vectors ( i.e. changes their length) and thus it proves that [link] holds true for the following two eigenvalues that you were asked to find: ${\lambda }_{1}=2$ ${\lambda }_{2}=4$ If you need more convincing, then one could also easilygraph the vectors and their corresponding product with $A$ to see that the results are merely scaled versions of our original vectors, ${v}_{1}$ and ${v}_{2}$ .

## Calculating eigenvalues and eigenvectors

In the above examples, we relied on your understanding of thedefinition and on some basic observations to find and prove the values of the eigenvectors and eigenvalues. However, as youcan probably tell, finding these values will not always be that easy. Below, we walk through a rigorous and mathematicalapproach at calculating the eigenvalues and eigenvectors of a matrix.

## Finding eigenvalues

Find $\lambda \in \mathbb{C}$ such that $v\neq 0$ , where $0$ is the "zero vector." We will start with [link] , and then work our way down until we find a way to explicitly calculate $\lambda$ . $Av=\lambda v$ $Av-\lambda v=0$ $(A-\lambda I)v=0$ In the previous step, we used the fact that $\lambda v=\lambda Iv$ where $I$ is the identity matrix. $I=\begin{pmatrix}1 & 0 & \dots & 0\\ 0 & 1 & \dots & 0\\ 0 & 0 & \ddots & ⋮\\ 0 & \dots & \dots & 1\\ \end{pmatrix}()$ So, $A-\lambda I$ is just a new matrix.

Given the following matrix, $A$ , then we can find our new matrix, $A-\lambda I$ . $A=\begin{pmatrix}a_{1, 1} & a_{1, 2}\\ a_{2, 1} & a_{2, 2}\\ \end{pmatrix}()$ $A-\lambda I=\begin{pmatrix}a_{1, 1}-\lambda & a_{1, 2}\\ a_{2, 1} & a_{2, 2}-\lambda \\ \end{pmatrix}()$

If $(A-\lambda I)v=0$ for some $v\neq 0$ , then $A-\lambda I$ is not invertible . This means: $\det (A-\lambda I)=0$ This determinant (shown directly above) turns out to be a polynomial expression (of order $n$ ). Look at the examples below to see what this means.

Starting with matrix $A$ (shown below), we will find the polynomial expression, where our eigenvalues will be the dependent variable. $A=\begin{pmatrix}3 & -1\\ -1 & 3\\ \end{pmatrix}()$ $A-\lambda I=\begin{pmatrix}3-\lambda & -1\\ -1 & 3-\lambda \\ \end{pmatrix}()$ $\det (A-\lambda I)=(3-\lambda )^{2}--1()^{2}=\lambda ^{2}-6\lambda +8$ $\lambda =\{2, 4\}()$

Starting with matrix $A$ (shown below), we will find the polynomial expression, where our eigenvalues will be the dependent variable. $A=\begin{pmatrix}a_{1, 1} & a_{1, 2}\\ a_{2, 1} & a_{2, 2}\\ \end{pmatrix}()$ $A-\lambda I=\begin{pmatrix}a_{1, 1}-\lambda & a_{1, 2}\\ a_{2, 1} & a_{2, 2}-\lambda \\ \end{pmatrix}()$ $\det (A-\lambda I)=\lambda ^{2}-(a_{1, 1}+a_{2, 2})\lambda -a_{2, 1}a_{1, 2}+a_{1, 1}a_{2, 2}$

If you have not already noticed it, calculating the eigenvalues is equivalent to calculating the roots of $\det (A-\lambda I)={c}_{n}\lambda ^{n}+{c}_{n-1}\lambda ^{(n-1)}+\dots +{c}_{1}\lambda +{c}_{0}=0$

Therefore, by simply using calculus to solve for the roots of our polynomial we can easily find the eigenvalues of ourmatrix.

## Finding eigenvectors

Given an eigenvalue, ${\lambda }_{i}$ , the associated eigenvectors are given by $Av={\lambda }_{i}v$ $A\left(\begin{array}{c}{v}_{1}\\ ⋮\\ {v}_{n}\end{array}\right)=\left(\begin{array}{c}{\lambda }_{1}{v}_{1}\\ ⋮\\ {\lambda }_{n}{v}_{n}\end{array}\right)$ set of $n$ equations with $n$ unknowns. Simply solve the $n$ equations to find the eigenvectors.

## Main point

Say the eigenvectors of $A$ , $\{{v}_{1}, {v}_{2}, \dots , {v}_{n}\}()$ , span $\mathbb{C}^{n}$ , meaning $\{{v}_{1}, {v}_{2}, \dots , {v}_{n}\}()$ are linearly independent and we can write any $x\in \mathbb{C}^{n}$ as

$x={\alpha }_{1}{v}_{1}+{\alpha }_{2}{v}_{2}+\dots +{\alpha }_{n}{v}_{n}$
where $\{{\alpha }_{1}, {\alpha }_{2}, \dots , {\alpha }_{n}\}()\in \mathbb{C}$ . All that we are doing is rewriting $x$ in terms of eigenvectors of $A$ . Then, $Ax=A({\alpha }_{1}{v}_{1}+{\alpha }_{2}{v}_{2}+\dots +{\alpha }_{n}{v}_{n})$ $Ax={\alpha }_{1}A{v}_{1}+{\alpha }_{2}A{v}_{2}+\dots +{\alpha }_{n}A{v}_{n}$ $Ax={\alpha }_{1}{\lambda }_{1}{v}_{1}+{\alpha }_{2}{\lambda }_{2}{v}_{2}+\dots +{\alpha }_{n}{\lambda }_{n}{v}_{n}=b$ Therefore we can write, $x=\sum {\alpha }_{i}{v}_{i}$ and this leads us to the following depicted system: Depiction of system where we break our vector, x , into a sum of its eigenvectors.

where in [link] we have, $b=\sum {\alpha }_{i}{\lambda }_{i}{v}_{i}$

By breaking up a vector, $x$ , into a combination of eigenvectors, the calculation of $Ax$ is broken into "easy to swallow" pieces.

## Practice problem

For the following matrix, $A$ and vector, $x$ , solve for their product. Try solving it using two differentmethods: directly and using eigenvectors. $A=\begin{pmatrix}3 & -1\\ -1 & 3\\ \end{pmatrix}()$ $x=\left(\begin{array}{c}5\\ 3\end{array}\right)()$

Direct Method (use basic matrix multiplication) $Ax=\begin{pmatrix}3 & -1\\ -1 & 3\\ \end{pmatrix}()\left(\begin{array}{c}5\\ 3\end{array}\right)()=\left(\begin{array}{c}12\\ 4\end{array}\right)()$ Eigenvectors (use the eigenvectors and eigenvalues we found earlier for this same matrix) ${v}_{1}=\left(\begin{array}{c}1\\ 1\end{array}\right)$ ${v}_{2}=\left(\begin{array}{c}1\\ -1\end{array}\right)$ ${\lambda }_{1}=2$ ${\lambda }_{2}=4$ As shown in [link] , we want to represent $x$ as a sum of its scaled eigenvectors. For this case, we have: $x=4{v}_{1}+{v}_{2}$ $x=\left(\begin{array}{c}5\\ 3\end{array}\right)()=4\left(\begin{array}{c}1\\ 1\end{array}\right)()+\left(\begin{array}{c}1\\ -1\end{array}\right)()$ $Ax=A(4{v}_{1}+{v}_{2})={\lambda }_{i}(4{v}_{1}+{v}_{2})$ Therefore, we have $Ax=4\times 2\left(\begin{array}{c}1\\ 1\end{array}\right)()+4\left(\begin{array}{c}1\\ -1\end{array}\right)()=\left(\begin{array}{c}12\\ 4\end{array}\right)()$ Notice that this method using eigenvectors required no matrix multiplication. This may have seemed more complicated here, but just imagine $A$ being really big, or even just a few dimensions larger!

how can chip be made from sand
is this allso about nanoscale material
Almas
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where is the latest information on a no technology how can I find it
William
currently
William
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!

#### Get Jobilize Job Search Mobile App in your pocket Now! By Cath Yu By OpenStax By OpenStax By OpenStax By Marion Cabalfin By Anonymous User By Richley Crapo By Ryan Lowe By Jonathan Long By Richley Crapo