What is the importance of eigenvalues/eigenvectors?

**Answer**

### Short Answer

*Eigenvectors make understanding linear transformations easy*. They are the “axes” (directions) along which a linear transformation acts simply by “stretching/compressing” and/or “flipping”; eigenvalues give you the factors by which this compression occurs.

The more directions you have along which you understand the behavior of a linear transformation, the easier it is to understand the linear transformation; so you want to have as many linearly independent eigenvectors as possible associated to a single linear transformation.

### Slightly Longer Answer

There are a lot of problems that can be modeled with linear transformations, and the eigenvectors give very simply solutions. For example, consider the system of linear differential equations

\begin{align*}

\frac{dx}{dt} &= ax + by\\\

\frac{dy}{dt} &= cx + dy.

\end{align*}

This kind of system arises when you describe, for example, the growth of population of two species that affect one another. For example, you might have that species x is a predator on species y; the more x you have, the fewer y will be around to reproduce; but the fewer y that are around, the less food there is for x, so fewer xs will reproduce; but then fewer xs are around so that takes pressure off y, which increases; but then there is more food for x, so x increases; and so on and so forth. It also arises when you have certain physical phenomena, such a particle on a moving fluid, where the velocity vector depends on the position along the fluid.

Solving this system directly is complicated. But suppose that you could do a change of variable so that instead of working with x and y, you could work with z and w (which depend linearly on x and also y; that is, z=\alpha x+\beta y for some constants \alpha and \beta, and w=\gamma x + \delta y, for some constants \gamma and \delta) and the system transformed into something like

\begin{align*}

\frac{dz}{dt} &= \kappa z\\\

\frac{dw}{dt} &= \lambda w

\end{align*}

that is, you can “decouple” the system, so that now you are dealing with two *independent* functions. Then solving this problem becomes rather easy: z=Ae^{\kappa t}, and w=Be^{\lambda t}. Then you can use the formulas for z and w to find expressions for x and y..

Can this be done? Well, it amounts *precisely* to finding two linearly independent eigenvectors for the matrix \left(\begin{array}{cc}a & b\\c & d\end{array}\right)! z and w correspond to the eigenvectors, and \kappa and \lambda to the eigenvalues. By taking an expression that “mixes” x and y, and “decoupling it” into one that acts independently on two different functions, the problem becomes a lot easier.

That is the essence of what one hopes to do with the eigenvectors and eigenvalues: “decouple” the ways in which the linear transformation acts into a number of independent actions along separate “directions”, that can be dealt with independently. A lot of problems come down to figuring out these “lines of independent action”, and understanding them can really help you figure out what the matrix/linear transformation is “really” doing.

**Attribution***Source : Link , Question Author : Ryan , Answer Author : Sanchit*