*T*to a vector gives a vector in the same space as the original, the resulting vector usually points in a completely different direction from the original, that is,

*T*(

**x**) is neither parallel nor antiparallel to

**x**. However, it can happen that

*T*(

**x**)

*is*a scalar multiple of

**x**—even when

**x ≠ 0**—and this phenomenon is so important that it deserves to be explored.

If *T* : **R** ^{n }→ **R** ^{n }is a linear operator, then *T* must be given by *T*( **x**) = *A* **x** for some *n x n* matrix *A*. If **x ≠ 0** and *T*( **x**) = *A* **x** is a scalar multiple of **x**, that is, if **eigenvalue** of *T* (or, equivalently, of *A*). Any *nonzero* vector **x** which satisfies this equation is said to be an **eigenvector** of *T* (or of *A*) corresponding to λ. To illustrate these definitions, consider the linear operator *T* : **R** ^{2} → **R** ^{2} defined by the equation

That is, *T* is given by left multiplication by the matrix

Consider, for example, the image of the vector **x** = (1, 3) ^{T} under the action of *T*:

Clearly, *T*( **x**) is not a scalar multiple of **x**, and this is what typically occurs.

However, now consider the image of the vector **x** = (2, 3) ^{T} under the action of *T*:

Here, *T*( **x**) *is* a scalar multiple of **x**, since *T*( **x**) = (−4, −6) ^{T} = −2(2, 3) ^{T} = −2 **x**. Therefore, −2 is an eigenvalue of *T*, and (2, 3) ^{T} is an eigenvector corresponding to this eigenvalue. The question now is, how do you determine the eigenvalues and associated eigenvectors of a linear operator?