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Eigenvectors and Eigenvalues (view source)
Revision as of 16:00, April 10, 2017
, 16:00, April 10, 2017Added explanation of how to calculate eigenvalues and eigenvectors and an example
{{math-h}}
In [[linear algebra]], when a transformation of a space is carried out, some vectors (points in the space) are not rotated, but only extended or shrunk (moved farther or closer to the origin). The vectors are called the '''eigenvectors''' of the transformation, and the amount of extension or shrinkage carried out on that eigenvector is called the '''eigenvalue''' of the transformation corresponding to the given eigenvector.
In [[linear algebra]], when a transformation of a space is carried out, some vectors (points in the space) are not rotated, but only extended or shrunk (moved farther or closer to the origin). The vectors are called the '''eigenvectors''' of the transformation, and the amount of extension or shrinkage carried out on that eigenvector is called the '''eigenvalue''' of the transformation corresponding to the given eigenvector.
== Eigenvectors and Eigenvalues in Physics ==
== Eigenvectors and Eigenvalues in Physics ==
In Physics, eigenvectors and eigenvalues play an important role. For example, solving [[Schrödinger's equation]] (one of the fundamental equations of [[Quantum mechanics]]) is a problem of finding Eigenvectors and Eigenvalues for a very complex linear operator. The ''quantization'' (which means that discrete values are acceptable in physics, but not all of them: for example, an electron can have spin 1/2 or -1/2, but never 1/3 or 1/10) that was previously a postulated of Quantum mechanics can be derived from the fact that the eigenvalues of the operators come in discrete sets.
In Physics, eigenvectors and eigenvalues play an important role. For example, solving [[Schrödinger's equation]] (one of the fundamental equations of [[Quantum mechanics]]) is a problem of finding Eigenvectors and Eigenvalues for a very complex linear operator. The ''quantization'' (which means that discrete values are acceptable in physics, but not all of them: for example, an electron can have spin 1/2 or -1/2, but never 1/3 or 1/10) that was previously a postulated of Quantum mechanics can be derived from the fact that the eigenvalues of the operators come in discrete sets.
==Finding Eigenvalues and Eigenvectors==
This section describes a method for finding the eigenvalues and eigenvectors of a matrix <math>A</math>.
===Eigenvalues===
The eigenvectors and eigenvalues satisfy the relationship: <math>A\boldsymbol{x}=\lambda\boldsymbol{x}</math>. This can be rearranged to give:
<math>
(A - \lambda I) \boldsymbol{x}=\boldsymbol{0}
</math>
where <math>I</math> is the [[identity matrix]]. Solutions where <math>\boldsymbol{x} = \boldsymbol{0}</math> are trivial (simple) and are ignored. Therefore non trivial solutions occur only if <math>A - \lambda I</math> is a [[singular matrix]]. A singular matrix does not have an inverse, so its [[determinant]] must be 0. Setting the determinant to 0 produces <math>|A - \lambda I| = 0</math>. Expanding this gives the [[characteristic polynomial]]. This is a polynomial in <math>\lambda</math> and equal in degree that of the matrix A (i.e. a 3x3 matrix produces a cubic (degree 3) characteristic polynomial). The solutions to this equation are the eigenvalues of <math>A</math>.
===Eigenvectors===
The eigenvectors are then found by substituting in each value of <math>\lambda</math> in turn and solving the system of linear equations that results. The equations will not be linearly independent, meaning there will be an infinite number of solutions and that the solution for the eigenvector will be in terms of a parameter. This is best seen in the following example.
===Example===
Consider the 2x2 matrix:
<math>
A = \begin{pmatrix}
4 & -1 \\
2 & 1
\end{pmatrix}
</math>
The eigenvalues come from solving <math>|A-\lambda I| =0</math> so the [[characteristic polynomial]] comes from:
<math>
\begin{vmatrix}
4 - \lambda & -1 \\
2 & 1 - \lambda
\end{vmatrix}
=(4 - \lambda)(1 - \lambda) + 2
=\lambda^2 - 5 \lambda + 6 = 0
</math>
This has solutions of 2 and 3 so the eigenvalues are 2 and 3. To find the eigenvectors, first consider the eigenvalue equal to 2. This produces the equations:
<math>
(A- \lambda I) \boldsymbol{x} = \begin{pmatrix}
2 & -1 \\
2 & -1
\end{pmatrix}
\begin{pmatrix}
x \\
y
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0
\end{pmatrix}
</math>
This is just the equation <math>y = 2x</math>, so the eigenvectors are:
<math>
\boldsymbol{x} =
\begin{pmatrix}
x \\
2x
\end{pmatrix}
</math>
for any <math>x</math>. Similarly, the eigenvalue of 3 produces eigenvectors of the form:
<math>
\boldsymbol{x} =
\begin{pmatrix}
x \\
x
\end{pmatrix}
</math>
==See Also==
* [[Linear algebra]]
* [[Quantum Mechanics]]
[[Category:Linear Algebra]]
[[Category:Linear Algebra]]