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In linear algebra an eigenvalue of a (square) matrix
is a number
that satisfies the eigenvalue equation,
![{\displaystyle {\text{det}}(A-\lambda I)=0\ ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c44150bec555a64de0ab5ffe08c8090bf31a8d24)
where det means the determinant,
is the identity matrix of the same dimension as
,
and in general
can be complex.
The origin of this equation, the characteristic polynomial of A, is the eigenvalue problem, which is to find the eigenvalues and associated eigenvectors of
.
That is, to find a number
and a vector
that together satisfy
![{\displaystyle A{\vec {v}}=\lambda {\vec {v}}\ .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6486dc7dd37a91410fec63d88a41eb59506b4c43)
What this equation says is that even though
is a matrix its action on
is the same as multiplying the vector by the number
.
This means that the vector
and the vector
are parallel (or anti-parallel if
is negative).
Note that generally this will not be true. This is most easily seen with a quick example. Suppose
and ![{\displaystyle {\vec {v}}={\begin{pmatrix}v_{1}\\v_{2}\end{pmatrix}}\ .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ec7a39946a2f1ea7ae0c5588e697404b405bee3)
Then their matrix product is
![{\displaystyle A{\vec {v}}={\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}}{\begin{pmatrix}v_{1}\\v_{2}\end{pmatrix}}={\begin{pmatrix}a_{11}v_{1}+a_{12}v_{2}\\a_{21}v_{1}+a_{22}v_{2}\end{pmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0820fa077010c2b1588b9e55bf7ff0115fae3ac9)
whereas the scalar product is
![{\displaystyle \lambda {\vec {v}}={\begin{pmatrix}\lambda v_{1}\\\lambda v_{2}\end{pmatrix}}\ .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ea57f4030837a1ca8f3f2ed51f067e951dbb14ec)
Obviously then
unless
and simultaneously
.
For a given
, it is easy to pick numbers for the entries of
and
such that this is not satisfied.
The eigenvalue equation
So where did the eigenvalue equation
come from? Well, we assume that we know the matrix
and want to find a number
and a non-zero vector
so that
. (Note that if
then the equation is always true, and therefore uninteresting.) So now we have
. It doesn't make sense to subtract a number from a matrix, but we can factor out the vector if we first multiply the right-hand term by the identity, giving us
![{\displaystyle (A-\lambda I){\vec {v}}={\vec {0}}\ .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/288b81c995d7114985f92eb56790f52ad5fc5c9b)
Now we have to remember the fact that
is a square matrix, and so it might be invertible.
If it was invertible then we could simply multiply on the left by its inverse to get
![{\displaystyle {\vec {v}}=(A-\lambda I)^{-1}{\vec {0}}={\vec {0}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8731f599be395028bc0891f186766fb88670b832)
but we have already said that
can't be the zero vector! The only way around this is if
is in fact non-invertible. It can be shown that a square matrix is non-invertible if and only if its determinant is zero. That is, we require
![{\displaystyle {\text{det}}(A-\lambda I)=0\ ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c44150bec555a64de0ab5ffe08c8090bf31a8d24)
which is the eigenvalue equation stated above.
A more technical approach
So far we have looked eigenvalues in terms of square matrices. As usual in mathematics though we like things to be as general as possible, since then anything we prove will be true in as many different applications as possible. So instead we can define eigenvalues in the following way.
Definition: Let
be a vector space over a field
, and let
be a linear map. An eigenvalue associated with
is an element
for which there exists a non-zero vector
such that
![{\displaystyle A({\vec {v}})=\lambda {\vec {v}}\ .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6bb623070bd6531412c804c7fa24964b15d6456a)
Then
is called the eigenvector of
associated with
.