Eigenvalue

In linear algebra an eigenvalue of a (square) matrix $A$ is a number $\lambda$ that satisfies the eigenvalue equation,

${\text{det}}(A-\lambda I)=0\ ,$

where $I$ is the identity matrix of the same dimension as $A$ and in general $\lambda$ can be complex. The origin of this equation is the eigenvalue problem, which is to find the eigenvalues and associated eigenvectors of $A$. That is, to find a number $\lambda$ and a vector ${\vec {v}}$ that together satisfy

$A{\vec {v}}=\lambda {\vec {v}}\ .$

What this equation says is that even though $A$ is a matrix its action on ${\vec {v}}$ is the same as multiplying it by the number $\lambda$. This means that the vector ${\vec {v}}$ and the vector $A{\vec {v}}$ are parallel (or anti-parallel if $\lambda$ is negative). Note that generally this will not be true. This is most easily seen with a quick example. Suppose

$A={\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}}$ and ${\vec {v}}={\begin{pmatrix}v_{1}\\v_{2}\end{pmatrix}}\ .$

Then their matrix product is

$A{\vec {v}}={\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}}{\begin{pmatrix}v_{1}\\v_{2}\end{pmatrix}}={\begin{pmatrix}a_{11}v_{1}+a_{12}v_{2}\\a_{21}v_{1}+a_{22}v_{2}\end{pmatrix}}$

whereas the scalar product is

$\lambda {\vec {v}}={\begin{pmatrix}\lambda v_{1}\\\lambda v_{2}\end{pmatrix}}\ .$

Obviously then $A{\vec {v}}\neq \lambda {\vec {v}}$ unless $\lambda v_{1}=a_{11}v_{1}+a_{12}v_{2}$ and simultaneously $\lambda v_{2}=a_{21}v_{1}+a_{22}v_{2}$, and it is easy to pick numbers for the entries of $A$ and ${\vec {v}}$ such that this cannot happen for any value of $\lambda$.

The eigenvalue equation

So where did the eigenvalue equation ${\text{det}}(A-\lambda I)=0$ come from? Well, we assume that we know the matrix $A$ and want to find a number $\lambda$ and a non-zero vector ${\vec {v}}$ so that $A{\vec {v}}=\lambda {\vec {v}}$. (Note that if ${\vec {v}}={\vec {0}}$ then the equation is always true, and therefore uninteresting.) So now we have $A{\vec {v}}-\lambda {\vec {v}}={\vec {0}}$. 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

$(A-\lambda I){\vec {v}}={\vec {0}}\ .$

Now we have to remember the fact that $A-\lambda I$ 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

${\vec {v}}=(A-\lambda I)^{-1}{\vec {0}}={\vec {0}}$

but we have already said that ${\vec {v}}$ can't be the zero vector! The only way around this is if $A-\lambda I$ 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

${\text{det}}(A-\lambda I)=0\ ,$

which is the eigenvalue equation stated above.