**Eigenvalue** ([edit](http://github.com/nicebyte/wiki/edit/master/Wiki/Eigenvalue.md)) # Definition If for a matrix $\boldsymbol{M}$ and a nonzero vector $\overline{a}$, $\boldsymbol{M}\overline{a} = \lambda \overline{a}$, then $\overline{a}$ is called $\boldsymbol{M}$'s _eigenvector_, and $\lambda$ is called $\boldsymbol{M}$'s _eigenvalue_. To put it another way, eigenvectors of a matrix are vectors that don't change direction when the transform represented by the said matrix is applied to them. # Finding Eigenvalues Let $\overline{a}$ be an eigenvector of a matrix $\boldsymbol{M}$: $$ \boldsymbol{M}\overline{a} = \lambda \overline{a} $$ $$ \boldsymbol{M}\overline{a} = \lambda \boldsymbol{I} \overline{a} $$ $$ \boldsymbol{M}\overline{a} - \lambda \boldsymbol{I} \overline{a} = \overline{0} $$ $$ (\boldsymbol{M} - \lambda \boldsymbol{I}) \overline{a} = \overline{0} $$ which is only possible when the matrix $(\boldsymbol{M} - \lambda \boldsymbol{I})$ is singular (i.e. has a Determinant equal to $0$). The determinant of said matrix is a polynomial function of $\lambda$ and the roots of that polynomial are the eigenvalues. # Eigendecomposition According to the defining property of eigenvectors, $$ \boldsymbol{M}\overline{a} = \lambda \overline{a} $$ Let $\boldsymbol{Q}$ be the matrix where each column is one of $\boldsymbol{M}$'s eigenvectors. The matrix $\boldsymbol{M}\boldsymbol{Q}$ then is the same as $\boldsymbol{Q}$, but with each column multiplied by the corresponding eigenvalue. We can write $\boldsymbol{M}\boldsymbol{Q}$ as $\boldsymbol{Q}\boldsymbol{\Lambda}$, where $\boldsymbol{\Lambda}$ is a diagonal matrix that has the eigenvalues of $\boldsymbol{M}$ on the main diagonal. If we assume that the eigenvectors of $\boldsymbol{M}$ are linearly independent, then $\boldsymbol{Q}$ is invertible, and therefore: $$ \boldsymbol{M}\boldsymbol{Q}\boldsymbol{Q^{-1}} = \boldsymbol{Q}\boldsymbol{\Lambda}\boldsymbol{Q^{-1}} $$ which gives us a factorization of $\boldsymbol{M}$ called the _eigendecomposition_: $$ \boldsymbol{M} = \boldsymbol{Q}\boldsymbol{\Lambda}\boldsymbol{Q^{-1}} $$ Categories: Mathematics, LinearAlgebra