Characteristic polynomial/2 5 -3 4/Eigenvalues/Eigenspaces/Example

For the matrix

${}M={\begin{pmatrix}2&5\\-3&4\end{pmatrix}}\,,$

the characteristic polynomial is

${}\chi _{M}=\det {\begin{pmatrix}X-2&-5\\3&X-4\end{pmatrix}}=(X-2)(X-4)+15=X^{2}-6X+23\,.$

Finding the zeroes of this polynomial leads to the condition

${}(X-3)^{2}=-23+9=-14\,,$

which has no solution over ${}\mathbb {R}$, so that the matrix has no eigenvalues over ${}\mathbb {R}$. However, considered over the complex numbers ${}\mathbb {C}$, we have the two eigenvalues ${}3+{\sqrt {14}}{\mathrm {i} }$ and ${}3-{\sqrt {14}}{\mathrm {i} }$. For the eigenspace for ${}3+{\sqrt {14}}{\mathrm {i} }$, we have to determine

${}{\begin{aligned}\operatorname {Eig} _{3+{\sqrt {14}}{\mathrm {i} }}{\left(M\right)}&=\operatorname {ker} {\left({\left({\left(3+{\sqrt {14}}{\mathrm {i} }\right)}E_{2}-M\right)}\right)}\\&=\operatorname {ker} {\left({\begin{pmatrix}1+{\sqrt {14}}{\mathrm {i} }&-5\\3&-1+{\sqrt {14}}{\mathrm {i} }\end{pmatrix}}\right)},\end{aligned}}$

a basis vector (hence an eigenvector) of this is ${}{\begin{pmatrix}5\\1+{\sqrt {14}}{\mathrm {i} }\end{pmatrix}}$. Analogously, we get

${}\operatorname {Eig} _{3-{\sqrt {14}}{\mathrm {i} }}{\left(M\right)}=\operatorname {ker} {\left({\begin{pmatrix}1-{\sqrt {14}}{\mathrm {i} }&-5\\3&-1-{\sqrt {14}}{\mathrm {i} }\end{pmatrix}}\right)}=\langle {\begin{pmatrix}5\\1-{\sqrt {14}}{\mathrm {i} }\end{pmatrix}}\rangle \,.$