Editing Eigenvalue algorithm (section)

===2×2 matrices===

For dimensions 2 through 4, formulas involving radicals exist that can be used to find the eigenvalues. While a common practice for 2×2 and 3×3 matrices, for 4×4 matrices the increasing complexity of the [[Quartic function#Ferrari's solution|root formulas]] makes this approach less attractive.

For the 2×2 matrix

:<math>A = \begin{bmatrix} a  & b \\ c & d \end{bmatrix},</math>

the characteristic polynomial is

:<math>\det \begin{bmatrix} \lambda - a & -b \\ -c & \lambda - d \end{bmatrix} = \lambda^2\, -\, \left( a + d \right )\lambda\, +\, \left ( ad - bc \right ) = \lambda^2\, -\, \lambda\, {\rm tr}(A)\, +\, \det(A).</math>

Thus the eigenvalues can be found by using the [[quadratic formula]]:

:<math>\lambda = \frac{{\rm tr}(A) \pm \sqrt{{\rm tr}^2 (A) - 4 \det(A)}}{2}.</math>

Defining <math display="inline"> {\rm gap}\left ( A \right ) = \sqrt{{\rm tr}^2 (A) - 4 \det(A)}</math>  to be the distance between the two eigenvalues, it is straightforward to calculate

:<math>\frac{\partial\lambda}{\partial a} = \frac{1}{2}\left ( 1 \pm \frac{a - d}{{\rm gap}(A)} \right ),\qquad \frac{\partial\lambda}{\partial b} =  \frac{\pm c}{{\rm gap}(A)}</math>

with similar formulas for {{math|''c''}} and {{math|''d''}}. From this it follows that the calculation is well-conditioned if the eigenvalues are isolated.

Eigenvectors can be found by exploiting the [[Cayley–Hamilton theorem]]. If {{math|''λ''<sub>1</sub>, ''λ''<sub>2</sub>}} are the eigenvalues, then {{math|1=(''A'' − ''λ''<sub>1</sub>''I'')(''A'' − ''λ''<sub>2</sub>''I'') = (''A'' − ''λ''<sub>2</sub>''I'')(''A'' − ''λ''<sub>1</sub>''I'') = 0}}, so the columns of {{math|(''A'' − ''λ''<sub>2</sub>''I'')}} are annihilated by {{math|(''A'' − ''λ''<sub>1</sub>''I'')}} and vice versa. Assuming neither matrix is zero, the columns of each must include eigenvectors for the other eigenvalue. (If either matrix is zero, then {{math|''A''}} is a multiple of the identity and any non-zero vector is an eigenvector.)

For example, suppose

:<math>A = \begin{bmatrix} 4 & 3 \\ -2 & -3 \end{bmatrix},</math>

then {{math|1=tr(''A'') = 4 − 3 = 1}} and {{math|1=det(''A'') = 4(−3) − 3(−2) = −6}}, so the characteristic equation is

:<math> 0 = \lambda^2 - \lambda - 6 = (\lambda - 3)(\lambda + 2),</math>

and the eigenvalues are 3 and -2. Now,

:<math>A - 3I = \begin{bmatrix} 1 & 3 \\ -2 & -6 \end{bmatrix}, \qquad  A + 2I = \begin{bmatrix} 6 & 3 \\ -2 & -1 \end{bmatrix}.</math>

In both matrices, the columns are multiples of each other, so either column can be used. Thus, {{math|(1, −2)}} can be taken as an eigenvector associated with the eigenvalue -2, and {{math|(3, −1)}} as an eigenvector associated with the eigenvalue 3, as can be verified by multiplying them by {{math|''A''}}.