Editing Eigenvalue algorithm (section)

===Symmetric 3×3 matrices===

The characteristic equation of a symmetric 3&times;3 matrix {{math|''A''}} is:

:<math>\det \left( \alpha I - A \right) = \alpha^3 - \alpha^2 {\rm tr}(A) - \alpha \frac{1}{2}\left( {\rm tr}(A^2) - {\rm tr}^2(A) \right) - \det(A) = 0.</math>

This equation may be solved using the methods of [[Cubic equation#Cardano's method|Cardano]] or [[Cubic equation#Lagrange's method|Lagrange]], but an affine change to {{math|''A''}} will simplify the expression considerably, and lead directly to a [[Cubic equation#Trigonometric and hyperbolic solutions|trigonometric solution]]. If {{math|1=''A'' = ''pB'' + ''qI''}}, then {{math|''A''}} and {{math|''B''}} have the same eigenvectors, and {{math|''β''}} is an eigenvalue of {{math|''B''}} if and only if {{math|1=''α'' = ''pβ'' + ''q''}} is an eigenvalue of {{math|''A''}}. Letting <math display="inline"> q = {\rm tr}(A)/3</math> and <math display="inline"> p =\left({\rm tr}\left((A - qI)^2\right)/ 6\right)^{1/2}</math>, gives

:<math>\det \left( \beta I - B \right) = \beta^3 - 3 \beta - \det(B) = 0.</math>

The substitution {{math|1=''β'' = 2cos ''θ''}} and some simplification using the identity {{math|1=cos 3''θ'' = 4cos<sup>3</sup> ''θ'' − 3cos ''θ''}} reduces the equation to {{math|1=cos 3''θ'' = det(''B'') / 2}}. Thus

:<math>\beta = 2{\cos}\left(\frac{1}{3}{\arccos}\left( \det(B)/2 \right) + \frac{2k\pi}{3}\right), \quad k = 0, 1, 2.</math>

If {{math|det(''B'')}} is complex or is greater than 2 in absolute value, the arccosine should be taken along the same branch for all three values of {{math|''k''}}. This issue doesn't arise when {{math|''A''}} is real and symmetric, resulting in a simple algorithm:<ref name=Smith>{{Citation |last=Smith |first=Oliver K. |title=Eigenvalues of a symmetric 3 × 3 matrix. |journal=[[Communications of the ACM]] |volume=4 |issue=4 |date=April 1961 |page=168 |doi=10.1145/355578.366316|s2cid=37815415 |doi-access=free }}</ref>

<syntaxhighlight lang="matlab">
% Given a real symmetric 3x3 matrix A, compute the eigenvalues
% Note that acos and cos operate on angles in radians

p1 = A(1,2)^2 + A(1,3)^2 + A(2,3)^2
if (p1 == 0) 
   % A is diagonal.
   eig1 = A(1,1)
   eig2 = A(2,2)
   eig3 = A(3,3)
else
   q = trace(A)/3               % trace(A) is the sum of all diagonal values
   p2 = (A(1,1) - q)^2 + (A(2,2) - q)^2 + (A(3,3) - q)^2 + 2 * p1
   p = sqrt(p2 / 6)
   B = (1 / p) * (A - q * I)    % I is the identity matrix
   r = det(B) / 2

   % In exact arithmetic for a symmetric matrix  -1 <= r <= 1
   % but computation error can leave it slightly outside this range.
   if (r <= -1) 
      phi = pi / 3
   elseif (r >= 1)
      phi = 0
   else
      phi = acos(r) / 3
   end

   % the eigenvalues satisfy eig3 <= eig2 <= eig1
   eig1 = q + 2 * p * cos(phi)
   eig3 = q + 2 * p * cos(phi + (2*pi/3))
   eig2 = 3 * q - eig1 - eig3     % since trace(A) = eig1 + eig2 + eig3
end
</syntaxhighlight>

Once again, the eigenvectors of {{math|''A''}} can be obtained by recourse to the [[Cayley–Hamilton theorem]]. If {{math|''α''<sub>1</sub>, ''α''<sub>2</sub>, ''α''<sub>3</sub>}} are distinct eigenvalues of {{math|''A''}}, then {{math|1=(''A'' − ''α''<sub>1</sub>''I'')(''A'' − ''α''<sub>2</sub>''I'')(''A'' − ''α''<sub>3</sub>''I'') = 0}}. Thus the columns of the product of any two of these matrices will contain an eigenvector for the third eigenvalue. However, if {{math|1=''α''<sub>3</sub> = ''α''<sub>1</sub>}}, then {{math|1=(''A'' − ''α''<sub>1</sub>''I'')<sup>2</sup>(''A'' − ''α''<sub>2</sub>''I'') = 0}} and {{math|1=(''A'' − ''α''<sub>2</sub>''I'')(''A'' − ''α''<sub>1</sub>''I'')<sup>2</sup> = 0}}. Thus the ''generalized'' eigenspace of {{math|''α''<sub>1</sub>}} is spanned by the columns of {{math|''A'' − ''α''<sub>2</sub>''I''}} while the ordinary eigenspace is spanned by the columns of {{math|1=(''A'' − ''α''<sub>1</sub>''I'')(''A'' − ''α''<sub>2</sub>''I'')}}.  The ordinary eigenspace of {{math|''α''<sub>2</sub>}} is spanned by the columns of {{math|(''A'' − ''α''<sub>1</sub>''I'')<sup>2</sup>}}.

For example, let

:<math>A = \begin{bmatrix} 3 & 2 & 6 \\ 2 & 2 & 5 \\ -2 & -1 & -4 \end{bmatrix}.</math>

The characteristic equation is

:<math> 0 = \lambda^3 - \lambda^2 - \lambda + 1 = (\lambda - 1)^2(\lambda + 1),</math>

with eigenvalues 1 (of multiplicity 2) and -1. Calculating,

:<math>A - I = \begin{bmatrix} 2 & 2 & 6 \\ 2 & 1 & 5 \\ -2 & -1 & -5 \end{bmatrix}, \qquad A + I = \begin{bmatrix} 4 & 2 & 6 \\ 2 & 3 & 5 \\ -2 & -1 & -3 \end{bmatrix}</math>

and

:<math>(A - I)^2 = \begin{bmatrix} -4 & 0 & -8 \\ -4 & 0 & -8 \\ 4 & 0 & 8 \end{bmatrix}, \qquad (A - I)(A + I) = \begin{bmatrix} 0 & 4 & 4 \\ 0 & 2 & 2 \\ 0 & -2 & -2 \end{bmatrix}</math>

Thus {{math|(−4, −4, 4)}} is an eigenvector for −1, and {{math|(4, 2, −2)}} is an eigenvector for 1. {{math|(2, 3, −1)}} and {{math|(6, 5, −3)}} are both generalized eigenvectors associated with 1, either one of which could be combined with {{math|(−4, −4, 4)}} and {{math|(4, 2, −2)}} to form a basis of generalized eigenvectors of {{math|''A''}}. Once found, the eigenvectors can be normalized if needed.

==== Eigenvectors of normal 3×3 matrices ====
If a 3×3 matrix <math>A</math> is normal, then the cross-product can be used to find eigenvectors. If <math>\lambda</math> is an eigenvalue of <math>A</math>, then the null space of <math>A - \lambda I</math> is perpendicular to its column space. The [[cross product]] of two independent columns of  <math>A - \lambda I</math> will be in the null space. That is, it will be an eigenvector associated with <math>\lambda</math>. Since the column space is two dimensional in this case, the eigenspace must be one dimensional, so any other eigenvector will be parallel to it.

If <math>A - \lambda I</math> does not contain two independent columns but is not {{math|'''0'''}}, the cross-product can still be used. In this case <math>\lambda</math> is an eigenvalue of multiplicity 2, so any vector perpendicular to the column space will be an eigenvector. Suppose <math>\mathbf v</math> is a non-zero column of <math>A - \lambda I</math>. Choose an arbitrary vector <math>\mathbf u</math> not parallel to <math>\mathbf v</math>. Then <math>\mathbf v\times \mathbf u</math>  and  <math>(\mathbf v\times \mathbf u)\times \mathbf v</math> will be perpendicular to <math>\mathbf v</math> and thus will be eigenvectors of  <math>\lambda</math>.

This does not work when <math>A</math> is not normal, as the null space and column space do not need to be perpendicular for such matrices.