In the previous section, we use the applet to find the eigenvectors and eigenvalues of any 2 x 2 matrix. In fact, we can compute them in a systematic way, as illustrated in the following example:[br][br][u]Example[/u]: Let [math]A=\begin{pmatrix}2&3\\3&-6\end{pmatrix}[/math]. Find all the eigenvalues and their eigenvectors.[br][br]By definition, if [math]x[/math] is an eigenvector corresponding to the eigenvalue [math]\lambda[/math] (yet to be determined), we have [math]Ax=\lambda x[/math]. We can rewrite this equation as follows:[br][br][math]Ax=\lambda x = \lambda I x\Rightarrow Ax-\lambda I x=0\Rightarrow (A-\lambda I)x=0[/math][br][br]It becomes a homogeneous equation and an eigenvector [math]x[/math] is exactly its non-trivial solution. Recall that such a homogeneous equation has non-trivial solutions if and only if [math]\det(A-\lambda I)=0[/math]. It turns out that the determinant can be expressed as a polynomial in [math]\lambda[/math]. It is called the [b]characteristic polynomial[/b] of [math]A[/math] and the polynomial equation [math]\det(A-\lambda I)=0[/math] is called the [b]characteristic equation[/b] of [math]A[/math].[br][br]In our example, we write down the characteristic equation explicitly as follows:[br][br][math]\det(A-\lambda I)=\begin{vmatrix}2-\lambda&3\\3&-6-\lambda\end{vmatrix}=0[/math][br][br]Computing the 2 x 2 determinant, we get[br][br][math]\begin{eqnarray}0&=&(2-\lambda)(-6-\lambda)-3\cdot 3\\ &=& \lambda^2+4\lambda-21\\ &=& (\lambda-3)(\lambda+7)\end{eqnarray}[/math][br][br]Hence, the eigenvalues of [math]A[/math] are [math]3[/math] and [math]-7[/math].[br][br]

To find the eigenvectors corresponding to each eigenvalue, we just need to put the eigenvalue into the homogeneous equation [math](A-\lambda I)x=0[/math] and solve for the solutions:[br][br]When [math]\lambda=3[/math], [br][br][math](A-3I)x=\begin{pmatrix}-1&3\\3&-9\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}[/math] [br][br]It can easily be row reduced into the following: [math]\begin{pmatrix}1&-3\\0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}[/math]. Hence, we have[br][math]\begin{pmatrix}x_1\\x_2\end{pmatrix}=s\begin{pmatrix}3\\1\end{pmatrix}[/math] for any real number [math]s[/math]. The set of all these eigenvectors and the zero vector form the subspace [math]\text{Nul}(A-3I)[/math], which is called the [b]eigenspace[/b] of [math]A[/math] corresponding to [math]\lambda=3[/math].[br][br]When [math]\lambda=-7[/math], [br][br][math](A-(-7)I)x=\begin{pmatrix}9&3\\3&1\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}[/math] [br][br]It can easily be row reduced into the following: [math]\begin{pmatrix}1&\frac13\\0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}[/math]. Hence, we have[br][math]\begin{pmatrix}x_1\\x_2\end{pmatrix}=s\begin{pmatrix}-\frac13\\1\end{pmatrix}[/math] for any real number [math]s[/math]. The set of all these eigenvectors and the zero vector form the subspace [math]\text{Nul}(A-(-7)I)[/math], which is the eigenspace of [math]A[/math] corresponding to [math]\lambda=-7[/math].[br]

Given an n x n matrix [math]A[/math], we can find its eigenvalues by the same method: Solve the characteristic equation [math]\det(A-\lambda I)=0[/math]. It is a polynomial equation in [math]\lambda[/math] of degree [math]n[/math]. Therefore, there are at most [math]n[/math] distinct eigenvalues of [math]A[/math].[br][br][u]Example[/u]: Find the characteristic equation of [math]A=\begin{pmatrix}4&-1&6&-3\\0&2&-4&9\\0&0&4&1\\0&0&0&-1\end{pmatrix}[/math].[br]Solution: [math]\det(A-\lambda I)=\begin{vmatrix}4-\lambda&-1&6&-3\\0&2-\lambda&-4&9\\0&0&4-\lambda&1\\0&0&0&-1-\lambda\end{vmatrix}=0[/math][br][br]Since it is an upper triangular matrix, the determinant is the product of all diagonal entries i.e. [br][br][math](4-\lambda)^2(2-\lambda)(-1-\lambda)=0[/math][br][br]Therefore, the eigenvalues of [math]A[/math] are [math]-1, 2[/math] and [math]4[/math]. The eigenvalue [math]\lambda=4[/math] is said to have [b]multiplicity[/b] 2 because the factor [math](\lambda-4)[/math] appears twice in the characteristic equation. In general the [b]multiplicity[/b] of an eigenvalue [math]\lambda[/math] is its multiplicity as a root of the characteristic equation.[br][br]