We consider the matrix \[A= \matrix{-2 & 1 & 1\\ -11 & 4 & 5\\ -1 & 1 & 0}\] If \(\vec{v}=\cv{1\\3\\1}\), then: \[A(\vec{v})= \matrix{-2 & 1 & 1\\ -11 & 4 & 5\\ -1 & 1 & 0}\!\cv{1\\3\\1}=\cv{2\\6\\2}=2\cv{1\\3\\1}\] We say: \(\vec{v}\) is an eigenvector \(A\) corresponding to the eigenvalue \(2\).

Check yourself:

• \(\cv{0\\-1\\1}\) is an eigenvector \(A\) corresponding to the eigenvalue \(-1\).
  
  
• \(\cv{1\\2\\1}\) is an eigenvector \(A\) corresponding to the eigenvalue \(1\).

We can place the three eigenvectors as columns in a transformation matrix \(T\): \[T=\matrix{1 & 0 & 1\\ 3 & -1 & 2\\ 1 & 1& 1}\] Then (check it yourself): \[T^{-1}=\matrix{-3 & 1 & 1 \\ -1 & 0 & 1 \\ 4 & -1 & -1}\quad\text{and}\quad T^{-1}\! A\, T = \matrix{2 & 0 & 0\\ 0 & -1& 0\\ 0 & 0 & 1}\] In other words, the matrix \(A\) is similar to the diagonal matrix \(D\) with the eigenvalues on the main diagonal.

The similarity of a matrix \(A\) to a diagonal matrix \(D\) by means of a transformation \(T\) has an important application. Suppose that you want to compute \(A^{10}\) for whatever reason. Instead of computing many matrix multiplications you better consider \[A^{10}=(T D T^{-1})^{10}= T D^{10} T^{-1}\] and realize that\(D^{10}\) is a diagonal matrix with the main diagonal \(2^{10}=1024\), \((-1)^{10}=1\) and \(1\). Then you get faster the result \[\begin{aligned}A^{10}& =\matrix{1 & 0 & 1\\ 3 & -1 & 2\\ 1 & 1& 1}\matrix{1024 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1}\matrix{-3 & 1 & 1 \\ -1 & 0 & 1 \\ 4 & -1 & -1} \\ \\ &= \matrix{-3068 & 1023 & 1023\\ -9207 & 3070 & 3069\\ -3069 & 1023 & 1024}\end{aligned}\] Now you may wonder when you would ever need such powers of matrices, but the answer to this question is simple: science fields such as population ecology and studies of stochastic processes (e.g. in neural networks) are full of high powers of matrices; Google's search algorithm PageRank is based on finding an eigenvector. One also encounters eigenvalues and eigenvectors when solving linear dynamic systems.