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.