The question stems from a problem i stumbled upon while working with eigenvalues. Asking to explain why $A^{100}$ is close to $A^\infty$
$$A=
\left[ \begin{array}{cc}
.6 & .2 \\
.4 & .8
\end{array} \right]
$$
$$A^\infty=
\left[ \begin{array}{cc}
1/3 & 1/3 \\
2/3 & 2/3
\end{array} \right]
$$
Answer being given that (skipping calculations) that $A$ has eigenvalues $\lambda_1=1$ and $\lambda_2=0.4$ with eigenvectors $x_1=(1,2)$ and $x_2=(1,-1)$, and $A^{\infty}$ has eigenvalues $\lambda_1=1$ and $\lambda_2=0$, with same eigenvectors, while $A^{100}$ has eigenvalues $\lambda_1=1$ and $\lambda_2=(0.4)^{100}$ with same eigenvectors as the others, concluding that as the eigenvectors are the same and the eigenvalues are close comparing $A^\infty$ and $A^{100}$ they must be close.
Creating the basis for my question, how can one conclude that two matrices with same eigenvectors and close/equal eigenvalues are close/equal to each other?
My initial thoughts is that two matrices with equal eigenvectors and eigenvalues founds the basis for the same transformation which is why they are equal – Am I completly off?
Best Answer
The two matrices $\left[ \begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array} \right] $ and $\left[ \begin{array}{cc} 1 & 0 \\ 200 & 1 \end{array} \right] $ both have the same eigenvalues and eigenvectors, but they are nowhere near equal to each other.
But if an $n\times n$ matrix has $n$ distinct eigenvalues or otherwise has a set of eigenvectors that form a basis of $\mathbb R^n,$ then the only matrix that has the same eigenpairs, i.e. the same eigenvectors, each with the same eigenvalue, is that same matrix. That is because a linear transformation is completely determined by what it does with a basis.