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Strang's Linear Algebra and Its Applications 4e, question 5.1.19 (p. 280) asks:

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It's clear that taking increasing powers of $$A = \begin{bmatrix} .8 & .3 \\ .2 & .7 \end{bmatrix}$$

yields the convergent behavior shown, and I can also see that the eigenvalues of $A^k$ are $1$ and $2^{-k}$, with the latter going to zero as $k$ approaches infinity and the matrix becomes singular.

However, I don't understand the explanation Strang provides:

answer

There are plenty of matrices whose eigenvalues are $1$ and $.25$ that don't lie halfway between $A$ and $A^\infty$. Is there a more specific way to answer the question as posed: Why does $A^2 = \frac{1}{2}(A + A^\infty)$?

I'm assuming a full explanation rests on the fact that the eigenvectors for all powers of $A$ are the same (how do I prove this?), and somehow connects that to $A^k$'s convergent behavior.

Thank you.

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  • $\begingroup$ diagonalise${}$? $\endgroup$ Commented Sep 10, 2019 at 3:45
  • $\begingroup$ That topic hasn't been introduced yet. $\endgroup$ Commented Sep 10, 2019 at 3:50

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The matrix $A$ has eigenvectors $v_1$ and $v_2$ with $Av_1=v_1$ and $Av_2=(1/2)v_2$. Then $A^2 v_1=v_1$, $A^2 v_2=(1/4)v_2$, $A^\infty v_1=v_1$ and $A^\infty v_2=0$. So $$\frac12(A+A^\infty)v_1=\frac12(v_1+v_1)=v_1=A^2v_1$$ and $$\frac12(A+A^\infty)v_2=\frac12(v_1+0)=\frac12 v_1=A^2v_2.$$ As every vector $v$ is a linear combination of $v_1$ and $v_2$, then $$\frac12(A+A^\infty)v=A^2v.$$

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  • $\begingroup$ That's it! Thank you. $\endgroup$ Commented Sep 10, 2019 at 4:04
  • $\begingroup$ The second set of equations should be $$\frac{1}{2}(A + A^{\infty})v_2 = \frac{1}{2}\left(\frac{1}{2}v_2 + 0\right) = \frac{1}{4}v_2 = A^2 v_2$$ $\endgroup$ Commented Sep 10, 2019 at 4:27

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