Compute the limit of this expression of norms:
Compute the limit, as n goes to infinity, of the quotient:
$$\frac{||A^{n+2}(x)||}{||A^n(x)||} $$, given the matrix $$ \begin{bmatrix} 0 & 3 \\ -2 & 5 \\ \end{bmatrix}$$
and the vector x = (1,0).
I diagonalized A, found its eigenvalues and eigenvectors, and arrive at the expression
$A^{n+2}$ = $SD^{n+2}S^{-1}$ = $$\begin{bmatrix} 3 & 1 \\ 2 & 1 \\ \end{bmatrix} \begin{bmatrix} 2^{n+2} & 0 \\ 0 & 3^{n+2} \\ \end{bmatrix}\begin{bmatrix} 1 & -1 \\ -2 & 3 \\ \end{bmatrix}$$,
where the columns of S form a basis of eigenvectors of A for the 2-dimensional space.
Now, how do I actually compute: $${||A^{n+2}(x)||}?$$
Can I do this (from following my intuition): $$\begin{bmatrix} 3 & 1 \\ 2 & 1 \\ \end{bmatrix} \begin{bmatrix} 2^{n+2} & 0 \\ 0 & 3^{n+2} \\ \end{bmatrix}\begin{bmatrix} 1 & -1 \\ -2 & 3 \\ \end{bmatrix}* (1,0)^T$$, and then just multiply everything through and arrive at some 2x1 column vector, with entries that depend on n, and then take the usual p-norm for vectors, say, the 2-norm, then take the limit as n goes to infinity?
I tried this method of "brute force", which seems valid, but I am not getting the desired answer.
What's going wrong?
linear-algebra matrices norm normed-spaces
$\begingroup$ Matrix norms in general aren't multiplicative. You only have $\Vert A B \Vert \leq \Vert A \Vert \cdot \Vert B \Vert$. $\endgroup$ – Zardo Jun 10 '15 at 22:20
$\begingroup$ This would the be the sub-multiplicative property of norms for $||AB|| \leq ||A|| ||B||$. $\endgroup$ – Xoque55 Jun 10 '15 at 22:20
$\begingroup$ ah -- you guys are right. When do we have equality? For norms of vectors / modulus of complex numbers / absolute value of real numbers? $\endgroup$ – user247323 Jun 10 '15 at 22:22
$\begingroup$ For norms of vectors we have the Cauchy Schwarz inequality, which is similar. And yes, we have equality for the modulus of complex numbers. $\endgroup$ – Omnomnomnom Jun 10 '15 at 22:56
$\begingroup$ You've shown that the limit is bounded from above by $\|A^2\|$ $\endgroup$ – Omnomnomnom Jun 10 '15 at 22:57
Hint: note that $$ \|A^n(x)\| = \left\| \pmatrix{ 3\cdot 2^n-2\cdot 3^n \\ 2 \cdot 2^n-2\cdot 3^n } \right\| = 3^{n}\left\| \pmatrix{ 3\cdot (2/3)^n-2 \\ 2\cdot (2/3)^n-2 } \right\| $$ and that as $n \to \infty$, we have $$ \left\| \pmatrix{ 3\cdot (2/3)^n-2 \\ 2\cdot (2/3)^n-2 } \right\| \to \left\| \pmatrix{ -2 \\ -2 } \right\| $$
OmnomnomnomOmnomnomnom
$\begingroup$ I think your second row should be $2*2^{n} - 2*3^{n}$ .. $\endgroup$ – user247323 Jun 11 '15 at 0:08
$\begingroup$ Well, either way, the general idea is exactly the same. $\endgroup$ – Omnomnomnom Jun 11 '15 at 0:09
$\begingroup$ Double checked: you're right, fixed it. $\endgroup$ – Omnomnomnom Jun 11 '15 at 0:13
$\begingroup$ EXCELLENT!! Thank you SO much, @Omnomnomnom. I arrived at $3^{2}$ = 9 :-). $\endgroup$ – user247323 Jun 11 '15 at 0:17
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