# 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?

Thanks,

• Matrix norms in general aren't multiplicative. You only have $\Vert A B \Vert \leq \Vert A \Vert \cdot \Vert B \Vert$. Commented Jun 10, 2015 at 22:20
• This would the be the sub-multiplicative property of norms for $||AB|| \leq ||A|| ||B||$. Commented Jun 10, 2015 at 22:20
• ah -- you guys are right. When do we have equality? For norms of vectors / modulus of complex numbers / absolute value of real numbers? Commented Jun 10, 2015 at 22:22
• For norms of vectors we have the Cauchy Schwarz inequality, which is similar. And yes, we have equality for the modulus of complex numbers. Commented Jun 10, 2015 at 22:56
• You've shown that the limit is bounded from above by $\|A^2\|$ Commented Jun 10, 2015 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 $$\left\| \pmatrix{ 3\cdot (2/3)^n-2 \\ 2\cdot (2/3)^n-2 } \right\| \to \left\| \pmatrix{ -2 \\ -2 } \right\|$$

• I think your second row should be $2*2^{n} - 2*3^{n}$ .. Commented Jun 11, 2015 at 0:08
• Well, either way, the general idea is exactly the same. Commented Jun 11, 2015 at 0:09
• Double checked: you're right, fixed it. Commented Jun 11, 2015 at 0:13
• EXCELLENT!! Thank you SO much, @Omnomnomnom. I arrived at $3^{2}$ = 9 :-). Commented Jun 11, 2015 at 0:17