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Consider the $k-$dimensional homogeneous linear difference system $ x(n+1) = A(n) x(n) $.

Define $ H(n):= A^T (n) A(n)$.

(a) Prove the Lagrange identity $$\displaystyle{ || x(n+1) ||_2 ^2 = x^T (n+1) x(n+1) = x^T (n) H(n) x(n) }$$

(b) Show that all eigenvalues of $H(n)$ are real and nonnegative.

(c) Let that all the eigenvalues of $H(n)$ be ordered as $ \lambda_1 (n) \leq \lambda_2 (n) \leq \cdots \leq \lambda_k (n) $. Show that for all $ \displaystyle{x \in \mathbb R ^{ k \times 1} }$,

$$ \lambda_1 (n) x^T x \leq x^T H(n) x \leq \lambda_k (n) x^T x $$

(d) Using (a) and (c) show that:

$$ \left( \prod_{i=n_0}^{n-1} \lambda_1 (i) \right ) x^T(n_0) x(n_0) \leq x^T (n) x(n) \leq \left( \prod_{i=n_0}^{n-1}\lambda_k (i) \right)x^T(n_0) x(n_0) $$

(e) Show that $$\displaystyle{\prod_{i=n_0}^{n-1} \lambda_1 (i) \leq || |\Phi( n, n_0) ||^2 \leq \prod_{i=n_0}^{n-1} \lambda_k (i)}$$

where $\Phi( n, n_0) = \Phi(n) \Phi^{-1} (n_0)$ and $ \Phi$ is a fundamental matrix of the system.

.........................................................

I have done (a) and (b) and I need some help for the others.

This is not homework. I am studying for a midterm exam.

Thank's in advance!

P.S This is an exercise from S.N.Elaydi book on difference equations.

edit: I have edit the question (d). I had a typo in one term of the inequality. Now is correct. I am really sorry for that.

Thank you!

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For $c$, $H(n)$ is symmetric and hence has an orthonormal basis of eigenvalues. In this case, $\|H\|_2=\lambda_k$, which is easy to show by expanding any vector in terms of the orthonormal basis. –  Alex R. Jan 10 '13 at 19:47
    
@Alex: I see that $ ||H||_2 = \lambda _k$ but I can't understand how (c) follows. –  passenger Jan 10 '13 at 19:54

1 Answer 1

up vote 2 down vote accepted

Point c) is a basic property of (symmetric) positive definite matrices. Once you know these matrices allow an orthogonal diagonalization $H = P \Lambda P^t$ where $P$ is the (orthogonal) eigenvenctors matrix and $\Lambda$ is diagonal (with positive eigenvalues in its diagonal), so, for any $x$

$$x^T H x = x^T P \Lambda P^T x = y^T \Lambda y = \lambda_1 y_1^2 + \lambda_2 y_2^2 + \cdots\lambda_k y_k^2 $$

where $y=P^t x$, with $|y| = |x|$.

The above is bounded from below by $\lambda_1 |y|^2=\lambda_1 |x|^2$ and from above by $\lambda_k |y|^2=\lambda_k |x|^2$ Hence, in conclusion

$$ \lambda_1 |x|^2 \le x^t H x \le \lambda_k |x|^2$$

For point d): Take for example the upper bound, and apply it, together with a), iteratively:

$$ x_n^T x_n = x_{n-1}^T H_{n-1} x_{n-1} \le \lambda^{(k)}_{n-1} x_{n-1}^T x_{n-1} = \lambda^{(k)}_{n-1} x_{n-2}^T H_{n-2} x_{n-2} \le \lambda^{(k)}_{n-1} \lambda^{(k)}_{n-2} x_{n-3}^T x_{n-3} $$

You shoud get inequalities e) choosing $x(n_0)$ as the vector that attains the supremum in the matrix norm definition, and relating $\Phi$ to $A$ and $H$.

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O.K I got it. Thank you very much! How can I use this to prove (d) and (e) ? –  passenger Jan 10 '13 at 20:11
    
What do you mean relating $ \Phi $ to $ A$ and $ H $ ? –  passenger Jan 10 '13 at 22:02
    
Done! I found it! Thank you very much for your time! Nice solution! –  passenger Jan 10 '13 at 22:44

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