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I need to find the norm of the matrix

$$ A=\left( \begin{array}{cc} e^{-x} \cos( \sin x) & e^{-x} \sin ( \sin x) \\ -e^{-x} \sin ( \sin x) & e^{-x} \cos (\sin x) \end{array} \right) $$

Here it is what I did:

$ A^{*} A = A^{T} A = \cdots = e^{-2x} \text{I}$

So the maximum eigenvalue of $A$ is $ e^{-2x} $. Thus, $ ||A||_2 =\sqrt{ e^{-2x} } = e^{-x}$.

Is the above correct? Because in my textbook gives a result $ 2 e^{-2x} $.

And one more question: When we are told to find the norm of a matrix which norm do we use? Are all the norms in $ K^{n \times n}$ equivelant with each other.

Thank's in advance!

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All norms on finite-dimensional vector spaces are equivalent (the same does not hold for infinite-dimensional spaces!). Now, the $n\times n$ matrices form a finite-dimensional vector space of dimension $n^2$. More generally, $n\times m$ matrices form a vector space of dimension $nm$. Conventionally, $\|A\|_2$ means the norm $\left(\sum_{i,j} a_{i,j}^2\right)^{1/2}$. – William Jan 1 '13 at 12:34
Here is a related problem. – Mhenni Benghorbal Jan 1 '13 at 12:34
@William: The form you write gives $||A||_2 = \sqrt{2} e^{-x}$ which is correct now? – passenger Jan 1 '13 at 12:39
No, I think you should be getting $\sqrt{4e^{-2x}} = 2e^{-x}$. – William Jan 1 '13 at 12:42
@William: Using which form? So is my first solution wrong? – passenger Jan 1 '13 at 12:50
up vote 2 down vote accepted

While all norms on a finite dimensional vector space are equivalent in the sense that there exist a constant $C > 0$ such that $$ \frac{1}{C} ||A||_1 \leq ||A||_2 \leq C ||A||_1 $$ this doesn't mean that they are all exactly the same. If $||\cdot||$ is a norm, so is $98||\cdot||$, etc. The book should be clear as to which matrix norm you need to use.

You calculated correctly the operator norm of $A$. For example, if you calculate the norm of $A$ using a different matrix norm, the Frobenius norm, you get $$ ||A|| = \left( \sum_{i,j} |a_{ij}|^2 \right)^{\frac{1}{2}} = \mathrm{tr}(A^*A)^{\frac{1}{2}} = \sqrt{2e^{-2x}}. $$

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Thank you for your answer! – passenger Jan 1 '13 at 14:38

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