Prove that [itex]A[/itex] is bounded with [itex]\|A\|=2[/itex]
Solution
Okay, here the first thing I did was notice that [itex]A[/itex] is a linear operator, so it takes a 2-element 'vector' [itex](x,y) \in \mathbb{R}^2[/itex] and maps it to another.
So by applying the transformation matrix [itex]A[/itex] to a vector we have
[tex]\left(\begin{array}{cc}
-2 & 0 \\
0 & 1 \\
\end{array}\right)\left(\begin{array}{c}
x \\
y \\
\end{array}\right)= \left(\begin{array}{c}
-2x\\
y \\
\end{array}\right)[/tex]
So we have that [itex]\|A\textbf{x}\| = 2x+y[/itex] which implies that
[tex]\|A\textbf{x}\|^2 = 4x^2 +y^2[/itex].
Now I am stuck. I need to prove that [itex]\|A\| \leq M\|x\|[/itex], but I have no methods of proceeding any further.
I have a idea that maybe we let
[tex]\|A\|^2 = \max\{4x^2+y^2\}[/tex]
subject to some sort of constraint. But I have no idea what constraint and even if this is the correct path. Any ideas or suggestions?
Subject this to the same constraint, ie [itex]-1\leq y\leq 1[/itex]. From this range of values [itex]\|A\|^2[/itex] takes the maximum value when [itex]y=\pm 1[/itex]. And
Why the unit circle!?!? If I knew this then I'd be a lot happier with the solution to these questions. (by the way, the answers are correct - so you dont have to worry about that).
Is it something to do with the fact that we want to normalize the vectors? Who knows?!
Let [itex](x_n)[/itex] be a sequence of elements in a Hilbert Space [itex]\mathcal{H}[/itex] for which [itex]\sum_{n=1}^{\infty} \|x_n\| < \infty[/itex].
So the first thing I deduced is that [itex](x_n)[/itex] is a Cauchy sequence.
This sequence then converges to some element [itex]x_j[/itex]. But I wasnt sure if this [itex]x_j[/itex] was actually in the Hilbert space, and whether or not the sequence [itex](x_n)[/itex] converges to THIS element.
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