Can Anyone Help Review My Final Exam Problems?

[Oxymoron]

I have a final exam coming up in a few days and I would like some help checking some problems that I have been working through.

 

[Oxymoron]

Question 1

Consider the linear operator A:\mathbb{R}^2\rightarrow\mathbb{R}^2 given by the matrix

\left(\begin{array}{cc}<br /> -2 &amp; 0 \\<br /> 0 &amp; 1 \\<br /> \end{array}\right)

Prove that A is bounded with \|A\|=2

Solution

Okay, here the first thing I did was notice that A is a linear operator, so it takes a 2-element 'vector' (x,y) \in \mathbb{R}^2 and maps it to another.

So by applying the transformation matrix A to a vector we have

\left(\begin{array}{cc}<br /> -2 &amp; 0 \\<br /> 0 &amp; 1 \\<br /> \end{array}\right)\left(\begin{array}{c}<br /> x \\<br /> y \\<br /> \end{array}\right)= \left(\begin{array}{c}<br /> -2x\\<br /> y \\<br /> \end{array}\right)

So we have that \|A\textbf{x}\| = 2x+y which implies that

\|A\textbf{x}\|^2 = 4x^2 +y^2[/itex].<br /> <br /> Now I am stuck. I need to prove that \|A\| \leq M\|x\|, but I have no methods of proceeding any further.<br /> <br /> I have a idea that maybe we let<br /> <br /> \|A\|^2 = \max\{4x^2+y^2\}<br /> <br /> 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?

 

[Oxymoron]

Okay, what about this...

I need to prove that A is bounded with \|A\| = 2. I know already that

\|A\|^2 = \max\{4x^2 + y^2\}

If I subject it to the constraint x^2+y^2 = 1 then after rearranging I have

y^2=1-x^2

Obviously this constaint is the unit circle with domain -1\leq x \leq 1, (since x can only vary between -1 and 1).

Therefore...

\|A\|^2 = \max\{4x^2 + (1-x^2)\}
\|A\|^2 = \max\{3x^2+1\}

Since -1\leq x\leq 1, this proves that A is bounded with norm 2.


If this is correct then my only gripe is why force the constraint? This method seems a bit illogical to me - but I may be wrong. Any ideas?

 

[Oxymoron]

I applied the same method to an alternative example.

If

A=\left(\begin{array}{cc}0 &amp; 2 \\ 0 &amp; 1\end{array}\right)

then

\left(\begin{array}{cc}0 &amp; 2 \\ 0 &amp; 1\\ \end{array}\right)\left(\begin{array}{c}x &amp; y \\ \end{array}\right) = \left(\begin{array}{c}2y &amp; y\\ \end{array}\right)

and

\|A\textbf{x}\|^2 = 4y^2 + y^2 = 5y^2

Hence

\|A\|^2 = \max\{5y^2\}

Subject this to the same constraint, ie -1\leq y\leq 1. From this range of values \|A\|^2 takes the maximum value when y=\pm 1. And

\|A\|^2 = 5
\|A\| = \sqrt{5}

 

[Oxymoron]

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 don't have to worry about that).

Is it something to do with the fact that we want to normalize the vectors? Who knows?!

 

[Oxymoron]

Another question...

Question 2

Let (x_n) be a sequence of elements in a Hilbert Space \mathcal{H} for which \sum_{n=1}^{\infty} \|x_n\| &lt; \infty.

Show that the sequence of partial sums

s_n = \sum_{k=1}^{n} x_n

converges to a point in \mathcal{H}.

 

[Oxymoron]

So the first thing I deduced is that (x_n) is a Cauchy sequence.

This sequence then converges to some element x_j. But I wasnt sure if this x_j was actually in the Hilbert space, and whether or not the sequence (x_n) converges to THIS element.

So

\sum |x_j|^2 = \sum |\lim_{n\rightarrow\infty}x_n|^2

= \lim_{n\rightarrow\infty}\sum |x_n|^2

= \lim_{n\rightarrow\infty}\|x_n\|^2 \in \mathcal{H}

\leq \lim_{n\rightarrow\infty} M

where \|x_n\|^2 &lt; M.

I know that this M exists because all Cauchy sequences are bounded. So now

\sum |x_j|^2 = \lim_{n\rightarrow\infty}\|x_n\|^2 \leq M

which is an increasing and bounded above. Hence

\sum \|x_n\|^2 &lt; \sum \|x_n\| &lt; \infty

and the sequence converges to a point in \mathcal{H}

 

[Oxymoron]

I guess no-one knows or has the time. *sigh*