# Homework Help: Questions for Final Exam

1. Jun 11, 2005

### 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.

2. Jun 11, 2005

### Oxymoron

Question 1

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

$$\left(\begin{array}{cc} -2 & 0 \\ 0 & 1 \\ \end{array}\right)$$

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

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

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

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

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?

3. Jun 12, 2005

### Oxymoron

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?

Last edited: Jun 12, 2005
4. Jun 12, 2005

### Oxymoron

I applied the same method to an alternative example.

If

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

then

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

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

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

5. Jun 12, 2005

### 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 dont have to worry about that).

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

6. Jun 12, 2005

### Oxymoron

Another question...

Question 2

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

Show that the sequence of partial sums

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

converges to a point in $\mathcal{H}$.

7. Jun 12, 2005

### 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 $\|x_n\|^2 < M$.

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

$$\sum \|x_n\|^2 < \sum \|x_n\| < \infty$$

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

8. Jun 13, 2005

### Oxymoron

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