Need Help with Final Exam Problems?

  • Thread starter Oxymoron
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    Exam Final
In summary, the sequence (x_n) converges to a point in \mathcal{H} as long as the sum of all the x_n's is less than or equal to M.
  • #1
Oxymoron
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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.
 
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  • #2
Question 1

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

[tex]\left(\begin{array}{cc}
-2 & 0 \\
0 & 1 \\
\end{array}\right)[/tex]

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?
 
  • #3
Okay, what about this...

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

[tex]\|A\|^2 = \max\{4x^2 + y^2\}[/tex]

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

[tex]y^2=1-x^2[/tex]

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

Therefore...

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

Since [itex]-1\leq x\leq 1[/itex], this proves that [itex]A[/itex] 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:
  • #4
I applied the same method to an alternative example.

If

[tex]A=\left(\begin{array}{cc}0 & 2 \\ 0 & 1\end{array}\right)[/tex]

then

[tex]\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)[/tex]

and

[tex]\|A\textbf{x}\|^2 = 4y^2 + y^2 = 5y^2[/tex]

Hence

[tex]\|A\|^2 = \max\{5y^2\}[/tex]

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

[tex]\|A\|^2 = 5[/tex]
[tex]\|A\| = \sqrt{5}[/tex]
 
  • #5
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?!
 
  • #6
Another question...

Question 2

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].

Show that the sequence of partial sums


[tex] s_n = \sum_{k=1}^{n} x_n[/tex]

converges to a point in [itex]\mathcal{H}[/itex].
 
  • #7
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.

So

[tex]\sum |x_j|^2 = \sum |\lim_{n\rightarrow\infty}x_n|^2[/tex]

[tex] = \lim_{n\rightarrow\infty}\sum |x_n|^2[/tex]

[tex] = \lim_{n\rightarrow\infty}\|x_n\|^2 \in \mathcal{H}[/tex]

[tex] \leq \lim_{n\rightarrow\infty} M[/tex]

where [itex]\|x_n\|^2 < M[/itex].

I know that this [itex]M[/itex] exists because all Cauchy sequences are bounded. So now

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

which is an increasing and bounded above. Hence

[tex]\sum \|x_n\|^2 < \sum \|x_n\| < \infty[/tex]

and the sequence converges to a point in [itex]\mathcal{H}[/itex]
 
  • #8
I guess no-one knows or has the time. *sigh*
 

1. What topics will be covered on the final exam?

The final exam will cover all of the material we have discussed throughout the course. This includes key concepts, theories, and methods that we have covered in lectures and readings.

2. How many questions will be on the final exam?

The final exam will have a total of 50 multiple-choice questions. There will also be a few short answer questions that require you to demonstrate your understanding of key concepts.

3. Will the final exam be cumulative?

Yes, the final exam will cover material from the entire course. However, the majority of the questions will focus on the most recent topics and readings.

4. Can we use a cheat sheet or notes during the final exam?

No, the final exam will be closed book and closed notes. This is to ensure that all students are evaluated based on their own understanding and knowledge of the material.

5. Will there be any extra credit questions on the final exam?

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