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Questions for Final Exam

  1. Jun 11, 2005 #1
    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. jcsd
  3. Jun 11, 2005 #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?
     
  4. Jun 12, 2005 #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: Jun 12, 2005
  5. Jun 12, 2005 #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]
     
  6. Jun 12, 2005 #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 dont have to worry about that).

    Is it something to do with the fact that we want to normalize the vectors? Who knows?!
     
  7. Jun 12, 2005 #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].
     
  8. Jun 12, 2005 #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]
     
  9. Jun 13, 2005 #8
    I guess no-one knows or has the time. *sigh*
     
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