J.P. McCarthy: Math Page

MA 2055: Some Linear Algebra Exam Questions

July 21, 2011 in MA 2055 | Leave a comment

Summer 2011: Question 3

Suppose that $n$ is a positive integer. Pick $n$ intervals $I_1,I_2,\dots,I_n$ on the real number line. Assume that any pair of these intervals are disjoint. Pick $n$ real numbers $a_1,a_2,\dots,a_n$ .

Using ideas of linear algebra, prove that there is a polynomial function $p(x)$ of degree at most $n-1$ , with real number coefficients, so that $\int_{I_j}p(x)\,dx=a_j$
How many such polynomial functions are there? Justify your answer?

[HINT: Make a linear map. If the integral of a continuous function on an interval vanishes, then the function vanishes somewhere on the interval].

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MS 2001: Autumn 2007 Q. 2

July 21, 2011 in MS 2001 | Leave a comment

Determine whether the following functions are defined and where they are continuous.

$f(x)=\frac{x+\sin \pi x}{x^2+3x+2}$ , $g(x)=\frac{x^2}{x+1}$ .

Solution

As $\sin \pi x$ is defined everywhere, $f(x)$ is defined as long as the denominator, $x^2+3x+2\neq0$ . Now

$x^2+3x+2=x^2+2x+x+2=x(x+2)+1(x+2)=(x+1)(x+2)$ ,

therefore $x^2+3x+2\neq0\Leftrightarrow x\neq -1,-2$ . So $f(x)$ is defined for all $x\in\mathbb{R}$ except $x=-1,-2$ . As $x+\sin\pi x$ is continuous (as the sum of a polynomial and a sine) and $x^2+3x+2$ is continuous (as a polynomial), $f(x)$ is continuous as long as $x^2+3x+2\neq 0\Leftrightarrow x\neq-1,-2$ .

Similarly $g(x)$ is defined for, and continuous at, all $x+1\neq0\Leftrightarrow x\neq -1$ .

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ST 1023: Histogram Question

July 19, 2011 in Grinds | Leave a comment

Taken from here.

A manufacturing company keeps records of the numbers of defective items it produces per day. A random sample of days was selected. From their records, the company calculated the proportion of defective items produced per day. The frequency distribution of of proportions defective is as follows:

Proportion Defective	Number of Days
0 – 1%	66
1 – 2%	44
2 – 3%	32
3 – 4%	19
4 – 5%	8
5 – 6%	5
6 – 8%	4
8% or more	2

(a)

Draw a histogram for the distribution on proportions defective. Comment on its shape.

The histogram is skewed to the left towards 0.

(b)

Explain how you determined the heights of the bars for the last two frequency classes in preparing the histogram.

A histogram is constructed such that the area under each bar is equal to the number in the class interval. For the 6 – 8% class interval the base was 2 and the number in the class interval was 4 so we wanted height $\times$ base = 2; i.e. the base to equal 2. In the 8 – 100% the height is effectively zero — by the same analysis we have height = $2/98$ .

(c)

Provide a numerical measure to describe the proportions defective. Justify your choice of numerical measure.

Using the formula for the mean, $\bar{x}$ , of a frequency distribution:

$\bar{x}=\frac{\sum_{i}f_ix_i^*}{\sum_{i}f_i}$ ,

where the sum is over all class intervals where $f_i$ is the number of elements in the $i$ th class interval and $x_i^*$ is the value of the midpoint of the $i$ th class interval.

We will however crop the 8 – 100% class to 8 – 10% to prevent a massive bias here. This will improve our calculation because any data above 10% is clearly an outlier (which is bad for the mean — see below).

Thus we get

$\bar{x}=1.972\%$ .

We use the mean as there are not too many outliers and while the data is skewed, the data is relatively spread out.

(d)

Calculate the first quartile and interpret it’s value. State any assumptions you make. Assess whether these assumptions are valid.

There are 180 days in total so one quarter is 45. Now the lowest 45 are in the class interval 0 – 1% which actually includes 66 elements. Hence we look to take just $45/66\approx0.682$ of the first bar. We do this as follows:

So the first quarter lies between 0 and 0.682%… therefore the first quartile is 0 — 0.682%. The interpretation is that the 25% best days have less than 0.682% defections.

To do this we assume that the distribution is uniform across the class interval 0 – 1%.

While this answer may well be accurate we can’t be too sure — particularly when the model suggest there are days with very little defections. For example, this assumption demands that there is a day when there are 0.015% defections; i.e. 1 in 6,666 — but does the manufacturer ever make this many a day?

(e)

The manufacturing company operates a policy which specifies that if the proportion defective exceeds 4.75% on any given day, an investigation of the manufacturing process must be undertaken. Estimate the percentage of days on which such an investigation would be undertaken.

Looking at the histogram there were 5+4+2=11 days when the proportion defective exceeded 5%. Now the 4.75 – 5% class sub-interval corresponds to 1/4 of the 4 – 5% class interval — that is 2 days. Hence we have 11+2=13/180 days when the proportion defective exceeded 4.75%. 13/180 translates to an estimation of 7.2% of days of proportions defective exceeding 4.75%.

(f)

The manufacturing company reviewed the policy referred to in part (e). They want to change the cut-off for the proportion defective that necessitates the investigation of the manufacturing process (currently 4.75%). Estimate the cut-off that should be used to ensure that the percentage of days on which the investigations would be undertaken is at most 5%.

We want to adjust the 7.2% above to 5%. 5% of 180 days is nine days. So which were the nine worst days? We have six days worse than 6% and then we want to take 3 of the five bad days in 5 -6% class interval to make up the worst nine days. Now we want to take, therefore, the class sub-interval 5.4 – 6%, which is 3/5 of that interval. Hence the new threshold is 5.4% (which estimates an investigation on 5% of days).

Von Neumann Algebras: The Kaplansky Density Theorem

July 7, 2011 in C*-Algebras & Operator Theory, Research | Leave a comment

Taken from C*-algebras and Operator Theory by Gerald Murphy.

We prepare the way for the density theorem with some useful results on strong convergence.

Theorem 4.3.1

If $H$ is a Hilbert space, the involution $T\mapsto T^*$ is strongly continuous when restricted to the set of normal operators of $B(H)$ .

Proof

Let $x\in H$ and suppose that $T,S$ are normal operators in $B(H)$ . Then

$\|(S^*-T^*)(x)\|^2=\langle S^*x-T^*x,S^*x-T^*x\rangle$

$=\|Sx\|^2-\|Tx\|^2+\langle TT^*x,x\rangle-\langle ST^*x,x\rangle$

$+\langle TT^*x,x\rangle-\langle TS^*x,x\rangle$

$=\|Sx\|^2-\|Tx\|^2+\langle (T-S)T^*x,x\rangle+\langle x,(T-S)T^*x\rangle$

$\leq \|Sx\|^2-\|Tx\|^2+2\|(T-S)T^*x\|\|x\|.$

If $\{T_\lambda\}_{\lambda\in\Lambda}$ is a net of normal operators strongly convergent to a normal operator $T$ , then the net $\|T_\lambda x\|^2$ is convergent to $\|Tx\|^2$ and the net $\{(T-T_\lambda)T^*x\}$ is convergent to $0$ , so $\{T_\lambda^*x-T^*x\}$ is convergent to $0$ . Therefore, $\{T_\lambda^*\}$ is strongly convergent to $T^*$ $\bullet$

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Von Neumann Algebras: The Weak and Ultraweak Topologies

July 6, 2011 in C*-Algebras & Operator Theory, Research | 2 comments

Taken from C*-algebras and Operator Theory by Gerald Murphy.

Preparatory to our introduction of the weak and ultraweak topologiesm we show now that $L^1(H)$ is the dual of $K(H)$ , and $B(H)$ is the dual of $L^1(H)$ .

Let $H$ be a Hilbert space, and suppose that $T\in L^1(H)$ . It follows from Theorem 2.4.16 (https://jpmccarthymaths.wordpress.com/2011/01/18/c-algebras-and-operator-theory-2-4-compact-hilbert-space-operators/) that the function

$\text{tr}(T\cdot):K(H)\rightarrow\mathbb{C}$ , $S\mapsto \text{tr}(TS)$ ,

is linear and bounded, and $\|\text{tr}(T\cdot)\|\leq \|T\|$ . We therefore have a map

$L^1(H)\rightarrow K(H)^\star$ , $T\mapsto \text{tr}(T\cdot)$ ,

which is clearly linear and norm-decreasing. We call this map the canonical map from $L^1(H)$ to $K(H)^\star$ .

Theorem 4.2.1

If $H$ is a Hilbert space, then the canonical map from $L^1(H)$ to $K(H)^\star$ is an isometric linear isomorphism.

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MS 2001: Autumn Examination

July 5, 2011 in MS 2001 | Leave a comment

Please find the solutions to the Summer exam here. Note that these also include the marking scheme — numbers in bold brackets indicate marks, i.e. [3] implies three marks.

You will need to do exercises to prepare for your repeat exam. All of the following are of exam grade. If you have any questions please do not hesitate to use the comment function on the bottom:

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Von Neumann Algebras: The Double Commutant Theorem

June 24, 2011 in C*-Algebras & Operator Theory, Research | 3 comments

Taken from C*-algebras and Operator Theory by Gerald Murphy.

A useful way of thinking of the theory of C*-algebras is as “non-commutative topology”. This is justified by the correspondence between abelian C*-algebras and locally compact Hausdorff spaces given by the Gelfand representation. The algebras studied in this chapter, von Neumann algebras, are a class of C*-algebras whose study can be thought of as “non-commutative measure theory”. The reason for the analogy in this case is that the abelian von Neumann algebras are (up to isomorphism) of the form $L^\infty(\Omega,\mu)$ , where $(\Omega,\mu)$ is a measure space.

The theory of von Neumann algebras is a vast and very well-developed area of the theory of operator algebras. We shall be able only to cover some of the basics. The main results of this chapter are the von Neumann double commutant theorem and the Kaplansky density theorem.

St Num	Test / 15	Maple / 15	CA / 30	Exam % for Pass
219	15	15	30	14
905	13	13	26	20
960	11	15	26	20
993	11	14	25	21
102	11	13	24	23
BL	9	15	24	23
787	9	13	22	26
675	8	12	20	29
904	6	14	20	29
484	6	13	19	30
271	5	14	19	30
047	5	14	19	30
MS	0.00	11	11	41
Ave	7.83	13.54	21.37	27

MATH6014: Exercise Sheets, etc.

May 6, 2011 in MATH6014 | Leave a comment

Here are a number of additional exercises for those who never got them in class:

The following are all of exam level difficulty:

	J.P. McCarthy on MATH6040: Spring 2022, Week…
	J.P. McCarthy on MATH6040: Spring 2022, Week…
	J.P. McCarthy on MATH7019: Winter 2020, Week…
	J.P. McCarthy on MATH7019: Winter 2020, Week…
	A Sufficient Conditi… on Almost All Trees have Quantum…

J.P. McCarthy: Math Page

MA 2055: Some Linear Algebra Exam Questions

Summer 2011: Question 3

MS 2001: Autumn 2007 Q. 2

Solution

ST 1023: Histogram Question

(a)

(b)

(c)

(d)

(e)

(f)

Von Neumann Algebras: The Kaplansky Density Theorem

Theorem 4.3.1

Proof

Von Neumann Algebras: The Weak and Ultraweak Topologies

Theorem 4.2.1

MS 2001: Autumn Examination

Von Neumann Algebras: The Double Commutant Theorem

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MS 2001: Summer 2011 Solutions.

MATH6037: Continuous Assessment Summary

MATH6014: Exercise Sheets, etc.

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Summer 2011: Question 3

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Solution

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(f)

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Theorem 4.3.1

Proof

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Theorem 4.2.1

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