Chaitanya's Random Pages

December 26, 2010

Major Straits of the World

Filed under: geography — ckrao @ 12:52 pm

I had a look at map of the world and decided to note down some of the straits (narrow stretches of water between two larger bodies of water) or relatively narrow stretches of ocean. The following list mainly has those between different countries. There are countless more in archipelagos such as in Scandinavia, northern Canada, the Philippines, Indonesia and the Caribbean. Most distances shown refer to the narrowest width of water. The information is taken from the relevant Wikipedia entries (hence not to be taken as definitive) as well as the references given at the bottom.


  • Cook Strait between the two main islands of New Zealand is 23km wide.
  • Bass Strait between mainland Australia and Tasmania is around 240km wide and only 50 metres deep. The Tasmanian Aborigines were isolated from mainland Australia for around 8000 years after sea levels rose.
  • Torres Strait between Australia and Papua New Guinea is 150km wide with at least 274 islands.

South East and South Asia

  • The Malacca Strait between peninsular Malaysia and Sumatra is 805km long and as narrow as 2.8km at Phillips Channel (south of Singapore). It apparently carries one quarter of world’s trade goods.
  • Sunda Strait between Java and Sumatra is 30km wide.
  • Palk Strait between India and Sri Lanka is 30km wide at Adam’s Bridge (containing a chain of islands).
  • Bab-el-Mandeb between Yemen in the Arabian Peninsula and the Horn of Africa is about 30km wide.
  • Hormuz Strait between Oman and Iran across the Persian Gulf, is 54km wide at its narrowest point.

East and North Eurasia

  • The Hainan (Qiongzhou) Strait between the island of Hainan and mainland China is 30km wide.
  • The Taiwan Strait is 131km wide between Taiwan and China.
  • Korea Strait between South Korea and southern Japan is about 200km wide.
  • The largest island of Japan, Honshu, is 19.5km from the northern island Hokkaido (via the Tsugaru Strait), 15km from Shikoku (via the Inland Sea) and just 600 metres from the southern island Kyushu (via the Kanmon Strait).
  • La Pérouse Strait between Sakhalin Island of Russia and Hokkaido Island of Japan is 40km wide.
  • Tartar Strait between east mainland Russia and the island of Sakahlin is 7.3km wide.
  • Matochkin Strait between the Severny and Yuzhny Islands of the Arctic (extreme north east of Europe) is 100km long but only 600m wide at its narrowest point. I had previously thought this was one island!
  • Kara Strait between mainland Russia and Yuzhny Island is 56km wide.


  • Øresund between Zealand in Denmark (containing Copenhagen) and Sweden, is 4km wide at one point.
  • The English Channel (between England and France) is 34km wide and the world’s busiest international seaway.
  • The North Channel between Scotland and Northern Ireland is 36km wide.
  • The Strait of Gibraltar at the entrance to the Mediterranean (between Spain and Morocco) is 14.3km wide and 300-900m deep.
  • The Bosporus (Istanbul, between Asia and Europe) is 704m wide at its narrowest point.
  • The Dardanelles in northwestern Turkey are 61km long but 1.2 to 6km wide and 55-82m deep.
  • The Strait of Messina is 3.1km wide between mainland Italy and the island of Sicily.
  • The Strait of Bonifacio between the Mediterranean islands of Sardinia (Italy) and Corsica (France) is 11km wide.
  • The Euripus Strait between mainland Greece and the island of Euboea (150km in length) is only 38 metres wide at its narrowest point!


  • The Strait of Magellan at the southern tip of South America is 570km long and 2km wide at its narrowest point.
  • The Yucatán Channel between Mexico and Cuba is about 196km wide.
  • The Windward Passage between Cuba and Hispaniola (Haiti) is 80km wide.
  • The Florida Straits between Cuba and Florida are about 160km wide.
  • The Strait of Belle Isle is 125km long and 15-60km wide and separates Newfoundland and mainland Canada (Labrador)
  • The Gulf of California in Mexico is 1126km long and 48-241km wide (actually a narrow bay rather than a strait)
  • The Strait of Georgia between mainland western Canada and Victoria Island is 18.5 to 55km wide and 240km long.
  • The Bering Strait between Alaska and Russia is 85km wide.





December 21, 2010

The Matrix Inversion Lemma

Filed under: mathematics — ckrao @ 4:32 am

I had thought the matrix inversion lemma was difficult to prove, but it is in fact not so tricky!

The lemma states that if A and C are square invertible matrices (and B, D are matrices so that A and BCD have the same dimensions), then

\displaystyle (A + BCD)^{-1} = A^{-1} - A^{-1}B(C^{-1} + DA^{-1}B)^{-1}DA^{-1} \quad (*)

Thanks to [1], it is now easier for me to derive this formula than to remember it, the way much of mathematics should be. Other ways of arriving at the formula are by matrix blockwise elimination or inversion. See the Wikipedia entry on the Woodbury matrix identity (another name for the lemma) for more information.


1. Start with the equation (A + BC)x = b. We find x in terms of b either as x = (A + BC)^{-1} b or as follows.

2. Let y = Cx, giving us the two equations

\begin{array}{rclr}Ax + By &=& b &\quad \text{(1)}\\y &=& Cx &\quad \text{(2)}\end{array}

3. From (1) we obtain

\displaystyle x = A^{-1}(b-By). \quad \text{(3)}

4. Substituting (3) into (2) gives y = CA^{-1} (b-By) and rearranging this gives

\displaystyle y = (I + CA^{-1}B)^{-1}CA^{-1}b. \quad \text{(4)}

5. From (3) and (4) we end up with

\displaystyle x = A^{-1}b - A^{-1}By = \left[A^{-1} - A^{-1}B(I + CA^{-1}B)^{-1}CA^{-1}\right]b.

6. Since b was arbitrary, from step 1 we conclude that

\displaystyle (A + BC)^{-1} = A^{-1} - A^{-1}B(I + CA^{-1}B)^{-1}CA^{-1}.

7. To arrive at the slightly more complicated form (*) we replace C with CD and note that

\displaystyle(I + CDA^{-1}B)^{-1}CD = (I + CDA^{-1}B)^{-1}\left(C^{-1}\right)^{-1}D = (C^{-1} + DA^{-1}B)^{-1}D

(using the result (XY)^{-1} = Y^{-1}X^{-1}).

The matrix inversion lemma is especially useful when it is easy to invert A and C, e.g. if they are diagonal or have small dimension. The latter may occur in recursive formulas such as recursive least squares or the Kalman filter. The lemma is actually a special case of the Binomial inverse theorem, which applies when C is not invertible:

\displaystyle (A + BCD)^{-1} = A^{-1} - A^{-1}BC(C + CDA^{-1}BC)^{-1}CDA^{-1} \quad (*)

A couple more special cases of the matrix inversion lemma follow:

1. Sherman-Morrison formula (B and D replaced by column vectors u and v, C replaced by the identity):

\displaystyle \left(A + uv^T\right)^{-1}= A^{-1} - \frac{A^{-1}uv^TA^{-1}}{1 + v^TA^{-1}u}


\displaystyle (A + B)^{-1} = A^{-1} - A^{-1}B\left(B + BA^{-1}B\right)^{-1}BA^{-1}


\displaystyle (I + A)^{-1} = I - (I + A)^{-1}A


[1] S. Boyd and L. Vandenberghe, Convex Optimization (Appendix C.4.3), Cambridge University Press, 2004

December 11, 2010

2010-11 Ashes – recap after the first two tests

Filed under: cricket,sport — ckrao @ 6:36 am

I had expected Australia to be leading 1-0 after 2 tests, instead they are down 0-1 to an England side playing very well indeed. England now only needs one win in the remaining three matches to retain the Ashes.

Australia started well in Brisbane after bowling out England on the first day for 260. The series was off to a dramatic start when captain Strauss was out in the very first over for a duck. Later in the day England were at 4/197 when Peter Siddle struck with a hat-trick on his birthday, the 11th by an Australian. He finished with 6 wickets. Australia in their first innings were in trouble at 5/143 before a partnership of 307 between Hussey (195) and Haddin (136) saw them amass a first innings lead of 221. Then it was England’s turn to fight back again as they made the game a safe draw declaring at 1/517 including a partnership of 329 between opener Cook (235*) and Trott (135*). Strauss was the lowest scorer for his side in both innings, but managed 110 second time around!

In the second test, Australia could not have had a worse start, losing 3 wickets for 2 runs in the first 13 balls of the game! They never recovered as they were bowled out for 245, thanks largely to Anderson’s 4/51. In Adelaide a first innings score of 500 is closer to par! Hussey again top scored with 93 as he and Haddin again had to rescue Australia. Then it was England’s turn to show how to bat. The Aussie bowlers were highly ineffective on a great batting track as England scored 5d/630. Cook scored 148 and Pietersen a sensational 227. Australia with a deficit of 375 could only manage 304 in reply after losing their last 5 wickets for 18 on a pitch finally offering some assistance for the bowlers. Swann bowled beautifully taking 5/91. Hence England won by an innings and 71 runs, taking 20 wickets while only losing 5 of their own for the match. It was one of the most devastating performances by an away team in Australia that I can recall.

Here are a few interesting statistical highlights coming out of the first two tests.

– At one point Australia took only 3 wickets in 17 hours of play spread over the two tests.

– Of the 9 Australians to have taken hat-tricks (two did it twice), remarkably 7 of them have been Victorians: Trumble (twice), T J Matthews (a hat-trick in each innings!), Kline, M G Hughes, Fleming, Warne (just 2 months after Fleming’s) and now Siddle. Glenn McGrath and Fred Spofforth had the other two, the latter represented Victoria six years after achieving the first hat-trick in test cricket. There was only one Australian hat-trick between 1912 and 1988 (Kline’s in 1958).

– Alastair Cook is the youngest player other than Sachin Tendulkar to have 15 test centuries to his name. More information about that here.  Here is a list of the players who have played the most test matches before the age of 26. Cook will add one more to his current total if he plays in Perth.

Player (team) # tests before 26th birthday
SR Tendulkar (India) 68
N Kapil Dev (India) 66
AN Cook (Eng) 62
RR Sarwan (WI) 61
DL Vettori (NZ) 59
RJ Shastri (India) 58
AB de Villiers (SA) 57
MV Boucher (SA) 56
Harbhajan Singh (India) 56
Mohammad Ashraful (Ban) 55
GC Smith (ICC/SA) 54
Saleem Malik (Pak) 52
JH Kallis (SA) 52
Javed Miandad (Pak) 52
CH Gayle (WI) 52

Cook currently has 4814 runs while Tendulkar had 5177 when he turned 26. It will be interesting to see where they both end up!

– Still on Cook, he scored 383 runs before finally being dismissed (235* + 148). This is the most by an Englishman, beating Wally Hammond’s record of 365 (336* + 29). The all-time record is 497 by Tendulkar. Cook also spent 1058 minutes at the crease before being dismissed (5th most by anyone). The record there is 1513 by Shivnarine Chanderpaul. His double century was the first scored by a visiting batsman in Brisbane. It was also the highest score by an Englishman anywhere since Gooch’s 333 way back in 1990.

– A total of 35 double centuries have been scored against Australia in test cricket. Of these, only Brian Lara’s 277 in Sydney and 226 in Adelaide were scored more rapidly than Pietersen’s 227. The full list of double centuries is here and more info is at the newsgroup here. Surprisingly few double centuries have been scored by Englishmen in Ashes tests. Prior to this series, the only ones since WW2 were by Barrington, Gower, Hussain and Collingwood. Only Foster, Hammond and Collingwood had done so in Australia. After Bradman’s time (he alone scored 8!) the Aussies to have scored double centuries in Ashes tests are Morris, Simpson (twice), Cowper, Stackpole, Border, Taylor and Langer. In recent times it has happened once a decade for each side.

– The Gabba test was only the second time that two 300 run partnerships have occurred in the one match. This previous occasion was between India and Pakistan in Lahore in January 2006, a game in which 8 wickets for 1089 runs were scored. In total there have been 79 triple hundred stands in test cricket, of which 38 have happened since 1999.

– The 329 was England’s highest partnership for any wicket in Australia.

– Jonathon Trott averages 100.5 against Australia in 3 tests (402 runs in 5 innings).

– So far no number 8 batsman has scored a run in the series. In the first test Broad made a first ball duck during Siddle’s hat-trick, Johnson was out for a 19-ball duck, and then in the second test Harris had a king pair (i.e. faced a solitary ball in each innings).

– Australia has now played 5 tests without a win, their last coming against Pakistan in England earlier this year. The last time this happened was 22 years ago against Pakistan (away) and the West Indies (at home). During the mid 1980s they had 14 tests without a win.

– England’s second innings at the Gabba was the first time in test cricket that at least three batsmen played and all scored centuries, i.e. the lowest scorer also made a century.

– Prior to the Cook/Strauss stand of 188 at the Gabba, the previous highest opening stand by England at Brisbane was only 114. The similar record for the second wicket was then extended from 134 to 329.

A big part of Australia’s fall from the number one position has been the decreased ability of individuals to convert 50s to 100s. In the first two tests of this series for example Australians have scored 2 100s and 8 50s, while England have 5 100s and 4 50s. There is more about that in a recent Cricinfo article here.  Another aspect has been that batting collapses have occurred more frequently, and their opposition has been let off the hook on more than one occasion. I shall provide some instances of that in a future post.

More records by S Rajesh from the first test are here and from the second test here. More interesting stats at The Confectionary Stall (a Cricinfo blog by Andy Zaltzman) are here.


December 7, 2010

Why circles map to ellipses under linear maps

Filed under: mathematics — ckrao @ 11:16 am

Here is an interesting fact worth pondering. Take a circle and stretch it along some direction. It becomes an ellipse. Now take the ellipse and stretch it again, this time in a different direction. Continue stretching or shrinking in other directions if you like. No matter what directions are used, the final image is an ellipse, having two perpendicular axes (ignoring the degenerate case of a line segment). The direction of these axes depend on the direction and amount of stretching, but the existence of perpendicular axes always remains. This did not seem immediately obvious to me – surely one can be clever about the way one stretches and remove the perpendicularity of the axes!

A linear transformation of a circle is a sequence of stretches along axes, and such a sequence can always be reduced to at most two perpendicular stretches. That is, circles map to ellipses under linear maps.

One way of seeing why this is the case is that the equation of an ellipse is a quadratic form, and applying linear transformations does not change its degree (i.e. a conic remains a conic). Since the only bounded conic sections are ellipses, the result follows. However I would be more satisfied with a coordinate-free proof. Here is one argument that I read recently and outline below. (See more via Trefethen and Bau, Planetmath, and also the nice explanation due to Aubrey Poor here)

Let T be a linear transformation (map) from n-dimensional to m-dimensional space. T is usually represented by an m by n matrix, but we will not need that here. We will assume complex vector spaces for generality, but a 2-dimensional real vector space is easier to visualise. We wish to show that a unit sphere maps to an ellipsoid, by which we mean there exist perpendicular axes of the sphere that each map to perpendicular axes in the m dimensional space.

Consider the image of the unit sphere S=\{x: ||x|| = 1\} under T. There exists a vector v_1 \in S such that ||Tv_1|| is maximal (by the extreme value theorem). Let Tv_1 = \sigma_1 u_1 where u_1 and v_1 are unit vectors and \sigma_1 > 0 (we ignore the degenerate case \sigma_1 = 0). We wish to show that any vector v_2 perpendicular to v_1 maps to a vector perpendicular to u_1. Suppose Tv_2 = \alpha u_1 + \beta u_2 where u_2 \perp u_1 (such an orthogonal decomposition is possible). We identify a vector that maps to something as least as long as \sigma_1 u_1. Consider the unit vector

\displaystyle v = \frac{\sigma_1 v_1 + \alpha^* v_2}{\left(\sigma_1^2 + |\alpha|^2\right)^{1/2}}.


\displaystyle T v = T\left(\frac{\sigma_1 v_1 + \alpha^* v_2}{\left(\sigma_1^2 + |\alpha|^2\right)^{1/2}}\right) = \frac{\sigma_1^2 u_1 + |\alpha|^2 u_1 + \alpha^* \beta u_2}{\left(\sigma_1^2 + |\alpha|^2\right)^{1/2}}.

The length of this vector is at least

\displaystyle \frac{\sigma_1^2 + |\alpha^2|}{\left(\sigma_1^2 + |\alpha|^2\right)^{1/2}} = \left(\sigma_1^2 + |\alpha|^2\right)^{1/2} \geq \sigma_1,

where equality holds when \alpha = 0. In other words, to prevent Tv from having length greater than \sigma_1, we require \alpha = 0. This means Tv_2 = \beta u_2 \perp u_1 as we wished to show.

By an inductive argument we may then show that there exist orthonormal vectors v_1, v_2, \ldots v_r in the n-dimensional space and orthonormal vectors u_1, u_2, \ldots u_r in the m-dimensional space such that

Tv_i = \sigma_i u_i.

In other words there exist orthonormal bases in the row space and column space of T which map to each other. This is a restatement of the singular value decomposition (SVD), that spheres map to ellipsoids (generalised to any dimension). The \sigma_i are known as singular values and are interpreted as the semi-axes lengths of the ellipsoidal image of the unit sphere. Stacking the v’s and u’s into matrices (and adding orthonormal vectors from nullspaces if necessary) gives us the alternate form

\displaystyle AV =U \Sigma, \quad \text{or } \ A = U \Sigma V^*,

where A is the matrix representation of the linear transformation T, and \Sigma has the singular values down its diagonal and 0 entries elsewhere.

The SVD (which applies to any matrix) is the generalisation of the spectral theorem (which applies to normal matrices) and has wide applications, from solving least squares problems (via the pseudo-inverse) to finding low-rank approximations to matrices (enabling compression).

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