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Computing orthogonal complements on finite tori
David Wilding
MRSC 2012
Many of us would be able to sketch the row space of a (sufficientlysmall!) real matrix, but what about the row space of a matrix whoseentries are congruence classes of integers? As you’ll see in the talk,it’s possible, and actually quite fun, to draw such spaces and theirorthogonal complements on the surface of a torus. I’ll explain whatan orthogonal complement is in this context, although you mightwell be able to guess, and I’ll describe a nice way to compute suchthings.
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These first four slides show the row spaces of various matrices inthe real plane R2. The origin is the black dot in the middle and aswe increase the top right entry in the matrix the row space (whichis a line because the matrix only has one non-zero row) increases insteepness.
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These five slides show the row spaces of the same matrices (plusone more) in the plane Z2
15, which has 152 = 225 points in total.All operations here are carried out modulo 15, so the left and rightedges of the plane should be identified. Similarly the bottom andtop edges should be identified, forming a torus.
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These five slides show the row spaces of the same matrices again,but this time in the plane Z2
60 drawn as a torus. The first coordinatedirection goes right from the origin (the black dot) round the torus,and the second coordinate direction goes up from the origin over thetorus. All operations here (and for the rest of the talk) are carriedout modulo 60.
Orthogonality [a b
]·[x y
]= ax + by = 0
For example
I[1 0
]and
[0 1
]are orthogonal
I so are[1 2
]and
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Orthogonality [a b
]·[x y
]= ax + by = 0
For example
I[1 0
]and
[0 1
]are orthogonal
I so are[1 2
]and
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]
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Two vectors are orthogonal if their dot product is zero (modulo 60).This allows more vectors to be orthogonal than we would usuallyexpect. The orthogonal complement of a set X of vectors is the setof vectors y with x · y = 0 for all x ∈ X . The purpose of this talk isto show you how to, given a matrix A, find a matrix B whose rowspace is the orthogonal complement of the row space of A.
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These two slides show a simple row space followed by its orthogonalcomplement. I use a squiggly arrow for the procedure “find a matrixwhose row space is the orthogonal complement” described above.
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A more complicated row space has, perhaps unsurprisingly, a morecomplicated orthogonal complement.
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By giving the matrix another non-zero row I can double the numberof points in the row space. I then get the same kind of orthogonalcomplement, but this time with half as many points. This is ageneral result: the product of the size of a row space and the sizeof its orthogonal complement is the size of the whole space. In thisexample 120 · 30 = 602.
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]
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]
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]
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[a 00 b
]
[60/a 00 60/b
]
[a 00 b
]
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[a 00 b
]
[60/a 00 60/b
]
[a 00 b
]
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Finding an orthogonal complement matrix for a diagonal matrix iseasy: just divide 60 by the diagonal entries. Doing this a second timeproduces the original matrix again, which suggests that the doubleorthogonal complement of a row space is (for diagonal matrices atleast) the same as the row space itself. In fact this result is true forarbitrary matrices.
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[a 00 b
]
[60/a 00 60/b
]
[a 00 b
]Hang on . . .
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The procedure for finding an orthogonal complement matrix for adiagonal matrix is not quite as simple as I suggested because thediagonal entries might not divide 60.
[a 00 b
]and
[gcd(a, 60) 0
0 gcd(b, 60)
]are equivalent
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Replacing the diagonal entries by their greatest common divisorswith 60 produces (without changing the row space) a matrix whosediagonal entries do divide 60. These new entries can then be dividedinto 60 to produce an orthogonal complement matrix.
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The greatest common divisor of 53 and 60 is 1, so 53 is replacedby 60/1, which is zero (modulo 60). Similarly the greatest commondivisor of 25 and 60 is 5, so 25 is replaced by 60/5 = 12. Applyingthis procedure again does not give the original matrix, but at leastit gives one with the original row space.
Arbitrary A
BNt
Diagonal MAN B
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Arbitrary A
BNt
Diagonal MAN
B
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Arbitrary A
BNt
Diagonal MAN B
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Arbitrary A BNt
Diagonal MAN B
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To find an orthogonal complement matrix for an arbitrary matrix A,first make it diagonal by multiplying it on the left and on the right byinvertible matrices M and N (this can always be done), then find anorthogonal complement matrix B for this diagonal matrix. Finallymultiply B on the right by the transpose of the invertible matrix N.
To get M
I permute rows
I negate rows
I add multiples of rows
To get N
I permute columns
I negate columns
I add multiples of columns
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To get M
I permute rows
I negate rows
I add multiples of rows
To get N
I permute columns
I negate columns
I add multiples of columns
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The matrices M and N can be found using row operations andcolumn operations, and they will be invertible because the allowedoperations are invertible. For example, adding a multiple of a rowto a different row can be undone by subtracting off that multipleagain.
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These final four slides illustrate the full procedure for finding anorthogonal complement matrix for an arbitrary matrix. The rowspace is first untwisted by multiplying the matrix on the right byan invertible matrix (in this example no matrix M is needed). Theorthogonal complement of this untwisted row space is then foundand, finally, twisted using the transpose of the invertible matrix.
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