Are a matrix and it's transpose similar?
Jason Lee
matrix to have complex entries.
Using the rational canonical form, I know this to be true for matrices of
size up to 3x3, but can't seem to find anything about matrices of larger size.
JLee
Robin Chapman
jl...@euclid.ucsd.edu (Jason Lee) wrote:
> Is it true that a square matrix and its transpose are similar? Assume the
> matrix to have complex entries.
Yes. It's true over any field.
> Using the rational canonical form, I know this to be true for matrices of
> size up to 3x3, but can't seem to find anything about matrices of larger size.
Using rational canonical form it reduces to showing that a companion
matrix is similar to its transpose. This reduces to noting that a
companion matrix has the same minimal polynomial as its transpose.
Robin Chapman + "They did not have proper
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> Using rational canonical form it reduces to showing that a companion
> matrix is similar to its transpose. This reduces to noting that a
> companion matrix has the same minimal polynomial as its transpose.
Well, a little more... they have the same elementary divisors (or
invariant factors)
Oscar Lanzi III
polynomial. Using elementary properties of the transpose, one can
easily prove that transposition has does not change the characteristic
polynomial. Therefore, a matrix and its transpose are always similar.
--OL
Robin Chapman
No. A COMPANION matrix has only one elementary divisor/invariant factor.
Robin Chapman
No. Counterexample
(0 0) (0 1)
(0 0) and (0 0).
Norris, Douglas Todd
>
> In article <16051-35...@newsd-153.iap.bryant.webtv.net>,
> o...@webtv.net (Oscar Lanzi III) wrote:
> > Two matrices are similar if they have the same characteristic
> > polynomial. Using elementary properties of the transpose, one can
> > easily prove that transposition has does not change the characteristic
> > polynomial. Therefore, a matrix and its transpose are always similar.
>
> No. Counterexample
> (0 0) (0 1)
> (0 0) and (0 0).
Could you explain the part again where your two matrices are
transposes of each other?
Doug
Harry Dewhurst
notice that expanding det(A) by rows is the same as expanding det(At) by
columns.
Robin Chapman wrote:
>
> In article <16051-35...@newsd-153.iap.bryant.webtv.net>,
> o...@webtv.net (Oscar Lanzi III) wrote:
> > Two matrices are similar if they have the same characteristic
> > polynomial. Using elementary properties of the transpose, one can
> > easily prove that transposition has does not change the characteristic
> > polynomial. Therefore, a matrix and its transpose are always similar.
>
> No. Counterexample
> (0 0) (0 1)
> (0 0) and (0 0).
>
Robin Chapman
"Norris, Douglas Todd" <norr...@euclid.colorado.edu> wrote:
> This is a multi-part message in MIME format.
>
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>
> Robin Chapman wrote:
> >
> > In article <16051-35...@newsd-153.iap.bryant.webtv.net>,
> > o...@webtv.net (Oscar Lanzi III) wrote:
> > > Two matrices are similar if they have the same characteristic
> > > polynomial. Using elementary properties of the transpose, one can
> > > easily prove that transposition has does not change the characteristic
> > > polynomial. Therefore, a matrix and its transpose are always similar.
> >
> > No. Counterexample
> > (0 0) (0 1)
> > (0 0) and (0 0).
>
> transposes of each other?
Is this a lame attempt at sarcasm? Lanzi asserted that two matrices
with the same characteristic polynomial are similar. Here are two matrices,
with the same characteristic polynomial, which are not similar. It should
be obvious to even the dullest mathematician that these matrices are
not transposes of each other.
Robin Chapman
Robin Chapman
dewh...@atl.mediaone.net wrote:
> A square matrix is always similar to it's transpose. To see this, just
> notice that expanding det(A) by rows is the same as expanding det(At) by
> columns.
Do you mean "expanding det(A) by columns"? Anyway this is a non sequitur:
two matrices being similar is a stronger property than having equal
determinants. See my first post in this thread for an outline of a proof.
Gerhard Niklasch
"Norris, Douglas Todd" <norr...@euclid.colorado.edu> writes:
|> Robin Chapman wrote:
|> > No. Counterexample
|> > (0 0) (0 1)
|> > (0 0) and (0 0).
|>
|> Could you explain the part again where your two matrices are
|> transposes of each other?
They aren't. Robin's Counterexample is a Counterexample to the claim
by Oscar Lanzi III in <16051-35...@newsd-153.iap.bryant.webtv.net>
that
|> > > Two matrices are similar if they have the same characteristic
|> > > polynomial.
The above two have the same characteristic polynomial X^2, but aren't
similar over _any_ field of coordinates (or indeed over any ring in
which 0 is distinct from 1).
Gerhard
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Edward C. Hook
"Norris, Douglas Todd" <norr...@euclid.colorado.edu> writes:
2> Robin Chapman wrote:
3> >
4> > In article <16051-35...@newsd-153.iap.bryant.webtv.net>,
5> > o...@webtv.net (Oscar Lanzi III) wrote:
6> > > Two matrices are similar if they have the same characteristic
7> > > polynomial. Using elementary properties of the transpose, one can
8> > > easily prove that transposition has does not change the characteristic
9> > > polynomial. Therefore, a matrix and its transpose are always similar.
a> >
b> > No. Counterexample
c> > (0 0) (0 1)
d> > (0 0) and (0 0).
e>
f> Could you explain the part again where your two matrices are
0> transposes of each other?
1>
Obviously, they aren't -- however, Oscar Lanzi III asserted
that matrices having the same characteristic polynomial are
necessarily similar. Robin Chapman's pair of matrices give a
counterexample to _that_ claim.
|> Doug
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Daniel Kressner
>
same eigenvalues. But the geometrical and algebraical multiples
of the eigenvalues must be equal. For the solution consider
the Jordan Decomposition: A is similar to J with
[ 0 1
J= diag(eigenvalues) + [ 0 1
[ 0 0
[ 0 1 ..
So A^T is similar to J'. With permutation matrices you can move the
ones under the diag above.
So A is similar to A^T.
-----------------------------------
Daniel Kressner
http://www.tu-chemnitz.de/~dkr/
Norris, Douglas Todd
> > This is a multi-part message in MIME format.
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> >
> > Robin Chapman wrote:
> > >
> > > In article <16051-35...@newsd-153.iap.bryant.webtv.net>,
> > > > polynomial. Using elementary properties of the transpose, one can
> > > > easily prove that transposition has does not change the characteristic
> > > > polynomial. Therefore, a matrix and its transpose are always similar.
> > >
> > > (0 0) (0 1)
> >
>
> Is this a lame attempt at sarcasm? Lanzi asserted that two matrices
> with the same characteristic polynomial are similar. Here are two matrices,
> with the same characteristic polynomial, which are not similar. It should
> be obvious to even the dullest mathematician that these matrices are
> not transposes of each other.
That's my point - you didn't say what you were providing a
counterexample
to.
If you don't prefer "lame sarcasm", perhaps you should be more specific.
Doug
deathscythe
polynomial even though they are not similar to each other.
And it's true that every complex matrix is similar to its transpose. You may
first prove that it's true for any nilpotent matrix and then investigate the
Jordan form of an arbitrary matrix.
However, i don't know if the same result holds if the entries of the matrix
are from other algebraic structure, e.g., finite fields,....etc.
deathscythe
Robin Chapman
I'm sorry but I was under the (no doubt naive) impression that on sci.math
I was addressing an intelligent audience.
Brian Stewart
>
> And it's true that every complex matrix is similar to its transpose. You may
> first prove that it's true for any nilpotent matrix and then investigate the
> Jordan form of an arbitrary matrix.
>
> However, i don't know if the same result holds if the entries of the matrix
> are from other algebraic structure, e.g., finite fields,....etc.
The proper canonical form to use for this question is not the Jordan
form but the Rational Canonical Form; every matrix A is similar (over
any field containing its entries) to a direct sum of companion matrices
C(g_1), C(g_2), ..., C(g_s) where each polynomial g_i divides the next
g_{i+1}. There is a standard algorithm to compute the g_i; just use row
and column operations to systematically reduce the matrix A-xI to Smith
normal form, that is to diagonal form, where the diagonal entries are
1,1, .., g_1, g_2, ..., g_s where the g_i divide each other as before.
Given that all this works (see any good book on linear algebra) it is
clear that A and its (sic) transpose A^t are similar since clearly if we
swap row and column operations a reduction of A becomes a reduction of
A^t.
B.
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Robin Chapman
"deathscythe" <wingdea...@mailexcite.com> wrote:
> Robin wanted to tell that two matrices may have the same characteristic
> polynomial even though they are not similar to each other.
> first prove that it's true for any nilpotent matrix and then investigate the
> Jordan form of an arbitrary matrix.
>
> However, i don't know if the same result holds if the entries of the matrix
> are from other algebraic structure, e.g., finite fields,....etc.
If two matrices over a field F are similar over a larger field (e.g.
if they have rational coefficients and are equivalent over the complexes)
then they are already similar over F. The proof of this is not entirely
trivial, and can be based on e.g., the rational canonical form. But
when one has the rational canonical form one can prove the similarity
of a matrix to its transpose directly.
On the other hand over a general ring a matrix need not be similar
to its transpose. For instance over the integers Z let A be the matrix
(0 1 0)
(0 0 2)
(0 0 0).
There is no matrix B in GL(3, Z) (i.e., with B and B^{-1} having integer
entries) with BAB^{-1} equalling the transpose of A.
Norris, Douglas Todd
>
>
> I'm sorry but I was under the (no doubt naive) impression that on sci.math
> I was addressing an intelligent audience.
That's funny. Maybe if this whole mathematician thing doesn't
work out, you could make it in American sit-coms.
Doug
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