What is the Tangent Space for a Given Matrix A?

[evalover1987]

Homework Statement

Homework Equations

The Attempt at a Solution

 

[arkajad]

I think that, excluding the zero matrix, for 2x2 matrices being rank 1 is equivalent to det(A)=0. That means tangent vectors are given by traceless matrices. That should be enough to solve your problem.

 

[evalover1987]

I think that, excluding the zero matrix, for 2x2 matrices being rank 1 is equivalent to det(A)=0. That means tangent vectors are given by traceless matrices. That should be enough to solve your problem.

tracelessness is clearly not required for det(A) = 0.

1 0
0 0 1 1
1 1

are not traceless.

I'm just trying to figure out what formula to use to find out the tangent space at the matrix A.
Does it mean that if B is in tangent space at matrix A, then
BA = 0 ?
or does it mean something else?

 

[arkajad]

I was talking about tangent vectors - they should be traceless. See "http://www.math.ntnu.no/~hanche/notes/diffdet/diffdet-600dpi.ps" ".

 

[evalover1987]

I was talking about tangent vectors - they should be traceless. See "http://www.math.ntnu.no/~hanche/notes/diffdet/diffdet-600dpi.ps" ".

sorry I misunderstood you.
unfortunately, i cannot open ps file.
(unless it's converted to pdf)
i guess i'll look for some other references.

 

[evalover1987]

I was talking about tangent vectors - they should be traceless. See "http://www.math.ntnu.no/~hanche/notes/diffdet/diffdet-600dpi.ps" ".

actually, in addition to traceless matrices, shouldn't the original matrix

0 0
1 1

be included in the basis for tangent space?

 

[arkajad]

Another possible approach: A is of rank 1 if and only if there are invertible U,V such that

A=U\begin{pmatrix}1&0\\0&0\end{pmatrix}V

Therefore a path in rank 1 matrices can be described as

A(t)=U(t)\begin{pmatrix}1&0\\0&0\end{pmatrix}V(t)

Differentiating:

\dot{A}=\dot{U}U^{-1}A+AV^{-1}\dot{V}.

This gives you the general form of a tangent vector at A.

P.S. This is not very useful though ...

 

[evalover1987]

thanks for help, and I again apologize for my rudeness in the previous post

Another possible approach: A is of rank 1 if and only if there are unitary U,V such that

A=U\begin{pmatrix}1&0\\0&0\end{pmatrix}V

Therefore a path in rank 1 matrices can be described as

A(t)=U(t)\begin{pmatrix}1&0\\0&0\end{pmatrix}V(t)

Differentiating:

\dot{A}=\dot{U}U^{-1}A+AV^{-1}\dot{V}.

Both \dot{U}U^{-1} and V^{-1}\dot{V}

are anti-hermitian. This gives you the general form of a tangent vector at A.

 

[arkajad]

thanks for help, and I again apologize for my rudeness in the previous post

I was not very helpful. I tried to be, but I was too fast, and what I wrote is not yet a solution. But I think the solution is pretty close.

 

[evalover1987]

I was not very helpful. I tried to be, but I was too fast, and what I wrote is not yet a solution. But I think the solution is pretty close.

yeah... while I believe that

0 1
0 0


0 0
1 0


two traceless matrices should be basis elements of tangent space,
since 2 by 2 matrices having rank 1 can be thought of as 3-dimensional submanifold,
the tangent space must also have 3 basis elements.

so I was wondering maybe the

0 0
1 1

must be the missing basis element in this case

 

[arkajad]

Take L=\begin{pmatrix}0&1\\-1&0\end{pmatrix}. Exponentiate it to get S(t)=exp(L(t)).
Calculate

S(t)\begin{pmatrix}0&0\\1&1\end{pmatrix}S(-t)

I think this gives you the needed path and the tangent vector L.

 

[evalover1987]

actually, I thnk I got it.
thanks for the help, thou :)

Take \begin{pmatrix}1&0\\0&-1\end{pmatrix}. Exponentiate it to get:
S(t)=\begin{pmatrix}e^t&0\\0&e^{-t}\end{pmatrix}

Calculate

S(t)\begin{pmatrix}0&0\\1&1\end{pmatrix}S(-t)=\begin{pmatrix}0&0\\e^{-2t}&1\end{pmatrix}

P.S. Still confused ...

 

[arkajad]

I have made several corrections in my example, but the last version may even work.

 

[evalover1987]

I have made several corrections in my example, but the last version may even work.

actually, i used a slightly different method to solve it, treating it as finding a null-space to
solution of a determinant function at the point (0, 0, 1, 1).

Do you mind taking a look at the new problem I posted?
I'm really struggling with that..

 

[arkajad]

You mean your immersion problem? I looked again in Sternberg trying to recall to myself this stuff, but I never really studied the proofs in this particular section, so I am giving up here.

 

[arkajad]

The formula \dot{A}=XA+AY when applied to your A=(0,0;1,1) gives a general form of the tangent vector: (a,a;c,d). This way you have your 3 dimensions.

The other stuff I first proposed, with trace, is good for det = 1 but not for det=0.