I Can a shear operation introduce a new linear dependency?

[swampwiz]

Specifically, an elemental shear operation.

I can prove how any existing linear dependency that exists persists with the new sheared vector, but I can't seem to prove that this shear operation does not introduce a linear dependency. I can see how the vector space between the 2 rows involved does not change since the situation of those vectors being non-parallel has the updated row still not being parallel to the other, and for the situation in which they are parallel, the updated row remains parallel. My goal is to be able to prove that the rank does not change in a shear operation.

EDIT: I think I can justify the contention that there is no new linear dependency since I do not introduce an equation of the form of the zero vector (or any constant vector) to a coefficient sum of row vectors - but this seems weak.

 

[fresh_42]

What happens if you shear a square by 45°? Now do the same operation with 90°.

 

[swampwiz]

What happens if you shear a square by 45°? Now do the same operation with 90°.

OK, I see how a shear of 90° is impossible. I just can't seem to prove that somehow by bad luck there isn't some new dependency being introduced in the total matrix.

 

[fresh_42]

Why is it impossible? It is a shear of a square to a flat line. Maybe I don't understand what you mean by a shear. How is it defined?

 

[swampwiz]

Why is it impossible? It is a shear of a square to a flat line. Maybe I don't understand what you mean by a shear. How is it defined?

I was interpreting it as the angle from the normal, so 0 degrees would be zero shear. In essence, the shear amount is the tangent, and of course the tangent of 90 degrees in infinity.

 

[swampwiz]

Specifically, an elemental shear operation.

I can prove how any existing linear dependency that exists persists with the new sheared vector, but I can't seem to prove that this shear operation does not introduce a linear dependency. I can see how the vector space between the 2 rows involved does not change since the situation of those vectors being non-parallel has the updated row still not being parallel to the other, and for the situation in which they are parallel, the updated row remains parallel. My goal is to be able to prove that the rank does not change in a shear operation.

EDIT: I think I can justify the contention that there is no new linear dependency since I do not introduce an equation of the form of the zero vector (or any constant vector) to a coefficient sum of row vectors - but this seems weak.

2nd EDIT: OK, I took the notion of not introducing a new linear dependency and went in reverse - which is easy to do since all it does is change the plus to a minus sign - and since they persist in both directions, the number of them must be invariant. I guess because a shear in invertible, I should have thought of it like that originally.

 

[WWGD]

AFAIK, a linear map $L: A \rightarrow B$ preserves subspaces iff DimA $\leq$DimB and L has trivial kernel.

 

[fresh_42]

AFAIK, a linear map $L: A \rightarrow B$ preserves subspaces iff DimA $\leq$DimB and L has trivial kernel.

So is it not a shear, if the angle is 90° which makes a square a flat line?

 

[WWGD]

So is it not a shear, if the angle is 90° which makes a square a flat line?

What is the matrix rep of a shear, and a shear by 90 deg?

 

[fresh_42]

What is the matrix rep of a shear, and a shear by 90 deg?

1. In general? I don't know. That's why I asked for the definition of a shear. Maybe I'm lost in translation.
2. $\begin{bmatrix}1&1 \\0&0\end{bmatrix}$

 

[WWGD]

I guess I am going by the standard Wiki:
https://en.wikipedia.org/wiki/Shear_mapping

Still, your map would not be a counter, since it has non-trivial kernel including, e.g. $\{(0,a): a \in \mathbb R \}$

 

[WWGD]

Specifically, an elemental shear operation.

I can prove how any existing linear dependency that exists persists with the new sheared vector, but I can't seem to prove that this shear operation does not introduce a linear dependency. I can see how the vector space between the 2 rows involved does not change since the situation of those vectors being non-parallel has the updated row still not being parallel to the other, and for the situation in which they are parallel, the updated row remains parallel. My goal is to be able to prove that the rank does not change in a shear operation.

EDIT: I think I can justify the contention that there is no new linear dependency since I do not introduce an equation of the form of the zero vector (or any constant vector) to a coefficient sum of row vectors - but this seems weak.

Would you please give us a precise definition of shear, e.g., the matrix with respect to the standard basis?

 

[fresh_42]

I guess I am going by the standard Wiki:
https://en.wikipedia.org/wiki/Shear_mapping

Still, your map would not be a counter, since it has non-trivial kernel including, e.g. $\{(0,a): a \in \mathbb R \}$

That was the idea. Just putting it to the extreme.