Understanding Transvections:T_W:V->V Tv=v+w

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This discussion centers on the mathematical concept of transvections in finite-dimensional vector spaces. A transvection is defined as an invertible linear map T: V → V that satisfies two conditions: (i) T|_W = 1_W, meaning T restricts to the identity on the codimension-1 subspace W, and (ii) for any vector v in V, T(v) = v + w, where w is an element of W. The participants debate the clarity of this definition and its implications, particularly regarding the matrix representation of transvections, which may not always conform to the structure of shear matrices.

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Understanding Transvections:T_W:V-->V Tv=v+w

Esteemed Algebraists:

Please help me understand better the definition of a transvection.

Let V be a finite-dimensional vector space, and let W

be a codimension-1 subspace of V . A transvection

is defined to be an invertible linear map T:V-->V

such that:

i) T|_W =1_W , i.e., the restriction of T to W

is the identity on W.

ii)For any v in V, T(v)=v+w ; w in W.

Condition i) is clear, but does condition ii) just say that

vectors in V-W are mapped to V-W?

Also: given a choice of basis for V, is the

matrix representation for V always that of

a shear matrix, i.e., a matrix with all diagonal

entries equal to 1, and all off-diagonal entries

except for exactly one equal to zero, i.e., a

matrix describing adding a multiple by k of one

row to another row?

I know this is the representation in vector spaces over R; is

it true for V.Spaces over any field F? ( I know all V.Spaces of same

dimension are isomorphic, but I don't know if that guarantees the result).


I was thinking of a simple example of a linear map from

R<sup>3</sup> to R<sup>3</sup>

preserving points of types (x,0,0) and (0,y,0). Then ii) above would say that, using the

standard basis {e_<sub>i</sub>; i=1,2,3}.

i) T(1,0,0)=(1,0,0)

ii) T(0,1,0)=(0,1,0)

iii) T(0,0,1)= (0,0,1)+(a,b,0) ; a,b in F


Is the intended meaning that for z in V-W, T(z) in V-W? Also, the representation of

this transvection does not seem to match that of a shear transformation, since it includes

the case of two non-zero entries a,b.

Any Ideas?

Thanks in Advance.

Thanks.
 
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Bacle said:
Esteemed Algebraists:

Please help me understand better the definition of a transvection.

Let V be a finite-dimensional vector space, and let W

be a codimension-1 subspace of V . A transvection

is defined to be an invertible linear map T:V-->V

such that:

i) T|_W =1_W , i.e., the restriction of T to W

is the identity on W.

ii)For any v in V, T(v)=v+w ; w in W.

Where did you find this definition. I don't say it's incorrect, but it's a bit weird. The definition on wiki is a little better:

To be more precise, if V is the direct sum of W and W′, and we write vectors as

v = w + w′

correspondingly, the typical shear fixing W is L where

L(v) = (w + Mw′) + w ′

where M is a linear mapping from W′ into W.

I like it better because it's little less ambiguous.

Condition i) is clear, but does condition ii) just say that

vectors in V-W are mapped to V-W?

No. Of course, the definition will still imply that V-W is mapped to V-W, but that's not quite enough. Take V=R^2 and W=R. Then T(1,0)=(1,0) and T(0,1)=(0,2) will satisfy that V-W is mapped to V-W, but it's not a transvection since there is no w in W such that T(0,1)=(0,2)+w.

Geometrically, a transvection maps W onto itself and translates points outside of W parallel to W.

Also: given a choice of basis for V, is the

matrix representation for V always that of

a shear matrix, i.e., a matrix with all diagonal

entries equal to 1, and all off-diagonal entries

except for exactly one equal to zero, i.e., a

matrix describing adding a multiple by k of one

row to another row?

I don't think this is true. If we follow wiki's definition, then a transvection always has the form

\left(\begin{array}{cc} I &amp; M\\ 0 &amp; I\end{array}\right)

In my (limited) understanding of the topic, a shear matrix will represent a transvection, but not vice versa.

You could help me enormously in providing the reference you're using...
 


Hi, Micromass:

Sorry for the delay. I am going from the book "Classical Groups

and Geometric Algebra", by Larry C. Groves; a GTM book; mostly

pages 7 and 22. He describes transvections on a fin.-dim v.space

V; with invariant codimension-1 subspace W, as maps T:V-->V ,

with T|<sub>W</sub> =1<sub>W</sub> (i.e., map T:V-->V

restricts to the identity on W ), and , for any v in V, T(v)=v+w for

some w in W.

Grove goes on to show that transvections generate

both SL(V):={M in GL(V), Det(M)=1}, as well as the

symplectic group of V, given a symplectic form.
 
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micromass said:
Where did you find this definition. I don't say it's incorrect, but it's a bit weird. The definition on wiki is a little better:



I like it better because it's little less ambiguous.



No. Of course, the definition will still imply that V-W is mapped to V-W, but that's not quite enough. Take V=R^2 and W=R. Then T(1,0)=(1,0) and T(0,1)=(0,2) will satisfy that V-W is mapped to V-W, but it's not a transvection since there is no w in W such that T(0,1)=(0,2)+w.

Yes, I understand that; but , since a transvection does map elements in V-W to elements in V-W, what additional condition do we need to characterize transvections? Maybe better to leave good-enough alone and accept T:v=v+w. So what else does the def. say that T:v-w is sent to v-w?

Geometrically, a transvection maps W onto itself and translates points outside of W parallel to W.

Well, actually, you would need a notion of orthogonality defined on your space; there are abstract vector spaces without an inner-product (thing homology over Z/2), without a standard ( if at all) notion of orthogonality.
.

I don't think this is true. If we follow wiki's definition, then a transvection always has the form

\left(\begin{array}{cc} I &amp; M\\ 0 &amp; I\end{array}\right)

In my (limited) understanding of the topic, a shear matrix will represent a transvection, but not vice versa.

You could help me enormously in providing the reference you're using...

Sorry to bring this up so late, I was just reviewing my posts.
 

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