Math Is Fun Forum

Janiffer

Member

Registered: 2008-04-26

Posts: 12

onto and one to one

hey... Suppose a linear transformation T is defined by a 4x3 matrix, can T be onto?

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: onto and one to one

I think you were trying to ask... can T be one to one?

One to one, after a quick googling, means injective. That means if your mapping is from V to W and v and u are both elements of V, then

in otherwords, if the input values are unequal, then the output values are unequal, and vice versa. This is what one to one means. The more formal term for this type of mapping is 'injective'

Assume the mapping is injective.
suppose we consider the natural basis elements of V

In order to be injective, we have that

must all be unique. However, we know that these are all values in a 3 dimensional vector space, and therefore they cannot be linearly independent. This means that there is one value which can be expressed as a linear combination of the others. Suppose, after a suitable relabeling of indexes,  that T(v[sub]4[/sub]) is this value. We have then, for some scalers

that

applying the properties of a linear transformation, this means 

since T is injective, we must have that

we know that because, if they weren't equal, then the mapping of each would be unequal.

BUT 

implies that the basis elements of the 4 dimensional vector space, are linearly dependant. Since they are a basis, we have a contradiction.

Therefore, the mapping cannot be injective, or 'one to one'.

Last edited by mikau (2008-04-26 08:38:30)

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A logarithm is just a misspelled algorithm.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: onto and one to one

Mikau, you didn’t answered the actual question: Can T be onto (i.e. surjective)? In any case, your analysis is wrong: T is a mapping from

to

, not the other way round.

There do indeed exist injective linear transformations defined by 4×3 matrices. For example

is injective. The transformation matrix is

Now try and prove that no linear mapping from

to

can be surjective.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: onto and one to one

I think you could define a surjective map with a 4x3 matrix as long as the domain and codomain were appropriate. Say,

, for example.

(I'm not sure if that one actually works, but something like it would.)

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Why did the vector cross the road?
It wanted to be normal.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: onto and one to one

I think not!

Definition: If V and W are vector spaces over the same field, a mapping T:V→W is called a linear transformation iff for all vectors u, v in V and all scalars a, b, T(au+bv) = aT(u)+bT(v).

The image of V under the linear transformation T, T(V), is a subspace of W.

Theorem: dim(T(V)) ≤ dim(V).

Proof. Let dim(V) = n and suppose w[sub]1[/sub], …, w[sub]n[/sub], w[sub]n+1[/sub] are n+1 vectors in T(V). Then there are vectors v[sub]1[/sub], …, v[sub]n[/sub], v[sub]n+1[/sub] in V such that T(v[sub]i[/sub]) = w[sub]i[/sub], i = 1, …, n+1. Since the v[sub]i[/sub]’s are linearly dependent, one of them is a linear combination of the others, say v[sub]n+1[/sub]. It follows that w[sub]n+1[/sub] = T(v[sub]n+1[/sub]) is a linear combination of the other w[sub]i[/sub]’s; hence the w[sub]i[/sub]’s are linearly dependent. So any set of n+1 vectors in T(V) are linearly dependent; therefore dim(T(V)) < n+1.

By the way,

is not a vector space.

Last edited by JaneFairfax (2008-04-26 12:43:38)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: onto and one to one

Here is the answer to Janiffer’s question in less abstract verbiage.

Janiffer

Member

Registered: 2008-04-26

Posts: 12

Re: onto and one to one

Thanks a lot guys;))

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: onto and one to one

Oh, I didn't realise linear transformations were restricted to being on vector spaces.
In that case, what about:

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Why did the vector cross the road?
It wanted to be normal.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: onto and one to one

Yup, that’s a linear transformation. If V and W are nontrivial vector spaces (i.e. not {0}) there are always at least two linear transformations between them: the identity map, and the one that maps everything in V to 0[sub]W[/sub].

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: onto and one to one

Mikau, you didn’t answered the actual question: Can T be onto (i.e. surjective)? In any case, your analysis is wrong: T is a mapping from

to

, not the other way round.

wow! I didn't even see this until just now.

argh! Row major format, of course! *face slap* silly me!

but I've never heard the term 'onto' to mean surjective. Thats weird!

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A logarithm is just a misspelled algorithm.