Subspaces of Vector Spaces

Let V be some vector space and S be a subset.

a. V=T(3,3) and S is the set of invertible linear transformations.

I know the V represents all linear transformations from R^3 -> R^3, and all linear transformations in S must be invertible, thus they must all be bijective. But I'm unsure how to go about this?

I know the three conditions that establish a subspace - S contains the 0 vector, S is closed under addition, and S is closed under scalar multiplication. Just not sure how to proceed.

Re: Subspaces of Vector Spaces

are you trying to show S is a subspace? Let's get that squared away first.

Re: Subspaces of Vector Spaces

Yes, either proving it is or disproving it.

Re: Subspaces of Vector Spaces

well you've outlined the conditions that need to be met for a subspace.

Is 0 in your space? .. DOH. 0 isn't very invertible is it? (it's the zero matrix by the way. your space consists of 3x3 matrices)

Re: Subspaces of Vector Spaces

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**turbozz** Yes, either proving it is or disproving it.

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Originally Posted by

**turbozz** Let V be some vector space and S be a subset.

a. V=T(3,3) and S is the set of invertible linear transformations.

I know the V represents all linear transformations from R^3 -> R^3, and all linear transformations in S must be invertible, thus they must all be bijective. But I'm unsure how to go about this?

I know the three conditions that establish a subspace - S contains the 0 vector, S is closed under addition, and S is closed under scalar multiplication. Just not sure how to proceed.

What would such a vector look like?

Re: Subspaces of Vector Spaces

I was unsure of what the 0 vector would be in T(2,2). Originally I thought it would be T(0), but I guess since every linear transformation can be expressed as T(x) = Ax, then the zero 3 by 3 matrix represents the zero matrix for T(2, 2). I assume this is true then for any vector space

T(m, n), where the zero vector is the n by m matrix.

Re: Subspaces of Vector Spaces

any dimensions.. what is **0**? It's the element such that (A + **0**) = A

There's only one matrix in the complex field that satisfies that and that's a matrix of all 0s.

Re: Subspaces of Vector Spaces

Equivalently, T is the set of linear transformations from $\displaystyle R^3$ to $\displaystyle R^3$ so that "0" is the linear transformation such that 0(x, y, z)= (0, 0, 0). Such a linear transformation does not have an inverse because it is not "one to one".

Re: Subspaces of Vector Spaces

Ok thanks and just to verify one last point, its fair to say that the dim(T(m,n)) =dim(R^n*m).

Re: Subspaces of Vector Spaces

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**turbozz** Ok thanks and just to verify one last point, its fair to say that the dim(T(m,n)) =dim(R^n*m).

If T must be invertible then T cannot be non-square. So m=n

If T is invertible then dim(T(n,n)) = n = dim(R^n)

Re: Subspaces of Vector Spaces

But if we just say vector space T(m, n) contains all linear transformations of the form T:R^m -> R^n (regardless of whether they are invertible or not) then dim(T(m,n)) =mn, right? And since dim(R^n*m) = nm, can't one infer the dimensions are the same?

Note, when I write R^n*m I mean the vector space that contains all n by m matrices.

Re: Subspaces of Vector Spaces

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**turbozz** But if we just say vector space T(m, n) contains all linear transformations of the form T:R^m -> R^n (regardless of whether they are invertible or not) then dim(T(m,n)) =mn, right? And since dim(R^n*m) = nm, can't one infer the dimensions are the same?

Note, when I write R^n*m I mean the vector space that contains all n by m matrices.

at best dim(T(m,n)) = min(m,n)

Re: Subspaces of Vector Spaces

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**romsek** If T must be invertible then T cannot be non-square. So m=n

If T is invertible then dim(T(n,n)) = n = dim(R^n)

No, that's not true. In the vector space, T(n, n), of linear transformations (NOT necessarily invertible) from $\displaystyle R^n$ to $\displaystyle R^m$, given a basis $\displaystyle \{v_k\}$ for $\displaystyle R^n$, we can have take as basis the collection of transformations $\displaystyle A_{ij}$ defined by $\displaystyle A_{ij}(v_i)= v_j$ if but $\displaystyle A_{ij}(v_k)= 0$ for $\displaystyle k\ne i$. That has dimension $\displaystyle n^2$, not n. More generally, T(n, m), the set of linear transformations (NOT necessarily invertible) from $\displaystyle R^n$ to $\displaystyle R^m$, given basis $\displaystyle \{u_i}$ for $\displaystyle R^n$ and basis [itex]\{v_j\}[/itex] for $\displaystyle R^m$, has basis the collection of transformations $\displaystyle A_{ij}$ deined b $\displaystyle A_{ij}(u_i)= v_j$ with $\displaystyle A_{ij}(u_k)= 0$ if $\displaystyle k\ne i$ and so has dimension nm.

Re: Subspaces of Vector Spaces

I'm going to have to study this.

Thank you for the correction.