Linear algebra. Problem with vector spaces dimension

AI Thread Summary
The discussion revolves around determining the dimension of the subspace V of R^2x2 matrices defined by the equation 3x + 8y + 5z + w = 0. It is established that the dimension of the full space R^2x2 is 4, and the restriction imposed by the equation suggests that dimV is likely 3, as one equation typically reduces the dimension by one. Participants discuss the possibility of finding other isomorphic subspaces and clarify that all subspaces of the same dimension are isomorphic. Additionally, it is noted that any subspace intersecting V only at the zero element must have dimension 1 if V is indeed 3-dimensional. The conversation emphasizes the importance of understanding vector space properties and the implications of linear independence.
sphlanx
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Homework Statement


First of all sorry if my terminology sounds a bit weird, i have never studied mathematics in english before.

So this is the problem: We have the space R^2x2 of all the tables with numbers in R. We also have a subspace V of R^2x2 of all the tables with the following property: "If x,y is the first row and z,w the second row then: 3x+8y+5z+w=0. The main question is what is the dimV? There are more questions but i think i can solve them if I know dimV!
//EDIT: I will add a few more questions that i see i have a hard time solving even if I take into account that dimV=3
a) How to prove that dimV=3(i believe it is 3 because the subspace of the solution set of the linear system I provided is 3dimensonial)
b) Is there any other subspace of R^2x2 different BUT isomorphic with V
c) If D a subspace of R^2x2 with dimD=4 then D "contains" V
d) Is there any other subspace of R^2x2, isoporphic to V, that intersected with V has only one element, the zero element.

Homework Equations





The Attempt at a Solution



a)I can see that with the given equation not all of the variables are linear independent. This makes me think that dimV=3 but i can't figure out a way to prove it!
b)I believe the answer is YES. In a previous homework there was the same question about 2-dimensional subspaces and I replied yes because I can think of 2 planes crossing. I am not sure if 3 dimensional spaces can "cross" though.
c) Not a clue :P

Thanks in advance! (the deadline is tommorow Saturday at 24:00 so i would appreciate a quick answer!)
 
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sphlanx said:

Homework Statement


First of all sorry if my terminology sounds a bit weird, i have never studied mathematics in english before.

So this is the problem: We have the space R^2x2 of all the tables with numbers in R. We also have a subspace V of R^2x2 of all the tables with the following property: "If x,y is the first row and z,w the second row then: 3x+8y+5z+w=0."
The correct word is "matrices", not "tables". "Arrays" and "tables" do not have the addition and multiplication operations that matrices do and addition and scalar multiplication are necessary for a vector space.

The main question is what is the dimV? There are more questions but i think i can solve them if I know dimV!
//EDIT: I will add a few more questions that i see i have a hard time solving even if I take into account that dimV=3
a) How to prove that dimV=3(i believe it is 3 because the subspace of the solution set of the linear system I provided is 3dimensonial)
The given vector space of 2 by 2 matrices is, of course, 4 dimensional and, generally, one equation restricting the components reduces the dimension by 1. More precisely, you can solve the given equation for one of the components, say w= -3x- 8y- 5z. Then you can write the matrices as
\begin{bmatrix}x & y \\ z & w\end{bmatrix}= \begin{bmatrix}x & y \\ z & -3x- 8y- 5z\end{bmatrix}
= \begin{bmatrix}x & 0 \\ 0 & -3x\end{bmatrix}+ \begin{bmatrix}0 & y \\ 0 & -8y\end{bmatrix}+ \begin{bmatrix}0 & 0 \\ z & -5z\end{bmatrix}
= x\begin{bmatrix}1 & 0 \\ 0 & -3\end{bmatrix}+ y\begin{bmatrix}0 & 1 \\ 0 & -8\end{bmatrix}+ z\begin{bmatrix}0 & 0 \\ 1 & -5\end{bmatrix}

b) Is there any other subspace of R^2x2 different BUT isomorphic with V
All subspaces of the same dimension are isomorphic. Try just moving the components around.

c) If D a subspace of R^2x2 with dimD=4 then D "contains" V
Unless I have completely misunderstood what "R^2x2", the only subspace of dimension 4 is R^2x2 itself.

d) Is there any other subspace of R^2x2, isoporphic to V, that intersected with V has only one element, the zero element.
If V has dimension 3 than any subspace that intersects V only at the zero element must have dimension 1.

Homework Equations





The Attempt at a Solution



a)I can see that with the given equation not all of the variables are linear independent. This makes me think that dimV=3 but i can't figure out a way to prove it!
b)I believe the answer is YES. In a previous homework there was the same question about 2-dimensional subspaces and I replied yes because I can think of 2 planes crossing. I am not sure if 3 dimensional spaces can "cross" though.
c) Not a clue :P

Thanks in advance! (the deadline is tommorow Saturday at 24:00 so i would appreciate a quick answer!)
 
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