Linear algebra. Problem with vector spaces dimension

[sphlanx]

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!)

 

[HallsofIvy]

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!)

 

[Architeuthis Dux]

I understand the importance of clear and precise terminology, so thank you for acknowledging that English is not your first language.

To solve this problem, we can use the concept of linear independence. In a vector space, a set of vectors is linearly independent if no vector in the set can be written as a linear combination of the other vectors. In other words, the vectors are not redundant and each one adds a unique direction to the space.

In this case, we have the subspace V where each table has a specific property based on the values in its rows. This property can be written as a linear equation, which means that the vectors in V are not linearly independent - they are all related to each other through this equation.

To determine the dimension of V, we can use the fact that the dimension of a subspace is equal to the number of linearly independent vectors it contains. Since the vectors in V are not linearly independent, we need to find a set of linearly independent vectors that span V.

One way to do this is to consider the different values that the vectors in V can take on. In this case, the first row has two variables, x and y, and the second row has two variables, z and w. Therefore, we can choose two linearly independent vectors, one with x=1 and y=0 and the other with x=0 and y=1. Similarly, for the second row, we can choose two linearly independent vectors with z=1 and w=0, and z=0 and w=1.

These four vectors form a basis for V, meaning that any vector in V can be written as a linear combination of these four vectors. However, since we only need two linearly independent vectors to span V, the dimension of V is 2.

To answer the additional questions, we can use the same approach. For b), we can think of different linear equations that describe the same subspace V. For c), we can use the fact that a subspace with a higher dimension can contain a subspace with a lower dimension. And for d), we can use the concept of isomorphism to find other subspaces that are equivalent to V but only have one element in common with V.

I hope this helps you solve the problem and understand the concept of dimension in vector spaces. Good luck with your homework!