Determining Basis and Dimension for Vector Subspaces $U_1$ and $U_2$

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Discussion Overview

The discussion revolves around determining the basis and dimension for the vector subspaces $U_1$ and $U_2$ in $\mathbb{R}^4$, as well as exploring their intersection $U_1 \cap U_2$ and their sum $U_1 + U_2$. Participants are examining the linear independence of the vectors that define these subspaces and considering alternative representations of the subspaces.

Discussion Character

  • Technical explanation
  • Exploratory
  • Debate/contested

Main Points Raised

  • Some participants assert that the vectors of $U_1$ and $U_2$ are linearly independent, leading to a basis consisting of the entire subspaces, each with dimension 3.
  • Participants propose a method to find the intersection $U_1 \cap U_2$ by setting up a system of equations based on the definitions of the subspaces.
  • There is a suggestion to describe $U_1$ and $U_2$ in alternative forms based on the patterns observed in their vectors, with some participants questioning the necessity of certain conditions (e.g., inequalities among variables).
  • One participant questions whether the bases identified for $U_1$ and $U_2$ need to be proven as linearly independent before being accepted as bases.
  • Another participant explores the dimension of the intersection and calculates it based on the derived equations, leading to a conjecture about the dimension being 1.
  • There is a discussion about how to find a basis for the sum $U_1 + U_2$ and whether it requires checking the linear independence of the resulting vectors.
  • Participants inquire about the conditions necessary for extending $U_1$ to a basis of $\mathbb{R}^4$ to find a suitable subspace $W$ such that $U_1 \oplus W = \mathbb{R}^4$.

Areas of Agreement / Disagreement

Participants generally agree on the linear independence of the vectors in $U_1$ and $U_2$, but there is no consensus on the best method to find the intersection or the sum of the subspaces. The dimension of the intersection remains a point of contention, with differing views on whether it is 1 or another value.

Contextual Notes

Some participants note that the representations of $U_1$ and $U_2$ could be simplified, but there is no agreement on the implications of these simplifications. The discussion includes unresolved mathematical steps regarding the intersection and sum of the subspaces.

Who May Find This Useful

This discussion may be useful for students and educators in linear algebra, particularly those interested in vector spaces, subspaces, and their properties in higher dimensions.

mathmari
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Hey! :o

Let $U_1,U_2$ be vector subspaces of $\mathbb{R}^4$ that are defined as $$U_1=\begin{bmatrix}
\begin{pmatrix}
3\\
2\\
2\\
1
\end{pmatrix}, \begin{pmatrix}
3\\
3\\
2\\
1
\end{pmatrix}, \begin{pmatrix}
2\\
1\\
2\\
1
\end{pmatrix}
\end{bmatrix}, \ \ U_2=\begin{bmatrix}
\begin{pmatrix}
1\\
0\\
4\\
0
\end{pmatrix}, \begin{pmatrix}
2\\
3\\
2\\
3
\end{pmatrix}, \begin{pmatrix}
1\\
2\\
0\\
2
\end{pmatrix}
\end{bmatrix}$$

I want to determine for $U_1$, $U_2$, $U_1\cap U_2$,$ U_1+U_2$ a basis and the dimension. I have done the following:

The vectors of $U_1$ are linearly indepenendent, so a basis is the whole $U_1$ and so the dimension is $3$.

The same for $U_2$ : The vectors are linearly indepenendent, so a basis is the whole $U_2$ and so the dimension is $3$. For the intersection: Let $v\in U_1\cap U_2$ then $v\in U_1$, so $v=a(3,2,2,1)+b(3,3,2,1)+c(2,1,2,1)$ and $v\in U_2$, so $v=x(1,0,4,0)+y(2,3,2,3)+z(1,2,0,2)$.
We have that $$v-v=0 \Rightarrow a(3,2,2,1)+b(3,3,2,1)+c(2,1,2,1)-x(1,0,4,0)-y(2,3,2,3)-z(1,2,0,2)=(0,0,0,0)$$ So, we have to solve the system to find such $a,b,c,x,y,z$.
Is this correct? What about the dimension? (Wondering) For the sum we have that $\text{dim} (U_1+U_2)=\text{dim}(U_1)+\text{dim}(U_2)-\text{dim}(U_1\cap U_2)=3+3-\text{dim}(U_1\cap U_2)=6-\text{dim}(U_1\cap U_2)$, right? (Wondering)

Then to find a basis, do we take all the sums $u_1+u_2$, where $u_1\in U_1$ and $u_2\in U_2$ ? (Wondering) I want to find also vector subspace $W$ of $\mathbb{R}^4$ such that $U_1\oplus W=\mathbb{R}^4$. To find such a $W$ do we have to extend $U_1$ to a basis of $\mathbb{R}^4$ ? (Wondering)
 
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mathmari said:
Hey! :o

Let $U_1,U_2$ be vector subspaces of $\mathbb{R}^4$ that are defined as $$U_1=\begin{bmatrix}
\begin{pmatrix}
3\\
2\\
2\\
1
\end{pmatrix}, \begin{pmatrix}
3\\
3\\
2\\
1
\end{pmatrix}, \begin{pmatrix}
2\\
1\\
2\\
1
\end{pmatrix}
\end{bmatrix}, \ \ U_2=\begin{bmatrix}
\begin{pmatrix}
1\\
0\\
4\\
0
\end{pmatrix}, \begin{pmatrix}
2\\
3\\
2\\
3
\end{pmatrix}, \begin{pmatrix}
1\\
2\\
0\\
2
\end{pmatrix}
\end{bmatrix}$$

I want to determine for $U_1$, $U_2$, $U_1\cap U_2$,$ U_1+U_2$ a basis and the dimension. I have done the following:

The vectors of $U_1$ are linearly indepenendent, so a basis is the whole $U_1$ and so the dimension is $3$.

The same for $U_2$ : The vectors are linearly indepenendent, so a basis is the whole $U_2$ and so the dimension is $3$. For the intersection: Let $v\in U_1\cap U_2$ then $v\in U_1$, so $v=a(3,2,2,1)+b(3,3,2,1)+c(2,1,2,1)$ and $v\in U_2$, so $v=x(1,0,4,0)+y(2,3,2,3)+z(1,2,0,2)$.
We have that $$v-v=0 \Rightarrow a(3,2,2,1)+b(3,3,2,1)+c(2,1,2,1)-x(1,0,4,0)-y(2,3,2,3)-z(1,2,0,2)=(0,0,0,0)$$ So, we have to solve the system to find such $a,b,c,x,y,z$.
Is this correct? What about the dimension? (Wondering)
What you have done is correct, but maybe not the best way to proceed. I think it would be better to look carefully at the given vectors in $U_1$ and $U_2$, to see if there is a more convenient way to describe these spaces.

What I notice is that the three basis vectors in $U_1$ all have $2$ and $1$ for their third and fourth coordinates. In $U_2$, the three vectors all have their second and fourth coordinates equal. Can you use these facts to give alternative descriptions of $U_1$ and $U_2$?
 
Opalg said:
What you have done is correct, but maybe not the best way to proceed. I think it would be better to look carefully at the given vectors in $U_1$ and $U_2$, to see if there is a more convenient way to describe these spaces.

What I notice is that the three basis vectors in $U_1$ all have $2$ and $1$ for their third and fourth coordinates. In $U_2$, the three vectors all have their second and fourth coordinates equal. Can you use these facts to give alternative descriptions of $U_1$ and $U_2$?

Can we write these maybe as follows: $$U_1=\{(x,y,2,1) \mid x, y\in \mathbb{R}, x\neq y\} , \ \ U_2=\{(x,y,z,y)\mid x,y,z\in \mathbb{R}, y\neq x,z\}$$ ? (Wondering)
 
mathmari said:
Can we write these maybe as follows: $$U_1=\{(x,y,2,1) \mid x, y\in \mathbb{R}, x\neq y\} , \ \ U_2=\{(x,y,z,y)\mid x,y,z\in \mathbb{R}, y\neq x,z\}$$ ? (Wondering)
Nearly right. It should be $U_1=\{(x,y,2z,z) \mid x, y, z\in \mathbb{R}\} , \ \ U_2=\{(x,y,z,y)\mid x,y,z\in \mathbb{R}\}$. No need for $x$ and $y$ to be unequal. And you need the $z$ in $U_1$, otherwise it will not contain scalar multiples.
 
Opalg said:
Nearly right. It should be $U_1=\{(x,y,2z,z) \mid x, y, z\in \mathbb{R}\} , \ \ U_2=\{(x,y,z,y)\mid x,y,z\in \mathbb{R}\}$. No need for $x$ and $y$ to be unequal. And you need the $z$ in $U_1$, otherwise it will not contain scalar multiples.

Ah ok... So, we have that $$U_1=\{(x,y,2z,z) \mid x, y, z\in \mathbb{R}\}=\{x(1,0,0,0)+y(0,1,0,0)+z(0,0,2,1)\mid x, y, z\in \mathbb{R}\}$$ and $$U_2=\{(x,y,z,y)\mid x,y,z\in \mathbb{R}\}=\{x(1,0,0,0)+y(0,1,0,1)+z(0,0,1,0)\mid x,y,z\in \mathbb{R}\}$$
So, we see that a basis of $U_1$ is $\{(1,0,0,0), (0,1,0,0), (0,0,2,1)\}$ and therefore, the dimension is $3$, and a basis of $U_2$ is $\{(1,0,0,0), (0,1,0,1), (0,0,1,0)\}$ and therefore, the dimension is $3$.

Is this correct? Can we just say that these are a basis? Or do we have to prove first the linearly independence? (Wondering)
 
Knowing these bases, do we know the dimension of the intersection?

To find a basis for the intersection in the way I did in my first post, we have the following:
$$v=a(1,0,0,0)+b(0,1,0,0)+c(0,0,2,1), \ \ v=x(1,0,0,0)+y(0,1,0,1)+z(0,0,1,0)$$
$$v-v=0 \Rightarrow a(1,0,0,0)+b(0,1,0,0)+c(0,0,2,1)-x(1,0,0,0)-y(0,1,0,1)-z(0,0,1,0)=(0,0,0,0) \\ \Rightarrow a-x=0, b-y=0, 2c-z=0, c-y=0 \Rightarrow c=y=b, z=2b, x=a$$
So, choosing $a=b=1$ we get $v=(1,0,0,0)+(0,1,0,0)+(0,0,2,1)=(1,1,2,1)$, right? (Wondering)

So, is the dimension of the intersection $1$ ? Does this always hold? (Wondering) Then we have $\text{dim} (U_1+U_2)=\text{dim}(U_1)+\text{dim}(U_2)-\text{dim}(U_1\cap U_2)=3+3-1=5$.

To find a basis for the sum, do we find all the sums and check if the result is linearly independent?

So, $$(1,0,0,0)+(1,0,0,0)=(2,0,0,0), \ \ (1,0,0,0)+(0,1,0,1)=(1,1,0,1), \ \ (1,0,0,0)+(0,0,1,0)=(1,0,1,0), \\ (0,1,0,0)+(1,0,0,0)=(1,1,0,0), \ \ (0,1,0,0)+(0,1,0,1)=(0,2,0,1), \ \ (0,1,0,0)+(0,0,1,0)=(0,1,1,0), \\ (0,0,2,1)+(1,0,0,0)=(1,0,2,1), \ \ (0,0,2,1)+(0,1,0,1)=(0,1,2,2), \ \ (0,0,2,1)+(0,0,1,0)=(0,0,3,1)$$

Now we have to find $5$ linearly independent vectors, right? (Wondering)
To find a vector subspace $W$ of $\mathbb{R}^4$ such that $U_1\oplus W=\mathbb{R}^4$, do we have to extend $U_1$ to a basis of $\mathbb{R}^4$ ? (Wondering)
 

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