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

[mathmari]

Hey!

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)

 

[Opalg]

Hey!

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 $U_2$?

 

[mathmari]

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)

 

[Opalg]

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.

 

[mathmari]

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)

 

[mathmari]

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)