# Checking if intersection is correct

• Kernul
In summary, the conversation revolves around finding the dimension and base vectors for the intersection and union of two subspaces, V and U, given certain vectors. The dimension of the intersection is found to be 0, meaning the sum of the two subspaces is a direct sum. The four given vectors are found to be independent and can be considered as a base for a four-dimensional space.
Kernul

## Homework Statement

The exercise gives me ##V = <(1, 0, 0, 1), (-7, 0, 1, 0)>## and ##U = <(0, 4, 1, -1), (1, 12, 0, 26)>## and I have to find dimension and a base of ##V \cap U## and ##V + U##.

## The Attempt at a Solution

They both have dimension ##2##. I find a vector ##\vec v \in V \implies \vec v = \alpha (1, 0, 0, 1) + \beta (-7, 0, 1, 0) = (\alpha - 7 \beta, 0, \beta, \alpha)##
Now I have to see if ##\vec v \in U## and this implies that the vector just found has to be linearly dependent to the other two of U. I can see for what parameters this is true by putting all in a matrix and implying the determinant to be ##0##.
$$\begin{pmatrix} \alpha - 7 \beta & 0 & \beta & \alpha \\ 0 & 4 & 1 & -1 \\ 1 & 12 & 0 & 26 \end{pmatrix}$$
I use Kronecker's theorem and put the two determinants in a system.
$$\begin{cases} \begin{vmatrix} \alpha - 7 \beta & 0 & \beta \\ 0 & 4 & 1 \\ 1 & 12 & 0 \end{vmatrix} = 0 \\ \begin{vmatrix} \alpha - 7 \beta & 0 & \alpha \\ 0 & 4 & -1 \\ 1 & 12 & 26 \end{vmatrix} = 0 \end{cases}$$
But doing the calculations I find myself with this:
$$\begin{cases} -4 \beta - 12 \alpha + 84 \beta = 0 \\ 116 \alpha - 812 \beta - 4 \alpha = 0 \end{cases}$$
$$\begin{cases} 12 \alpha = 80 \beta \\ 112 \alpha = 812 \beta \end{cases}$$
And if I keep on going, I'll have ##\alpha = \frac{20}{3} \beta##. Should I stop here and go substituting this in the vector and get the intersection? What about the other equation in the system? Is it okay to leave it like that? Because if I substitute the ##\alpha## just found in there, I would get ##\beta = 0##.

Are the four vectors (1, 0, 0, 1), (-7, 0, 1, 0),(0, 4, 1, -1), (1, 12, 0, 26) independent?

ehild said:
Are the four vectors (1, 0, 0, 1), (-7, 0, 1, 0),(0, 4, 1, -1), (1, 12, 0, 26) independent?
I put them in a matrix and did the determinant. I have a determinant different from ##0## (##-196##), so they are independent.

Kernul said:
I put them in a matrix and did the determinant. I have a determinant different from ##0## (##-196##), so they are independent.
So what is the intersection of the subspaces U and V? Can be any linear combination of the base vectors of U element of V?

Can you consider the four independent vectors as base of a 4-dimensional space? So what is the union of U and V ?

ehild said:
So what is the intersection of the subspaces U and V? Can be any linear combination of the base vectors of U element of V?
Oh, so the intersection is void? This means that the sum will be a direct sum, right?

Kernul said:
Oh, so the intersection is void? This means that the sum will be a direct sum, right?
Yes, the intersection contains only the zero vector.
I do not know what you mean on U+V, but the "Union" of the two subspaces would be a sensible question.

ehild said:
Yes, the intersection contains only the zero vector.
I do not know what you mean on U+V, but the "Union" of the two subspaces would be a sensible question.
The exercise asks me about the intersection and the sum of the two subspaces. I know that when the intersection is the zero vector, then the sum of those two subspaces is a sum called "direct sum".
By the way, the union wouldn't simply have the dimension as the sum of the two dimension's subspaces and a base as the union of the two bases?

Kernul said:
The exercise asks me about the intersection and the sum of the two subspaces. I know that when the intersection is the zero vector, then the sum of those two subspaces is a sum called "direct sum".
By the way, the union wouldn't simply have the dimension as the sum of the two dimension's subspaces and a base as the union of the two bases?
Yes, I think so. All the four vectors have four components, so they are elements of a four-dimensional space. As they are independent, they can be considered as an other base of that space. You can consider it as the direct sum of the two two-dimensional spaces U and V.

Last edited:
Kernul

## 1. How can I check if the intersection of two shapes is correct?

The most reliable way to check the correctness of an intersection is to use mathematical calculations or algorithms. This involves finding the intersection points of the two shapes and comparing them to see if they match up.

## 2. What if the intersection points do not match up?

If the intersection points do not match up, then the intersection is not correct. This could be due to a mistake in the calculations or a flaw in the algorithm used. It is important to double check the calculations and make sure the algorithm is accurate.

## 3. Can I visually check if the intersection is correct?

It is possible to visually check the intersection of two shapes, but this method may not always be accurate. It is best to use mathematical calculations or algorithms to ensure the correctness of the intersection.

## 4. How do I know if the intersection is a true intersection or just an overlap?

A true intersection occurs when two shapes share a common boundary or overlap, while an overlap occurs when two shapes simply share some common points. To determine if the intersection is a true intersection, you will need to compare the coordinates of the intersection points and see if they match up exactly.

## 5. What if my shapes are complex and have multiple intersections?

In cases where there are multiple intersections, it is important to check each intersection point and make sure they all match up. This can be a more complex process, but it is necessary to ensure the correctness of the intersection.

• Calculus and Beyond Homework Help
Replies
8
Views
869
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
904
• Calculus and Beyond Homework Help
Replies
7
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
373
• Calculus and Beyond Homework Help
Replies
13
Views
2K
• Calculus and Beyond Homework Help
Replies
16
Views
1K
• Calculus and Beyond Homework Help
Replies
13
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
841
• Calculus and Beyond Homework Help
Replies
8
Views
906