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Suppose \(V\) has dimension \(p\) and \(W\) has dimension \(q\) and they are each contained in a subspace, \(U\) which has dimension equal to \(n\) where \(n>\max (p, q) .\) What are the possibilities for the dimension of \(V \cap W\) ? Hint: Remember that a linearly independent set can be extended to form a basis.

Short Answer

Expert verified
The possible dimensions of V \cap W \ are determined by \max(0, p + q - n) \leq k \leq \min(p, q)\.

Step by step solution

01

- Understand the given dimensions

Let the dimensions of the subspaces be: dimension of V = p, dimension of W = q, and dimension of U = n. It is also given that n > max(p, q).
02

- Define the dimension of the intersection

Let the dimension of the intersection of V and W be k, so we need to find the possible values for k.
03

- Apply the formula for dimensions of subspaces

The dimension formula relating the intersection of two subspaces V and W is given by: \[\text{dim}(V + W) = \text{dim}(V) + \text{dim}(W) - \text{dim}(V \cap W)\]Since V and W are subspaces of U, \text{dim}(V + W) \leq \text{dim}(U)\.
04

- Solve for the dimension of intersection

Rearranging the formula: \[k = \text{dim}(V) + \text{dim}(W) - \text{dim}(V + W)\] We have \[p + q - \text{dim}(V + W) \leq \text{dim}(U) = n\] Hence, assuming the minimal overlap, maximum k can be: \[k_{max} = \text{min}(p, q) \] Because the intersection cannot exceed the dimension of the smaller subspace.
05

- Constrain by subspace U

Since both V and W are contained in U, the maximal dimension \text{dim}(V + W) \cannot exceed \text{dim}(U). So: \(V + W \subseteq U\, \text{dim}(V + W) = \leq n\). So the possible value for \text{dim}(V \cap W) = k\ must satisfy: \max(0, p + q - n) \leq k \leq \min(p, q)\.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Algebra
Linear Algebra forms the foundation for understanding many advanced concepts in mathematics. It deals with vectors, vector spaces, linear transformations, and systems of linear equations. In the context of our exercise, understanding dimensions of subspaces is crucial.

A subspace is a set of vectors that is closed under addition and scalar multiplication. The dimension of a subspace is the maximum number of linearly independent vectors it can contain. For instance, if subspace V has a dimension of p, it contains exactly p linearly independent vectors, which form a basis for V.

To fully grasp this, remember:
  • The basis of a vector space or subspace is a set of vectors that are linearly independent and span the space.
  • The number of vectors in the basis is the dimension of the subspace.
  • Dimension is a measure of the 'freedom' you have within the space, or how many independent directions you can move.
It's essential to become comfortable with these ideas, as they are the building blocks for more complex topics in linear algebra.
Subspace Intersection
In linear algebra, the intersection of two subspaces, V and W, is a key concept. The intersection, denoted as V ∩ W, consists of all the vectors that both V and W contain.

Here's an important point to consider:
  • The dimension of the intersection of two subspaces is tied to their individual dimensions.
  • The intersection has a minimum dimension of 0, meaning V and W share no common vectors except the zero vector.
  • On the other side, the maximum dimension is the smaller of the dimensions of V and W, as V ∩ W cannot have more vectors than either V or W.
Let's apply this to our problem. Given subspaces V and W within a larger subspace U, we need to determine the dimension of V ∩ W. Using linear algebra principles, if V has dimension p, W has dimension q, and U has dimension n, where n > max(p, q), the possible dimension of V ∩ W ranges from max(0, p + q - n) to min(p, q). This formula combines the constraints from both subspaces and the bounding subspace U.
Linear Independence
Linear independence is a cornerstone concept in linear algebra. A set of vectors is linearly independent if no vector in the set is a linear combination of the others. This is crucial for determining the basis and dimension of a subspace.

To understand linear independence, keep these points in mind:
  • If a set of vectors is linearly independent, the removal of any vector should reduce the span of the set.
  • Linearly independent vectors provide the maximal 'spread' in a space without redundancy.
  • The number of linearly independent vectors in a subspace defines its dimension.
In our exercise, the hint suggests extending a linearly independent set to form a basis. This means that by starting with a small set of independent vectors and adding more, you can build up a full basis for a subspace.

This method helps when dealing with intersections like V ∩ W. We want to find all vectors that simultaneously exist in both subspaces while ensuring they form an independent set. Recognizing the independence of these vectors aids in calculating precise dimensions, particularly when working with intersection formulas.

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Most popular questions from this chapter

Consider the following scalar equation of a plane. $$ x+3 y+z=0 $$ Find the orthogonal complement of the vector \(\vec{v}=\left[\begin{array}{l}1 \\\ 2 \\ 1\end{array}\right] .\) Also find the point on the plane which is closest to (3,4,1) .

In general, you have a point \(\left(x_{0}, y_{0}, z_{0}\right)\) and a scalar equation for a plane \(a x+b y+c z=d\) where \(a^{2}+b^{2}+c^{2}>0\). Determine a formula for the closest point on the plane to the given point. Then use this point to get a formula for the distance from the given point to the plane. Hint: Find the line perpendicular to the plane which goes through the given point: \((x, y, z)=\left(x_{0}, y_{0}, z_{0}\right)+t(a, b, c) .\) Now require that this point satisfy the equation for the plane to determine \(t\).

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Here are some vectors in \(\mathbb{R}^{4}\). $$ \left[\begin{array}{r} 1 \\ 4 \\ -2 \\ 1 \end{array}\right],\left[\begin{array}{r} 1 \\ 5 \\ -3 \\ 1 \end{array}\right],\left[\begin{array}{l} 1 \\ 1 \\ 1 \\ 1 \end{array}\right],\left[\begin{array}{l} 2 \\ 1 \\ 3 \\ 2 \end{array}\right],\left[\begin{array}{r} 1 \\ 5 \\ -2 \\ 1 \end{array}\right] $$ Thse vectors can't possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

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