Linear Algebra: Basis and Dimension

In summary, for the first problem, the coordinates of b with respect to the given basis are (3,-1) and for the second problem, the basis for V is {(1,-1,0), (1,0,-1)} and the dimension is 1.
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bananasplit
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Homework Statement


In each case, check that (v1,...vk) is a basis for Rn and give the coordinates of the given vector b belongs to Rn with respect to that basis.


Homework Equations


a) v1=(2,3) v2=(3,5) b=(3,4)
b) v1=(1,0,1) v2=(1,1,2) v3=(1,1,1) b=(3,0,1)

The Attempt at a Solution


I put it into a matrix and row reduced the matrix and got:
a) column 1 = (1,0) column 2 = (0,1) equals column 3 (3, -1)
Then I said the coordinates were thus (3,-1)
b) I did the same thing and got:
column 1 = (1,0,0) column 2 = (0,1,0) column 3=(0,0,1) equals column 4 =(-1, -2, 2)
The coordinates are (-1,-2,2)


Question 2

Homework Statement


Find a basis for each of the given subspaces and determine its dimension.


Homework Equations


a) V={x belonging to R4: x1+x2+x3+x4=0, x2+x4=0} c R4
b) V=(Span(1,2,3))orthogonal c R3

The Attempt at a Solution


I put it into a matrix and row reduced the matrix and got:
a) column 1 = (1,0, 0, 0) column 2 = (0,1,0,0) So I said the dimension was 2 and the basis was {(1,0)(1,1)}
b) For part two I don't know how to start this because I do not know what the matrix will look like.
 
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  • #2
The orthogonal vector will be perpendicular to the span of (1,2,3) but I am not sure how to find it. Is there a specific method I should use to find the basis and dimension?

Your Solution

For the first problem, your solution is correct. However, for the second problem, there is a mistake in your answer. The dimension of V should be 1, not 2. This is because the given subspace only has one restriction, which is x1+x2+x3=0. The basis for this subspace would be {(1,-1,0), (1,0,-1)}.

For the second part, you can find a vector that is orthogonal to (1,2,3) by taking the cross product of (1,2,3) with any other vector. For example, let's take the vector (1,0,0). The cross product of (1,2,3) and (1,0,0) is (-2,1,-2). Therefore, the basis for V would be {(-2,1,-2)}. The dimension would still be 1.

Overall, your approach was correct, but make sure to double check your answers and make sure they align with the given information and restrictions.
 

1. What is a basis in linear algebra?

A basis is a set of linearly independent vectors that span a vector space. This means that any vector in the space can be expressed as a linear combination of the basis vectors. It is similar to the idea of a coordinate system, where the basis vectors act as the axes of the space.

2. How is a basis different from a basis set?

A basis is a specific set of vectors that satisfy the conditions of linear independence and spanning a vector space. A basis set, on the other hand, refers to any collection of vectors that can be used as a basis for a given vector space. A basis set may contain more or less vectors than the actual basis of a space.

3. What is the dimension of a vector space?

The dimension of a vector space is the number of vectors in its basis. It is also equal to the number of coordinates needed to uniquely describe any vector in the space. For example, a 2-dimensional vector space would have a basis of 2 linearly independent vectors, and any vector in this space can be represented using 2 coordinates.

4. How is the dimension of a vector space related to its basis?

The dimension of a vector space is directly related to the number of vectors in its basis. If a vector space has a basis of n vectors, then its dimension is n. Conversely, if the dimension of a vector space is n, then it must have a basis of n vectors.

5. How can the concept of basis and dimension be applied in real-world scenarios?

Linear algebra, specifically the concepts of basis and dimension, have numerous applications in real-world scenarios. For example, they are used in computer graphics to represent and manipulate 3-dimensional objects, in data compression algorithms, and in solving systems of linear equations in engineering and physics problems. They also have applications in machine learning and data analysis.

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