Linear Algebra - Topic: Basis & Subspaces

In summary: Now, if you can also express the vectors in the second group as linear combinations of those in the first, what does that tell you about the subspace spanned by the first group with respect to the second?Now, take a look at the basis for W given in the question.
  • #1
spring_rolls
1
0
I have a few questions here, my main problem is not understanding the notations used, hence not understanding the questions.

Homework Statement



1. Do the vectors [tex]j_{1}[/tex]= (1,0,-1,2) and [tex]j_{2}[/tex]= (0,1,1,2) form a basis for the space W = {(a,b,c,d) l a - b + c = 0, -2a - 2b + d =0} ?

2. Find a basis for W = { (a,b,c,d) : a - b + 2d = 0 , 3a + c + 3d = 0 }

3. Find the dimension of the following subspaces of [tex]\Re^{3}[/tex]
a) span { (1,0,1),(0,1,1),(2,0,0) }
b) span { (1,0,1),(2,2,4),(2,1,7),(-1,-1-2) }

4. Suppose that a subspace W [tex]\subset[/tex] [tex]\Re^{3}[/tex] has a basis { (1,2,3),(1,0,1) }.
a)Is { (-1,-2,-3), (3,2,5) } a basis for W? Why?


Homework Equations


How do you do this? Is there a working? or can you solved this by inspection.


The Attempt at a Solution



1. i substitute each vector in a,b,c,d equation and find the equality, so happen when i substitute in all returns 0, which implies that it does form a basis. Is there a proper working for this, cause i can do this by inspection.

2. I was wondering forming a matrix and row reduced the matrix and get a linear dependence equation and plug in some value for c & d to find a & b.

1 -1 0 2 0 ~ 1 0 1/3 1 0
3 0 1 3 0 0 1 1/3 -1 0

So: a = -1/3c - d & b= d - 1/3c

Let c =1 and d =1

a= -1/3 (1) - 1 = -4/3 & b = 1 - 1/3 (1) = 2/3

so basis (-1/3,2/3,1,1)

3 & 4. No idea what to do.
 
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  • #2
spring_rolls said:
1. i substitute each vector in a,b,c,d equation and find the equality, so happen when i substitute in all returns 0, which implies that it does form a basis. Is there a proper working for this, cause i can do this by inspection.
This is not a valid method. Imagine instead of giving you both j1 and j2, I gave you just j1 and ask if this vector alone spans the subspace W. Your method would indicate so, but in fact that is incorrect. To do this the 'proper way', note that if j1 and j2 span W, then all vectors in W can be expressed as linear combinations of j1 and j2. In other words,

[tex]k_1 \left(\begin{array}{c}1\\0\\-1\\2\end{array}\right) + k_2 \left(\begin{array}{c}1\\0\\-1\\2\end{array}\right) = \left(\begin{array}{c}a\\b\\b-a\\2(a+b)\end{array}\right) [/tex]

where [tex]k_1,k_2,a,b \in \Re[/tex]

What you have to do is to express the above as a matrix, and then by row-reduction operations show that for any real a,b there will exist k1,k2 such that the above is satisfied.

2. I was wondering forming a matrix and row reduced the matrix and get a linear dependence equation and plug in some value for c & d to find a & b.

1 -1 0 2 0 ~ 1 0 1/3 1 0
3 0 1 3 0 0 1 1/3 -1 0

So: a = -1/3c - d & b= d - 1/3c

Let c =1 and d =1

a= -1/3 (1) - 1 = -4/3 & b = 1 - 1/3 (1) = 2/3

so basis (-1/3,2/3,1,1)p
I have no idea what you are doing here.

3 & 4. No idea what to do.
For qn 3. First, recall the definition of dimension. The dimension of a subspace is the number of linearly independent vectors which span the given subspaces, ie. the number of vectors in the basis of these subspaces. The vectors in any given basis are all linearly independent. Do you see what to do now?

Qn 4. Suppose you are given a basis and another basis which may or may not span the same subspace as that of the first. Informally, if you can express every vector in the first group of vectors as a linear combination of the vectors in the second group, what does that tell you about the corresponding subspace spanned by the latter group with respect to the first?
 

1. What is a basis in linear algebra?

A basis in linear algebra is a set of linearly independent vectors that can be used to represent any vector in a vector space. This means that any vector in the space can be written as a linear combination of the basis vectors. The number of vectors in a basis is known as the dimension of the vector space.

2. How do you determine if a set of vectors is a basis?

To determine if a set of vectors is a basis, you must first check if the vectors are linearly independent. This means that none of the vectors can be written as a linear combination of the others. Next, you must check if the vectors span the entire vector space. This means that any vector in the space can be written as a linear combination of the basis vectors. If both of these conditions are met, then the set of vectors is a basis.

3. What is the difference between a basis and a spanning set?

A basis is a set of linearly independent vectors that can be used to represent any vector in a vector space. A spanning set is a set of vectors that can be used to represent all possible vectors in a vector space, but they may not be linearly independent. A basis is the smallest possible spanning set for a vector space, as it contains the minimum number of vectors needed to represent all possible vectors.

4. Can a vector space have more than one basis?

Yes, a vector space can have multiple bases. For example, in two-dimensional space, the standard basis is {e1 = (1,0), e2 = (0,1)}. However, another possible basis could be {v1 = (1,1), v2 = (-1,2)}. Both of these bases are valid and can be used to represent any vector in two-dimensional space.

5. What are subspaces in linear algebra?

A subspace is a subset of a vector space that is also a vector space in its own right. This means that it follows all the properties of a vector space, such as closure under addition and scalar multiplication. Subspaces can be formed by taking linear combinations of vectors in the original vector space, and they can have their own basis and dimension.

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