Spanning Set Theorem: Determining Basis and Rank of a Matrix A - Assignment Help

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SUMMARY

The discussion centers on the Spanning Set Theorem as applied to the matrix A = [1 0 0 1 -1; 0 1 1 2 0; 1 0 0 1 2; -1 2 2 3 1]. The rank of matrix A is determined to be 3, with the dimension of Null A calculated as 2. A basis for the column space B(Col A) is identified as sp{a1, a2, a5}, while the basis for the null space B(Null A) is found to be span{(0,-1,1,0), (1,-2,0,1)}. The discussion concludes with methods to express a vector x as a linear combination of the columns of A and to find its representation with respect to the basis B.

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  • Understanding of linear algebra concepts, specifically the Spanning Set Theorem.
  • Familiarity with matrix rank and null space calculations.
  • Proficiency in row reduction techniques for matrices.
  • Knowledge of linear combinations and basis representation in vector spaces.
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  • Study the process of row reducing matrices to find rank and null space.
  • Learn about the change of basis formula in linear algebra.
  • Explore the implications of the Spanning Set Theorem in different vector spaces.
  • Practice problems involving linear combinations and basis transformations.
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Spaning set theorem (new Question)

Hi

I have a couple of questions regarding an assignment which deals with the spanning set theorem.

Hope You can help

The matrix [tex]A = [a1 \ a2 \ a3 \ a4 \ a5] = \left[ \begin{array}{ccccc} 1 & 0 & 0 & 1 & -1 \\ 0 & 1 & 1 & 2 & 0 \\ 1 & 0 & 0 & 1 & 2 \\ -1 & 2 & 2 & 3 & 1 \end{array} \right][/tex]

a) First I determin the rank of A rank(A) = 3

The dimension of Null A: dim (Null A) = 2

b) Determin a basis for A's column space.

I do this using the spanning set theorem.

Since a3 = a2, a4 = 2a2 + a1 then

B(ColA) = sp{a1,a2, a5} According to the theorem.

c) Next the basis for the B(Null A).

First I row reduce A and then up and write up the set of solutions for A which results in the set B(Null A) = span{(0,-1,1,0) , (1,-2,0,1)}

Is that the correct approach ??

d) There is a vector x = a1 + a2 + a3 + a4 + a5. I'm tasked with showing that this vector belongs to Col A. Finally I'm tasked with finding the vector x with respect to the basis B.

I need some assistance is solving c) and d) therefore I hope there is somebody out there who can guide me :-)

Sincerely and Best Regards,

Fred



/Fred
 
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: For c), yes, that is the correct approach. To show that x belongs to Col A, you need to show that it can be written as a linear combination of the vectors a1, a2, and a5. Can you do that? For d), once you have found the basis B, you can express x in terms of this basis. To do this, you need to solve a system of equations with B as the columns of the coefficient matrix. Let me know if you need any more help.
 
,

It looks like you are on the right track with your approach. For part c), finding the basis for B(Null A) involves finding the solutions to the homogeneous system of equations Ax = 0. This is equivalent to finding the nullspace of A. So after row reducing A, the set of solutions will give you a basis for the nullspace, which is the same as B(Null A). So your approach is correct.

For part d), you can show that x belongs to Col A by expressing it as a linear combination of the columns of A. Since x = a1 + a2 + a3 + a4 + a5, you can write it as x = 1*a1 + 1*a2 + 0*a3 + 1*a4 + 1*a5. This means that x is a linear combination of the columns of A with coefficients 1, 1, 0, 1, 1. Therefore, x belongs to Col A.

To find the vector x with respect to the basis B, you can use the change of basis formula. This formula states that if you have a vector x written with respect to one basis, say the standard basis, you can find its representation with respect to another basis, say B, by multiplying x by the change of basis matrix T. This matrix T is formed by taking the basis vectors of B as columns, and the coefficients of x with respect to the basis B as rows. In this case, T = [1 1 0 1 1; 0 1 1 2 0; 1 0 0 1 2; -1 2 2 3 1]. So to find x with respect to basis B, you simply multiply x by T.

I hope this helps. Best of luck with your assignment!


 

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