Find linearly independent vectors for AVectors for A

In summary, In part (b) of the question, the student is asked to find linearly-independent vectors in a given matrix such that any column vector is a linear combination of those vectors. The student was able to find the three vectors by inspection in their reduced matrix.
  • #1
roam
1,271
12

Homework Statement



Let [tex]A = \left[\begin{array}{ccccc} 2&0&-2&1&-3 \\ 1&1&-3&0&-2\\1&0&-1&-1&3 \end{array}\right][/tex]

a) Find two linearly independent vectors u and v in R5 such that span{u,v}={[tex]x \in R^5[/tex] : Ax=0}.

b) Find three linearly independent vectors u,v and w such that col(A) = span{u,v,w}.

Homework Equations




3. The Attempt at a Solution


a) I row reduced A:

[tex]\left[\begin{array}{ccccc} 1&0&-1&0&0 \\ 0&1&-2&0&-2\\0&0&0&1&-3 \end{array}\right][/tex]

I'm not sure how to answer this question. I think the first and second column are linearly independent and if we add them we get the third column. I need some help, I don't know what to do from here.

 
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  • #2
How about you start solving the equation Ax = 0?

I.e. reduce
[tex]
A = \left[\begin{array}{ccccc|c} 2&0&-2&1&-3 &0 \\ 1&1&-3&0&-2 &0 \\1&0&-1&-1&3 &0 \end{array}\right][/tex]
Then you can take, for example, x4 = r, x5 = s and write the solution as
xi = xi(r, s)
 
  • #3
CompuChip said:
How about you start solving the equation Ax = 0?

I.e. reduce
[tex]
A = \left[\begin{array}{ccccc|c} 2&0&-2&1&-3 &0 \\ 1&1&-3&0&-2 &0 \\1&0&-1&-1&3 &0 \end{array}\right][/tex]
Then you can take, for example, x4 = r, x5 = s and write the solution as
xi = xi(r, s)

Hi

I reduced A and I'm left with:

[tex]= \left[\begin{array}{ccccc|c} 1&0&-1&0&0&0\\ 0&1&-2&0&-2&0 \\0&0&0&1&-3&0 \end{array}\right][/tex]

The linear system corresponding to the row-reduced matrix is:

x1-x3=0
x2-2x3-2x5=0
x4-3x5=0

variables corresponding to leading 1's are x1, x2 and x4, the rest I think are free variables, solving for the leading variables:

x1=x3
x2=2x5+2x3
x4=3x5


Now, I don't know where to go from here. I'm not exactly sure how to represent the solutions for this. How does this help us to solve part (a)?
 
  • #4
x1, x2, and x4 can be found if you know or are given x3 and x5.

Code:
x_1 = x_3
x_2 = 2x_3 + 2x_5
x_3 = x_3
x_4 =         3x_5
x_5 =          x_5
So any vector (x1, x2, x3, x4, x5) can be written as a linear combination of two vectors. I'm hopeful that you will notice these two vectors lurking in the arrangement above.
 
  • #5
Thanks, I notice it now, x3 and x5 are the two linearly independent vectors so u=(-1,-2,0) & v=(0,-2,-3).

Could you please explain to me part (b) of the question? I don't understand what the question wants :rolleyes:
 
  • #6
I don't think you understand the a part just yet. x3 and x5 are just plain old numbers, not vectors, let alone linearly independent vectors. And your vectors u and v have to be in R5, not R3 as the ones you show are.

Take a look again at the code block I put in my last response. There are two vectors in R5 hiding in there on the right side. If you stare at it long enough, you might see them.
 
  • #7
Hint:

in post #4 set the free variables x_3 and x_5 to constants c_1 and c_2.

you can separate the resulting vector into the sum of two vectors, one containing the c_1's and the other containing only the c_2's
 
  • #8
I see it!

[tex]\left[\begin{array}{ccccc} x_{3}\\2x_{3} + 2x_{5}\\x_{3}\\3x_{5}\\x_{5} \end{array}\right]=[/tex] x3[tex]\left[\begin{array}{ccccc} 1\\2\\1\\0\\0 \end{array}\right]+[/tex]x5[tex]\left[\begin{array}{ccccc} 0\\2\\0\\3\\1 \end{array}\right][/tex]

Is this what you meant? so u=(1,2,1,0,0) and v=(0,2,0,3,1), both in R5

Now, could you please explain how to solve part (b)?
 
  • #9
good!

b is much easier than a

think about what the column space means. they want you to find the 3 linearly independent vectors in the given matrix which constitute the basis of the column space. You already did the first step in this by finding the row reduced matrix. What can you pull from that new matrix that might tell you about the column space?
 
  • #10
Here's your row-reduced matrix:
[tex]\left[\begin{array}{ccccc} 1&0&-1&0&0 \\ 0&1&-2&0&-2\\0&0&0&1&-3 \end{array}\right][/tex]
For part b, you want three linearly independent vectors in R3, that span the columns of the original matrix. IOW, you want three linearly independent vectors u, v, w, in R3 such that any column vector in your matrix is a linear combination of u, v, and w.

You can get those three vectors by inspection in your reduced matrix.
 
  • #11
If I had to guess I'd say the first, second and the fourth column which have pivot elements ( leading 1's) & they are linearly independent and the other two vectors can be written in terms of them. Am I right?
 
  • #13
Thanks Mark. I have another question (my last question) about the same matrix, it looks similar to the part (b) but it's confusing:

Find linearly-independent vctors {u1, u2,...,um} such that: row(A)=span{u1, u2,...,um}
Justify that vectors are linearly independent.


How can I find this row vectors by looking at the reduced matrix?
 
  • #14
roam said:
How can I find this row vectors by looking at the reduced matrix?

Finding the Row Space is similar to the Column Space. First you find the transpose of the original matrix, and perform your row reductions to find the pivots as with before
 
  • #15
roam said:
Thanks Mark. I have another question (my last question) about the same matrix, it looks similar to the part (b) but it's confusing:

Find linearly-independent vctors {u1, u2,...,um} such that: row(A)=span{u1, u2,...,um}
Justify that vectors are linearly independent.


How can I find this row vectors by looking at the reduced matrix?

Here's your reduced matrix (taken from what you posted earlier)
[tex]= \left[\begin{array}{ccccc} 1&0&-1&0&0\\ 0&1&-2&0&-2 \\0&0&0&1&-3 \end{array}\right][/tex]

The rows in this reduced matrix span the rowspace of A, and are clearly (to my eye at least) linearly independent. Call these vectors u1, u2, and u3. To justify that they are linearly independent, show that the equation c1*u1 + c2*u2 + c3*u3 = 0 has only one solution for the constants c1, c2, and c3.
 

1. What are linearly independent vectors?

Linearly independent vectors are a set of vectors that cannot be expressed as a linear combination of each other. In other words, no vector in the set can be written as a combination of the other vectors multiplied by some scalar.

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

A set of vectors is linearly independent if the only solution to the equation a1v1 + a2v2 + ... + anvn = 0 is a1 = a2 = ... = an = 0. In other words, the only way to produce a linear combination of the vectors that equals 0 is by multiplying each vector by 0.

3. What is the importance of linearly independent vectors?

Linearly independent vectors have many important applications in mathematics and science, particularly in linear algebra and vector calculus. They are used to create a basis for vector spaces and to solve systems of linear equations.

4. Can a set of 3 or more vectors be linearly independent?

Yes, a set of 3 or more vectors can be linearly independent as long as they meet the criteria for linear independence. This means that no vector in the set can be written as a linear combination of the other vectors in the set.

5. What is the difference between linearly independent and linearly dependent vectors?

Linearly dependent vectors can be expressed as a linear combination of each other, while linearly independent vectors cannot. In other words, linearly dependent vectors are not unique and can be replaced by a linear combination of other vectors, while linearly independent vectors are unique and cannot be replaced by a linear combination of other vectors.

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