Linearly Dependent/Independent Matrices

[geminii]

Hi everyone!

I'm not sure how to tell if a set of matrices is linearly dependent or linearly independent. The textbook doesn't help and the online explanations talk about scalar multiples, and I have no idea what those are.

Here is the question I'm working on:

Determine whether the following set of vectors is linearly dependent.

a = [4]
      [1]
      [3]

b = [2]
      [-1]
      [3]

c = [-4]
      [2]
      [6]

It'd be great if someone could help with this! TIA

[dooyeon1998]

If a,b,c are linearly dependent  a=mb+nc can be established. m and n are constants

I will use the horizontal matrices for convenience

[4  1  3]=m [2  -1  3] + n [-4  2  6]

[4  1  3]= [2m  -m  3m] +  [-4n  2n  6n]

From this 3 equations can be derived
(1) 4=2m-4n
(2) 1= -m+2n
(3) 3=3m+6n

from (1) m=2+2n can be derived
If we substitute this m for (2)
1=-(2+2n) +2n
1=-2 -2n+2n
1=-2
This equation is not true and does not make sense
Therefore there are no solutions for m and n
therefore a,b,c are linearly independent
Hope this helps

[lzxnl]

If a,b,c are linearly dependent  a=mb+nc can be established. m and n are constants

I will use the horizontal matrices for convenience

[4  1  3]=m [2  -1  3] + n [-4  2  6]

[4  1  3]= [2m  -m  3m] +  [-4n  2n  6n]

From this 3 equations can be derived
(1) 4=2m-4n
(2) 1= -m+2n
(3) 3=3m+6n

from (1) m=2+2n can be derived
If we substitute this m for (2)
1=-(2+2n) +2n
1=-2 -2n+2n
1=-2
This equation is not true and does not make sense
Therefore there are no solutions for m and n
therefore a,b,c are linearly independent
Hope this helps

For the purpose of completeness, I will slightly alter your statement. If a = mb + nc for some scalar m and n, then the three vectors are linearly dependent. However, the converse is not necessarily true. The formal definition actually only requires that a linear combination of the vectors in consideration sum to zero. How is this different?

One definition of a dimension is the maximum number of linearly independent vectors. By definition you cannot present me 3 linearly independent vectors that exist in R² (try it). So, the three vectors 2i, i and j should be linearly independent. They are, because the first two are parallel, so 1(2i)+(-2)i + 0j = 0. However, you cannot write j as a sum of the other two.

Tl;dr, if you can write a as a linear combination of b and c, they must be linearly dependent. But that method won't find all the cases.

[geminii]

If a,b,c are linearly dependent  a=mb+nc can be established. m and n are constants

I will use the horizontal matrices for convenience

[4  1  3]=m [2  -1  3] + n [-4  2  6]

[4  1  3]= [2m  -m  3m] +  [-4n  2n  6n]

From this 3 equations can be derived
(1) 4=2m-4n
(2) 1= -m+2n
(3) 3=3m+6n

from (1) m=2+2n can be derived
If we substitute this m for (2)
1=-(2+2n) +2n
1=-2 -2n+2n
1=-2
This equation is not true and does not make sense
Therefore there are no solutions for m and n
therefore a,b,c are linearly independent
Hope this helps

Thank you very much!! This helped clarify the process. I understand it now!

For the purpose of completeness, I will slightly alter your statement. If a = mb + nc for some scalar m and n, then the three vectors are linearly dependent. However, the converse is not necessarily true. The formal definition actually only requires that a linear combination of the vectors in consideration sum to zero. How is this different?

One definition of a dimension is the maximum number of linearly independent vectors. By definition you cannot present me 3 linearly independent vectors that exist in R² (try it). So, the three vectors 2i, i and j should be linearly independent. They are, because the first two are parallel, so 1(2i)+(-2)i + 0j = 0. However, you cannot write j as a sum of the other two.

Tl;dr, if you can write a as a linear combination of b and c, they must be linearly dependent. But that method won't find all the cases.

Ah ok, I understand what you're saying. So even if two are parallel, and don't add up to the remaining matrix, they are still linearly dependent. But if none of them are parallel, and none of them add up to another matrix, then they are linearly independent.

On that note, how do you tell if two matrices are parallel? Are two matrices parallel when one can be multiplied by a scalar quantity to produce the second matrix?

Eg. [3 6 2] is parallel to [9 18 6]? If this is the case, is this matrix combination parallel?

[3 6 2] and [6 12 3]
(aka do all the numbers need to be multiplied by the same number to be parallel with the first matrix? Here, both 6 and 12 of the second matrix have been multiplied by 2 from the first matrix, but the 3 from the same matrix has been multiplied by 1.5 from the 2 in the first matrix.)

[lzxnl]

Thank you very much!! This helped clarify the process. I understand it now!


Ah ok, I understand what you're saying. So even if two are parallel, and don't add up to the remaining matrix, they are still linearly dependent. But if none of them are parallel, and none of them add up to another matrix, then they are linearly independent.

On that note, how do you tell if two matrices are parallel? Are two matrices parallel when one can be multiplied by a scalar quantity to produce the second matrix?

Eg. [3 6 2] is parallel to [9 18 6]? If this is the case, is this matrix combination parallel?

[3 6 2] and [6 12 3]
(aka do all the numbers need to be multiplied by the same number to be parallel with the first matrix? Here, both 6 and 12 of the second matrix have been multiplied by 2 from the first matrix, but the 3 from the same matrix has been multiplied by 1.5 from the 2 in the first matrix.)

It's more like, if any subset of the large group of vectors is linearly dependent, the large group is too. So to get around this, you're supposed to solve
ax + by + cz = 0 or whatever
If a solution exists where a, b and c are not all zero, they're dependent. That's the formal definition anyway.