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Linear independence

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
I got one question. If we got this vector \(\displaystyle V=(3,a,1)\),\(\displaystyle U=(a,3,2)\) and \(\displaystyle W=(4,a,2)\) why is it linear independence if determinant is not equal to zero? (I am not interested to solve the problem, I just wanna know why it is)

Regards,
\(\displaystyle |\pi\rangle\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Hello MHB,
I got one question. If we got this vector \(\displaystyle V=(3,a,1)\),\(\displaystyle U=(a,3,2)\) and \(\displaystyle W=(4,a,2)\) why is it linear independence if determinant is not equal to zero? (I am not interested to solve the problem, I just wanna know why it is)

Regards,
\(\displaystyle |\pi\rangle\)
If you put your vectors in a matrix, you get a linear function identified by the matrix.
If you can "reach" all of $\mathbb R^3$ with this function, your vectors are linearly independent.
The determinant shows if this is possible. Zero means it's not.
 

Petrus

Well-known member
Feb 21, 2013
739
If you put your vectors in a matrix, you get a linear function identified by the matrix.
If you can "reach" all of $\mathbb R^3$ with this function, your vectors are linearly independent.
The determinant shows if this is possible. Zero means it's not.
Thanks! I start to understand now!:)

Regards,
\(\displaystyle |\pi\rangle\)
 

Petrus

Well-known member
Feb 21, 2013
739
Is this state true.
1. We can check if a vector is linear Independence by checking if the determinant is not equal to zero
2. For a linear independence matrix there exist a inverse
3.
I did find this theorem in internet that don't say in my book.
"If amounts of vector \(\displaystyle v_1,v_2,v_3....v_p\) is in a dimension \(\displaystyle \mathbb R^n\)
then it is linear dependen if \(\displaystyle p>n\)

Regards,
\(\displaystyle |\pi\rangle\)
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Is this state true.
1. We can check if a vector is linear Independence by checking if the determinant is not equal to zero
2. For a linear independence matrix there exist a inverse
3.
I did find this theorem in internet that don't say in my book.
"If amounts of vector \(\displaystyle v_1,v_2,v_3....v_p\) is in a dimension \(\displaystyle \mathbb R^n\)
then it is linear dependen if \(\displaystyle p>n\)

Regards,
\(\displaystyle |\pi\rangle\)
You can only calculate the determinant of a square matrix.
That means you can only use a determinant to check independence of n vectors, each of dimension n.

An inverse can only exist for a square matrix.
If the matrix is square and the vectors in it are linearly independent, then there exists an inverse.

If you have n linearly independent vectors, they span an n-dimensional space, like $\mathbb R^n$.
If you have one more vector, that won't fit in that n-dimensional space anymore in an independent manner.
So a set of n+1 vectors in $\mathbb R^n$ will have to be dependent.
 

Petrus

Well-known member
Feb 21, 2013
739
You can only calculate the determinant of a square matrix.
That means you can only use a determinant to check independence of n vectors, each of dimension n.

An inverse can only exist for a square matrix.
If the matrix is square and the vectors in it are linearly independent, then there exists an inverse.

If you have n linearly independent vectors, they span an n-dimensional space, like $\mathbb R^n$.
If you have one more vector, that won't fit in that n-dimensional space anymore in an independent manner.
So a set of n+1 vectors in $\mathbb R^n$ will have to be dependent.
Thanks, I meant a square matrix :)

Regards,
\(\displaystyle |\pi\rangle\)