Difference between revisions of "????" - Rhea

Latest revision as of 06:21, 3 July 2012

Bonus point project for MA265.

Tricks for checking Linear Independence, Span and Basis

Note: For this article, I am assuming number of vectors is equal to the dimension of the vector space for calculating the determinant. If it is not, you need to do rref.

Linear Independence

• If end result of the rref(vectors) gives an identity matrix, it is linearly independent
  
• If end result of the rref(vectors) gives you a parameter in the matrix, the vectors are linearly dependent.

Tricks:

If #No of vectors > Dimension ⇔ it is linearly dependent
If det(vectors) != 0 ⇔ linearly independent
If det(vectors) = 0 ⇔ linearly dependent

Example: $ rref(\left( \begin{smallmatrix} 1&2&3\\ 2&3&4 \end{smallmatrix} \right)) = \left( \begin{smallmatrix} 1&0&-1\\ 0&1&2 \end{smallmatrix} \right) $ is linearly dependent in R² because the last column [-1 2]^T i.e z is a parameter as there can be no leading 1 for that column. You can express x = z and y = -2z

Span

• If end result of the rref(vectors) gives you a matrix with all rows having leading 1's, it spans.
• If end result of the rref(vectors) gives you a matrix with not all rows having a leading 1, it does not span.

Tricks:

If Dimension > #No of vectors -> it CANNOT span
If det(vectors) != 0 ⇔ it spans
If det(vectors) = 0 ⇔ does not span

For example: $ rref(\left( \begin{smallmatrix} 1&2&3\\ 2&3&4 \end{smallmatrix} \right)) = \left( \begin{smallmatrix} 1&0&-1\\ 0&1&2 \end{smallmatrix} \right) $ spans R2 because both rows have leading 1's.

Basis

If Dimension > #No of vectors ⇔ cannot span ⇔ is not a basis
If #No of vectors > Dimension -> it is not a basis.
If #No of vectors = Dimension -> it has to be linearly independent to span

Back to MA265, Fall 2010, Prof. Momin