| Line 9: | Line 9: | ||
Tricks: | Tricks: | ||
| − | + | If #No of vectors > Dimension ⇔ it is '''linearly dependent''' | |
If det(vectors) != 0 ⇔ '''linearly independent''' | If det(vectors) != 0 ⇔ '''linearly independent''' | ||
| − | |||
If det(vectors) = 0 ⇔ '''linearly dependent''' | If det(vectors) = 0 ⇔ '''linearly dependent''' | ||
| − | |||
| − | |||
Example: <math>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)</math> is '''linearly dependent''' in R<sup>2</sup> because the last column [-1 2]<sup>T</sup> i.e z is a parameter as there can be no leading 1 for that column. You can express x = z and y = -2z | Example: <math>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)</math> is '''linearly dependent''' in R<sup>2</sup> because the last column [-1 2]<sup>T</sup> i.e z is a parameter as there can be no leading 1 for that column. You can express x = z and y = -2z | ||
<u>'''Span'''</u> | <u>'''Span'''</u> | ||
| − | |||
| − | |||
*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 all rows having leading 1's, '''it spans'''.  | ||
| Line 26: | Line 21: | ||
Tricks: | Tricks: | ||
| − | If det(vectors) != 0 ⇔ it spans | + | If Dimension > #No of vectors -> '''it CANNOT span''' |
| − | If det(vectors) = 0 ⇔ '''does not span''' | + | If det(vectors) != 0 ⇔ '''it spans''' |
| + | If det(vectors) = 0 ⇔ '''does not span''' | ||
For example: <math>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)</math> spans R<sup>2</sup><br> | For example: <math>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)</math> spans R<sup>2</sup><br> | ||
Revision as of 10:42, 1 May 2011
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
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
