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| − | '''Tricks for checking Linear Independence, Span and Basis''' | + | [[Bonus_point_projects_how_to_and_why|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. | ||
<br> <u>'''Linear Independence'''</u> | <br> <u>'''Linear Independence'''</u> | ||
| − | If | + | *If end result of the rref(vectors) gives an identity matrix, it is '''linearly independent'''<br> |
| + | *If end result of the rref(vectors) gives you a parameter in the matrix, the vectors are '''linearly dependent.''' | ||
| + | |||
| + | Tricks: | ||
| − | If det(vectors) = 0 ⇔ '''linearly | + | If #No of vectors > Dimension ⇔ it is '''linearly dependent''' <br>If det(vectors) != 0 ⇔ '''linearly independent''' <br> 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 | |
<u>'''Span'''</u> | <u>'''Span'''</u> | ||
| − | If | + | *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 det(vectors) = 0 ⇔ ''' | + | If Dimension > #No of vectors -> '''it CANNOT span''' <br>If det(vectors) != 0 ⇔ '''it spans'''<br>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> because both rows have leading 1's. |
| + | |||
| − | < | + | <u>'''Basis'''</u> |
| − | If #No of vectors > Dimension -> it has to be linearly | + | <br>If Dimension > #No of vectors ⇔ cannot span ⇔ is not a basis<br>If #No of vectors > Dimension -> it is not a basis.<br>If #No of vectors = Dimension -> it has to be linearly independent to span |
| − | + | ---- | |
| + | [[2010_Fall_MA_265_Momin|Back to MA265, Fall 2010, Prof. Momin]] | ||
| − | [[Category: | + | [[Category:math]] |
| + | [[Category:linear algebra]] | ||
| + | [[Category:MA265]] | ||
| + | [[Category:MA265Fall2010Momin]] | ||
| + | [[Category:bonus point project]] | ||
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
