Line 1: Line 1:
'''<math>$M =
+
'''Tricks for checking Linear Independence, Span and Basis'''
\begin{array}{cc}
+
x & y \\
+
z & w \\
+
\end{array}$</math>Tricks for checking Linear Independence, Span and Basis'''  
+
 
+
<br>
+
  
 
<br> <u>'''Linear Independence'''</u>  
 
<br> <u>'''Linear Independence'''</u>  
  
If det(vectors)&nbsp;!= 0 ⇔ '''linearly independent'''<br> If end result of the rref(vectors) gives you a matrix with all rows having leading 1's it is '''linearly independent'''.  
+
If det(vectors)&nbsp;!= 0 ⇔ '''linearly independent'''<br> If end result of the rref(vectors) gives you a matrix with all rows having leading 1's it is '''linearly independent'''. <math>$M = \left( \begin{smallmatrix} a&b\\ c&d \end{smallmatrix} \right)$</math>
  
 
If det(vectors) = 0 ⇔ '''linearly dependent'''<br>If end result of the rref(vectors) gives you a parameter in the equation, the vectors are '''linearly dependent.'''  
 
If det(vectors) = 0 ⇔ '''linearly dependent'''<br>If end result of the rref(vectors) gives you a parameter in the equation, the vectors are '''linearly dependent.'''  

Revision as of 07:22, 1 May 2011

Tricks for checking Linear Independence, Span and Basis


Linear Independence

If det(vectors) != 0 ⇔ linearly independent
If end result of the rref(vectors) gives you a matrix with all rows having leading 1's it is linearly independent. $ $M = \left( \begin{smallmatrix} a&b\\ c&d \end{smallmatrix} \right)$ $

If det(vectors) = 0 ⇔ linearly dependent
If end result of the rref(vectors) gives you a parameter in the equation, the vectors are linearly dependent.

Tip: If #No of vectors > Dimension ⇔ it is linearly dependent

Span

If Dimension > #No of vectors -> it CANNOT span

If det(vectors) != 0 ⇔ it spans
If end result of the rref(vectors) gives you a matrix with all rows having leading 1's, it spans.

If det(vectors) = 0 ⇔ does not span
If end result of the rref(vectors) gives you a matrix with not all rows having a leading 1, it does not span.

Basis


If Dimension > #No of vectors ⇔ cannot span ⇔ is not a basis

If #No of vectors > Dimension -> it has to be linearly dependent to span (check the tip)

If #No of vectors = Dimension -> it has to be linearly independent to span
I'n's'e'r't'f'o'r'm'u'l'a'h'e'r'e

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009