commit e797a1685fc9250664a8f1700a4bda33c6f99d60
parent 3bf9866d6417d86f425ed82d8fcbf6684f74ea33
Author: Andrew <andrewlaack1@gmail.com>
Date: Thu, 4 Jul 2024 12:02:08 -0500
Lin alg notes
Diffstat:
1 file changed, 50 insertions(+), 0 deletions(-)
diff --git a/LinearIndependence.md b/LinearIndependence.md
@@ -17,3 +17,53 @@ Ie.
[3 8 1]
Given that [2,3] and [7,8] consumes all of R^2, we know there are no degrees of freedom provided by [10,1] despite it not being on either of the lines created by the first two columns.
+
+### Test For Dependence
+
+If c_1*a + c_2*b = 0 is true for some constants c_1 and c_2 then we have dependence assuming at least one coeficcient is not zero. This is true for an arbitrary number of vectors and constants. If this is only possible with coeficcients that are equal to zero then we have independence.
+
+
+### Intuitive Definition
+
+Linear independence means each vector in a set of vectors (possibly matrix) adds something to the matrix such that the [[Span.md]] of the set of vectors is larger.
+
+
+### Solving
+
+A simple way to solve this is using our knowledge that c_1 * a + ... + c_n * z = I where I is the identity matrix. Knowing this, we can create an augmented matrix that represents this information as follows:
+
+[1 3 4 | 0]
+[2 5 5 | 0]
+[2 4 7 | 0]
+
+Above the final column (which makes this an augmented matrix) represents the target value of the equation on the left. The left side is represented by columns whereby each is associated with a different coeficcient.
+
+If we show that this is only true when c_1,c_2, and c_3 are zero then they are independent. If there is a solution where all coeficcients are not zero then they are not independent and thus dependent. Observe the next steps of elimination by multiplying columns to fill the in zeroes to get to row echelon form:
+
+First let's multiply the first row by -2 and add it to the second row. This can be done because they are equivalent in value (0) thus the addition does not break the system.
+
+[1 3 4 | 0]
+[0 -1 -3 | 0]
+[2 4 7 | 0]
+
+Let's now multiply row 2 by 4 to cancel out the first and second zeroes of the final equation:
+
+[1 3 4 | 0]
+[0 -1 -3 | 0]
+[0 0 -5 | 0]
+
+Now we see each coeficcient has been singled out. Let's solve the system now where column 1 is c_1, column 2 is c_2, and column 3 is c_3:
+
+-5c_1 = 0
+c_1 = 0
+
+-1c_2 + -3c_1 = 0
+-1c_2 = 0
+c_2 = 0
+
+c_1 + 3c_2 + 4c_3 = 0
+c_1 = 0
+
+As we can see, the only solution when each variable is singled out is 0,0, and 0 thus each vector is independent.
+
+If this were not the case then we would not be able to single out each variable. When this is the case we should get as many zeroes as possible and then identify what coeficcients make the statement true without them all being 0. This is suficcient proof to declare the vectors are not independent.
