LinearIndependence.md (3187B)
1 # Linear Independence 2 3 **Source:** Linear Algebra Done Right 4 5 **Chapter:** 2 6 7 **Definition:** Linear independence means that every column in a given matrix gives another degree of freedom. 8 9 This can also be thought of as there only being one way to make every vector with linear combinations of vectors in the span. 10 11 Conversely, linear dependent vectors are vectors that are on the same line (or plane) as some other vector (or combination of vectors) thus not giving the matrix another degree of freedom. 12 13 Interesting thing; if you are in less dimensions than the number of vectors it is guaranteed their is linear dependence because you can't go beyond the current dimension. 14 15 Ie. 16 17 [2 7 10] 18 [3 8 1] 19 20 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. 21 22 ### Test For Dependence 23 24 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. 25 26 ### Intuitive Definition 27 28 Linear independence means each vector in a set of vectors (possibly matrix) adds something to the matrix such that the [Span](Span.md) of the set of vectors is larger. 29 30 ### Solving 31 32 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: 33 34 [1 3 4 | 0] 35 [2 5 5 | 0] 36 [2 4 7 | 0] 37 38 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. 39 40 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: 41 42 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. 43 44 [1 3 4 | 0] 45 [0 -1 -3 | 0] 46 [2 4 7 | 0] 47 48 Let's now multiply row 2 by 4 to cancel out the first and second zeroes of the final equation: 49 50 [1 3 4 | 0] 51 [0 -1 -3 | 0] 52 [0 0 -5 | 0] 53 54 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: 55 56 -5c_1 = 0 57 c_1 = 0 58 59 -1c_2 + -3c_1 = 0 60 -1c_2 = 0 61 c_2 = 0 62 63 c_1 + 3c_2 + 4c_3 = 0 64 c_1 = 0 65 66 As we can see, the only solution when each variable is singled out is 0,0, and 0 thus each vector is independent. 67 68 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.