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Kernal
Linear Independence
Basis
Dimension
Rank-Nullity Theorem
Coordinates How to Examples Exercise
How to:
For a set of vectors to be linearly independent means that they are each unique and any one vector cannot be expressed as a combination of the others.
In an MxN matrix, A, a set of vectors is linearly independent if and only if there is a leading one in each column. That is if the rank(A) = M.
By the same token, if and only ifrank(A) = M, then kernal of A is the zero-vector. So, if you've already found the kernal, you can also judge its linear independence.
Any column without a leading one is a linear combination of the others. To find out exactly what this combination is, we can set up an augmented matrix with all the original vectors of the columns with leading one's on one side, and any of the other vectors on the other side. This is made more clear in the example below.
Examples:
Let's start with the 3x3 matrix, R = [4 8 4; 1 2 2; 2 4 3]. We want to test the linear independence of the column vectors in R, so we need first to find its rank:
r = rank(R)
We find out that our rank is equal to 2. But R has 3 columns, so already we know that these vectors are not linearly independent. To find out which vector is a linear combination of the others (we know there is only one since our rank is 2 and we have 3 columns, 3 - 2 = 1), we need to put our matrix into reduced row-echelon form:
R = rref(R)
In RREF our matrix looks like this: R = [1 2 0; 0 0 1; 0 0 0]. If we look down the second column, we see it is A(:, 2) = [2; 0; 0] and is lacking a leading one. So, to find out the exact combination of vectors it takes to make up our 3rd column, we can put it into a set of linear equations in augmented matrix form (leaving out the colon in MATLAB of course, but not necessarily on paper):
[4 4; 1 2; 2 3]*[x1; x2] = [8; 2; 4]
S = [4 4 8; 1 2 2; 2 3 4]
So here, we should think of the final column of S (our original 2nd vector) as being on the other side of the colons.
After we put this new matrix into RREF, we find that it equals: rref(S) = [1 0 2; 0 1 0; 0 0 0]. This means that x1 = 2 and x2 = 0, or in other words:
2*[4; 1; 2] + 0*[4; 2; 3] = [8; 2; 4]
So we were right, our 2nd vector is a combination of the others. It is actually twice the first vector.
Exercise: