Quick linear algebra Questions...
- Thread starter demon8991
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Yes No and No I think.
Immediately it should be obvious that (1,2) and (3,4) are linearly independent. There are two of them and dimension(R^2) is obviously 2 so you have a linearly independent span and therefore a basis.
If you look at the vectors in R^3 you can see that they aren't linearly independent so they can't span R^3 and therefore cant be a basis. To see they aren't linearly independent you can do the following:
(1,-1,2)+(-1,1,2)-4*(0,0,1) = 0
Immediately it should be obvious that (1,2) and (3,4) are linearly independent. There are two of them and dimension(R^2) is obviously 2 so you have a linearly independent span and therefore a basis.
If you look at the vectors in R^3 you can see that they aren't linearly independent so they can't span R^3 and therefore cant be a basis. To see they aren't linearly independent you can do the following:
(1,-1,2)+(-1,1,2)-4*(0,0,1) = 0
In general, a system of vectors are linearly dependent if any one vector can be recreated by a combination of the others (and tokenbrit showed for the second problem).
The quick and easy way to show this, for larger systems, is to use the vectors as columns (or equivalently rows) of a matrix, and find the determinant. If the determinant is zero, the matrix is singular and the system is linearly dependent. A non-zero det shows non-singularity and so linear independence in the vectors.
A linearly independent set of n vectors will be form a basis for R^n. It will also span R^m, where m < n, but cannot form a basis as there will be redundancy in the set of vectors.
The quick and easy way to show this, for larger systems, is to use the vectors as columns (or equivalently rows) of a matrix, and find the determinant. If the determinant is zero, the matrix is singular and the system is linearly dependent. A non-zero det shows non-singularity and so linear independence in the vectors.
A linearly independent set of n vectors will be form a basis for R^n. It will also span R^m, where m < n, but cannot form a basis as there will be redundancy in the set of vectors.
Only problem is, the method most people are taught for finding determinants isn't at all quick - it's O(n!). I'd say you're generally better off using Gaussian Elimination / row reduction. (In fact, this is one way of calculating large determinants reasonably quickly - see http://www-gatago.com/sci/math/num-analysis/35699235.html, for example).Duff-Man said: