Chapter 3

1. Chapter 3 Section 3.4 Bases for Subspaces

2. Spanning Sets Let W be a subspace of , and a set of vectors in W (i.e). The set S is a spanning set for W (or SspansW) if and only if every vector in W is in the span of S(i.e. every vector w in W is a linear combination of ) For every w in W there exists numbers such that: Example Determine if the following sets are spanning sets for the subspace W that is the plane given to the right. The set does not span W. Take for example the vector w. , but has no solution The set does span W. Take a general vector win W. -R1+R2 -3R1+R3 -R2 2R2+R3 This system will be consistent as long as or equivalently that or that the vector

3. The set does not span W since the vector is not in the subspace W. Linearly Independent Vectors A set of vectors is linearly independent if the only solution is the zero vector, or there are no independent (free) variables in the system of equations to the right. Means A Basis for a Subspace A linearly independent spanning set of vectors for a subspace W is called a basis for the subspace W. These are the essential vectors needed to describe W. is a basis of W if: is linearly independent Maximal Independent Set A basis is the greatest number of linearly independent vectors that are required to span the subspace W. (Keep adding independent vectors to the set until you span all of W.) Minimal Spanning Set A basis is the smallest number of vectors required to span the subspace W that are linearly independent. (Keep removing dependent vectors from a spanning set until you get a linearly independent set.)

4. Example Find a basis for the subspace W which is span of the set of vectors in given to the right. The idea is to find a subset of the vectors that is linearly independent. This means that the variables must all be zero in the equation to the right. Row reducing and solving will give a relation between the variables. Row reduces to: Particular Solutions: Gives Relations: Each of the particular solutions is chosen by taking each dependent variable setting it equal to -1 and all the other dependent variables we set to zero. Rewrite the dependent variables on the left side of the equation. This gives the following: Both the 2nd and 4th vectors are linear combinations of the 1st and 3rd vectors!

5. The subspace has basis . This method has a much more direct way of characterizing it. To determine the linearly independent vectors in the set of vectors do the following. Form the matrix and row reduce it. The vectors that correspond to the columns of the dependent variables in the original matrix will be the linearly independent vectors. It is important you use the columns of the original matrix. The row operations change the span of column vectors. This comparison between the original matrix and the row reduced matrix is useful for determining a basis. Basis for the column space of a matrix Since the column space of a matrix is the span of the columns of A. A basis for the column space can be found by row reducing A and taking the columns in the original matrix A corresponding to the original matrix A. Basis for the range of a matrix Since the range of a matrix is equal to its column space (i.e. ) this same method can be applied.

6. Basis for the Null Space of a Matrix One of the most frequently occurring basis to find is a basis for the null space of a matrix. Like before we row reduce the matrix and look for a relation that ensures linear independence. Example Find a basis for the null space for the matrix A given to the right. (i.e. find a basis for ) Vector form of Solution: Row reduce A: Basis for the null space () Notice that each independent variable corresponds to a basis vector. The number of vectors in the basis for the null space is the same as the number of independent variables.

7. Basis for the Row Space of a Matrix Row operations do not change the span of the rows of a matrix since each one is a linear combination of the previous rows. Consider the example we had before. Example Find a basis for the row space for the matrix A given to the right. Row reduce A: Basis for Row Space: The non zero rows of the reduced matrix form the basis of the row space. • Row Operations and Basis • For a matrix A row operations change the range and column space • For a matrix A row operations do not change the row space and null space. In the standard basis is and Standard Basis Vectors for The standard basis vectors for are the vectors where the vector is a vector with a 1 in the position and zeros in every other position

8. Example For the matrix A given to the right find a basis for the row space, column space, range and null space. Begin by row reducing the matrix A to reduced echelon form. We see that the 1st and 3rd columns have the leading 1’s and the first two rows are non zero. Row Reduced Basis of Column Space and range: By comparing the original and row reduced matrices we can find a basis for the row space column space and range. Basis of Row Space: To find a basis for the null space we write the solution in vector form and expand it according to the independent variables. Null Space Basis:

9. Bases and Non Zero Scalar Multiples If we multiply a basis by a set of non zero scalars the new set of vectors is again a basis. This is useful for eliminating fractions from a basis. If is a basis for W and are non zero scalars then the set: Is also a basis for W. Example Find a basis for the null space of matrix A to the right that consists of all integer values. Row reduced Row reduce the matrix A and write the solution in vector form. Basis for Basis for Multiply the first vector by 3 and the second vector by 4 to get an integer basis.