An Introduction To Linear Algebra by Kenneth Kuttler

By Kenneth Kuttler

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Then V has a basis. 1 This is the plural form of basis. We could say basiss but it would involve an inordinate amount of hissing as in “The sixth shiek’s sixth sheep is sick”. This is the reason that bases is used instead of basiss. 5. AN APPLICATION TO MATRICES 57 Proof: Let v1 ∈ V where v1 = 0. If span {v1 } = V, stop. {v1 } is a basis for V . Otherwise, there exists v2 ∈ V which is not in span {v1 } . 10 {v1 , v2 } is a linearly independent set of vectors. If span {v1 , v2 } = V stop, {v1 , v2 } is a basis for V.

However, if we did know what it was, we could find ak+1 as follows. ij ak+1 = ij akir arj r This is because if you go from i to j in k + 1 steps, you first go from i to r in k steps and then for each of these ways there are arj ways to go from there to j. Thus akir arj gives the number of ways to go from i to j in k + 1 steps such that the k th step leaves you at location r. Adding these gives the above sum. Now you recognize this as the ij th entry of the product of two matrices. Thus a2ij = air arj r a3ij = a2ir arj r and so forth.

0 0 . 1 you are like some of us, you will usually have made a  2 2 . Find A−1 . −1 Set up the augmented matrix, (A|I)  1 2  1 0 3 1  2 1 0 0 2 0 1 0  −1 0 0 1 Next take (−1) times the first row and add to the row added to the last. This yields  1 2 2 1  0 −2 0 −1 0 −5 −7 −3 second followed by (−3) times the first  0 0 1 0 . 0 1 48 MATRICES AND LINEAR TRANSFORMATIONS Then take 5 times the second row and add to  1 2 2  0 −10 0 0 0 14 -2 times the last row.  1 0 0 −5 5 0  1 5 −2 Next take the last row and add to (−7) times the top row.

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