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New PDF release: Linear Algebra Done Wrong

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By Sergei Treil

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Am,k )T , k = 1, 2, . . , n. Note, these three examples are essentially just different representations of the same mathematical object. Before explaining how to solve a linear system, let us notice that it does not matter what we call the unknowns, xk , yk or something else. So, all the information necessary to solve the system is contained in the matrix A, which is called the coefficient matrix of the system and in the vector (right side) b. Hence, all the information we need is contained in the following matrix   a1,1 a1,2 .

E. the rows with all entries equal 0), if any, are below all non-zero entries. For a non-zero row, let us call the leftmost non-zero entry the leading entry. Then the second property of the echelon form can be formulated as follows: 2. For any non-zero row its leading entry is strictly to the right of the leading entry in the previous row. Pivots: leading (rightmost non-zero entries) in a row. The leading entry in each row in echelon form is also called pivot entry, or simply pivot, because these entries are exactly the pivots we used in the row reduction.

Let us use symbol y instead of b. We know that given y ∈ W the equation Ax = y Doesn’t this remind you of a basis? 26 1. Basic Notions has a unique solution x ∈ V . Let us call this solution B(y). Let us check that B is a linear transformation. We need to show that B(αy1 +βy2 ) = αB(y1 )+βB(y2 ). e. Axk = yk , k = 1, 2. Then A(αx1 + βx2 ) = αAx1 + βAx2 = αy2 + βy2 , which means B(αy1 + βy2 ) = αB(y1 ) + βB(y2 ). 7. 9. An m × n matrix is invertible if and only if its columns form a basis in Rm .

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Linear Algebra Done Wrong by Sergei Treil


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