A Practical Approach to Linear Algebra by Prabhat Choudhary

By Prabhat Choudhary

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Vm be a basis in IR n and let A be the n x m matrix with columns VI' v2' ... , vm . The fact that the system is a basis, means that the equation Ax = b has a unique solution for any (all possible) right side b. The existence means that there is a pivot in every row (of a reduced echelon form of the matrix), hence the number of pivots is exactly n. The uniqueness mean that there is pivot in every column of the coefficient matrix (its echelon form), so m = number of columns = number of pivots = n Proposition.

To calculate this, we need the I-st row of A and the I-st column of B, so let us cover up all unnecessary information, so that 2 4 3 -I] ~: [qtlx x qt2] x . 3 = 2 + 8 + 0 -3 = 7. Consider next q 12. To calculate this, we need the I-st row of A and the 2-nd column of B, so let us cover up all unnecessary information, so that 2 4 3-1]: ; x [ x x x x x x x x x -2 [X ql21 x xJ =X 1 x. 1 = 8 + 12 - 6 -1 = 13. Consider next q2I. 2 + 5. 3 = 3 + 2 + 0 + 6 = 11 55 Matries Consider next q22. To calculate this, we need the 2-nd row of A and the 2-nd column of B, so let us cover up all unnecessary information, so that x 4 x x x x 5 3 3 x 2 x -2 = x x x x x x X q22 X X j .

Generating) i echelon Proof The system VI' v2' •.. , vm E ~m is linearly independent ifand only if the equation XlvI +x2v2 +··· +xmvm=O has the unique (trivial) solution XI = x 2 = ... = xm = 0, or equivalently, the equation Ax = 0 has unique solution x = O. By statement 1 above, it happens if and only if there is a pivot in every column of the matrix. Similarly, the system VI' v 2' •.. , vm E ~m is complete in ~n ifand only if the equation +x2v2 +··· +xmvm=b has a solution for any right side b E ~ n .

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