HW6_solution

# HW6_solution - EE441 HW#6 Solution Spring 2007 Instructor...

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EE441 HW#6 Solution Spring 2007 Instructor : Dr. Jonckheere TA: Ben Raskob Due Date 2/22/2007 2.3.30 The pivots are in columns 2 and 3, which tells us that these columns are linearly independent and can therefore form a basis for the column space of U . Performing row operations can change the column space, but keep the same echelon form. For example, A = EU = 1 2 0 0 - 2 3 0 0 2 2 0 0 4 - 1 0 0 0 1 4 3 0 0 2 2 0 0 0 0 0 0 0 0 = 0 1 8 7 0 - 2 - 2 0 0 2 12 10 0 4 14 10 where E here performs row operations on U . Note that columns 2 and 3 of A are still linearly independent (since pivots appear there in echelon form) and can therefore form a basis for the column space of A . However, the column space of A is NOT equal to the column space of

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Unformatted text preview: A is NOT equal to the column space of U . 2.4.8 If n > m then there would be more columns than rows (a fat matrix). Since the dimension of the row space is equal to the dimension of the column 1 space, there would be at least n-m free variables, and thus inFnite solu-tions. Therefore, this cannot be the case. If m > n then there would be more rows than columns (a tall matrix). If all n columns were linearly independent, then the only solution would be x = 0, and the rank would be n . Similarly, if m = n and all columns were linearly independent (and therefore rows as well) then the rank would be n as well. 2...
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• Spring '08
• Neely
• Linear Algebra, column space, Dr. Jonckheere TA, Ben Raskob, U. Note

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