# hw3 - a 5 = 2 a 1-a 2 3 a 4 This allows us to read oﬀ...

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Linear Algebra Homework 3 Selected Solutions Note: Any mistakes are the sole responsibility of the author–T.Peters 3.6.12 Let A be a 4 × 5 matrix. If a 1 , a 2 , a 4 are linearly independent and a 3 = a 1 + 2 a 2 and a 5 = 2 a 1 - a 2 + 3 a 4 determine RREF( A ). Here’s an example of a matrix in RREF that satisﬁes the properties above: A 0 = 1 0 1 0 2 0 1 2 0 - 1 0 0 0 1 3 0 0 0 0 0 . Let’s argue that RREF( A ) = A 0 . We’re told that a 1 , a 2 , a 4 are linearly independent so if a 0 1 , a 0 2 , a 0 3 , a 0 4 , a 0 5 denote the columns of RREF( A ) then a 0 1 , a 0 2 , a 0 4 must also be linearly independent. Why is this? This is because dependence/independence relations of columns are pre- served under elementary row operations–we discussed this in class. Therefore RREF( A ) must be of the form 1 0 * 0 * 0 1 * 0 * 0 0 0 1 * 0 0 0 0 * . Since a 3 = a 1 +2 a 2 we get that a 0 3 = a 0 1 + 2 a 0 2 since dependence relations among columns are preserved under elementary row operations. Similarly we get
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Unformatted text preview: a 5 = 2 a 1-a 2 +3 a 4 . This allows us to read oﬀ RREF( A ) = A . 3.6.19 Let A and B be n × n matrices. (a) Show that AB = 0 if and only if the column space of B is a subspace of the nullspace of A . (b) Show that if AB = 0, then the sum of the ranks of A and B cannot exceed n . Solution: (a) Partition B into columns B = [ b 1 | b 2 | ... | b n ]. Then AB = [ Ab 1 | Ab 2 | ... | Ab n ]. So if AB = 0 then we see that Ab i = 0 for all i and hence that the column space of B ⊂ N ( A ). Similarly, if we had R ( B ) ⊂ N ( A ), we would have Ab i = 0 for all i and hence that AB = 0. (b) If AB = 0 then from (a) we have R ( B ) ⊂ N ( A ) hence rank B ≤ nullity A so that n =rank A + nullity A ≥ rank A + rank B . 1...
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