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Unformatted text preview: Sample Problems for Midterm 2 1. Find a basis for ker( A ) and Im( A ), where A = 1 2 3 4 0 1 2 3 0 0 0 0 . Solution. For Im( A ), there is a standard procedure for doing this. First rowreduce: A = 1 2 3 4 0 1 2 3 0 0 0 0 ∼ 1 0 1 2 0 1 2 3 0 0 . The third and fourth columns do not contain leading ones. The procedure tells us that it is these columns in the original matrix that are redundant, and so a basis for Im( A ) is 1 , 2 1 . It is worth noting that there other choices of column vectors that form a basis for Im( A ). For ker( A ), we can solve for the kernel using the reducedrow echelon form above. Let x 3 = t and x 4 = s . Then x 1 x 2 x 3 x 4 = x 3 + 2 x 4 2 x 3 3 x 4 x 3 x 4 = t + 2 s 2 t 3 s t s = t 1 2 1 + s 2 3 1 . These last two vectors span the kernel and you can see that they are linearly independent (they always will be, so you do not need to check this). 1 2. Let W = ( x,y ) ∈ R 2 x ≥ 0 and y ≤ . Is W is a subspace of R 2 ? If yes, prove it. If no, explain why not. Solution. Notice that (1 , 1) ∈ W . But 1(1 , 1) = ( 1 , 1) 6∈ W . This shows that W is not closed under scalar multiplication. Therefore W is not a subspace....
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This note was uploaded on 04/26/2009 for the course MATH 33a taught by Professor Lee during the Winter '08 term at UCLA.
 Winter '08
 lee

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