MATH 235 Final: MATH 235 UMass Amherst final-solution
Document Summary
Spring 1999: (15 points) the matrices a and b below are row equivalent (you do not need to check this fact). 4: the rank of a is 3. 0: a basis for the null space of a: the reduced echelon form is. The general solution is x1 x2 x3 x4 x5. 0 and the two column vectors on the right hand side are a basis for n ull(a): a basis for the column space of a: the rst, third and fth columns of a are its pivot columns. Justify your answer: dim(n ull(a)) + dim(row(a)) = 6. ( + 1)( 1)( 2) 1 (b) a basis of r3 consisting of eigenvectors of a: The three eigenvalues are 1, 1 and 2. The 1-eigenspace is n ull(a + i) and it is spanned by . The 1-eigenspace is n ull(a i) and it is spanned by . The 2-eigenspace is n ull(a 2i) and it is spanned by .
