Mathematics 2B - Linear Algebra 2020
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Mathematics 2B - Linear Algebra
2020
1. (i) Find a basis for the null space of
A = ┌ 
┐
!1 2 0 0 ! .
┌ -36┐
(ii) Show that v = 
5 is an element of null(A). Then compute the coordinate ! 5 !
vector [v]B of v with respect to the basis B that you found in part (i).
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Unseen computation, but similar to those seen in lectures. |
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(i) The RREF is ┌0(1) !0 |
2 0 0 |
0 0 ! , |
|
┌ x ┐
so x = !w!
null(A) = Put v1 = [-2, 1, 0, 0]T and v2 = [0, 0, 1, 1]T ; then βv1 , v2 } is a basis. |
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(ii) By inspection, v = 3v1 + 5v2 , so v is in the nullspace of A and [v]B = ┌ ┐5(3) . |
2.
Determine whether the linear transformation
T : R2 → R2 , T ╱┌ ┐、y(x) = 
┐
is invertible. If it is invertible, compute its inverse.
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Unseen, but similar to examples from lectures. T is represented by the
matrix A = too. The transformation T − 1 is represented by
A− 1 = So
T − 1 ┌ ┐v(u) = |
3.
Let A and B be n ! n matrices and let λ 
0 be an eigenvalue of AB . Prove that λ is an eigenvalue of BA.
(Hint: be careful about the fact that an eigenvalue must have a nonzero eigenvec- tor.)
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Unseen. Let 0 BABv = Bλv = λBv. Put u = Bv; then BAu = λu.
It remains to check that u |
4. Prove that A = ┌0(1) 1(1)┐ and I2 are not similar.
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Unseen. The only eigenvalue of A is 1, with algebraic multiplicity 2. Now (A - I)v = 0 ←李 ┌0(0) 0(1)┐ ┌ ┐v(v)2(1) = ┌ ┐0(0) ←李 v2 = 0, so the 1-eigenspace of A is ,t ┌ ┐0(1) : t è R、, which is 1-dimensional. Hence A is not diagonalisable, and so not similar to the diagonal matrix I2 . |
5. Find an orthonormal basis for the subspace W = span(u1 , u2 ) of R4 , where
┌ 0(1) ┐ u1 = 
-1 ,
! 2 !
┌ 1(0) ┐
u2 = ((!-11))! .
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Unseen, but similar to an example seen in lectures. We first show that u1 and u2 are linearly independent. Writing the linear combination c1 u1 + c2 u2 = 0 in matrix form and row reducing gives ┌( 0(1) 1(0) ┐) ┌(0(1) 1(0)┐) (-1 -1 ) ↓ (0 0 ) , ! 2 1 ! !0 0!
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2022-07-28