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Mathematics 2B - Linear Algebra

2021

1. In the real vector space R4 , let S be the set

S = {(1, 0, 2, 0), (3, 2, 0, 4), (2, 2, _2, 4), (2, 0, 4, 0), (0, 2, _6, 4)}. Find a subset of S which is a basis for the subspace Span(S).

Unseen computation, but similar to those seen in lectures. Let A be a matrix whose rows are the vectors in S .

┐' ┐' ┐'

A = '(')2 2 _2 4 '(') ~ '(')0 2 _6 4 '(') ~ '(')0 0 0 0 '(')

'2 0 4 0 ' '0 0 0 0 ' '0 0 0 0 '

The RREF of A is

0 0 0 0 0

2. Show that B = '(┌) '(┐) is similar to C = '(┌)

0(0)┐

1'.

Unseen, similar examples seen in lectures and tutorial.

Note:. Theorem 4.22 in the notes states that if A is similar to B then the two matrices share some common properties such as the the same determinant, etc., however the converse is not true, shared properties does not imply matrices are similar and so invoking Theorem 4.22 is not a correct approach to this question .

Most common approach

B is similar to C if there exists an invertible matrix P such that C = P一1 BP . Since both B and C are diagonal matrices we can use diagonalisation to ﬁnd P . The eigenvalues of B are the diagonal entries, 1, 2 and 3. Calculating the associated eigenspaces we have:

λ 1 = 1 :

(A _ I)v = 0 ←→ '(┌) '(┐) '(┌)'(┐) = '(┌)'(┐) ←→ v2 = v3 = 0,

so the 1-eigenspace of B is given by the span of ┌ ┐0(1)

'0'.

λ2 = 2 :

(A _ 2I)v = 0 ←→ '(┌)_001 '(┐) '(┌)'(┐) = '(┌)'(┐) ←→ v1 = v3 = 0,

so the 2-eigenspace of B is given by the span of ┌ ┐1(0)

'0'.

λ3 = 3 :

(A _ 3I)v = 0 ←→ '(┌)_002 0_01 '(┐) '(┌)'(┐) = '(┌)'(┐) ←→ v1 = v2 = 0,

so the 3-eigenspace of B is given by the span of ┌ ┐0(0)

'1'.

Since B and C have the same eigenvalues, the matrix P has columns given by the eigenvectors of B placed in the order that the eigenvalues of C occur.

P = ┌┐

'0 1 0'

We ﬁnd P一1 using the Gauss-Jordan method:

[P |I] = '(┌) '(┐) ~ '(┌) '(┐) ~ '(┌) '(┐) = [I|P 1]

Lastly, we check:

P一1 BP = '(┌) '(┐) '(┌) '(┐) '(┌) '(┐) = '(┌) '(┐) '(┌) '(┐) = C

Example alternative solution

B is similar to C if there exists an invertible matrix P such that C = P一1 BP . Since both B and C are diagonal matrices and B and C clearly have the same eigenvalues, namely 1,2,3 we can diagonalise B and C as follow, there exists and invertible matrix V such that V一1 BV = D and an invertible matrix U such that U一1 CU = D, where D is diagonal and we can take U = I , D = C . Then C = P一1 BP with P = VU一1 .

3. Let V be a 3-dimensional vector space with ordered basis B : w1 , w2 , w3 . Let 岱 : v1 , v2 , v3 , be a second ordered basis for V , where v1 = w1 + w2 + w3 , v2 = w2 _ w3 and v3 = w3 .

(i) Find the change of basis matrix PB←岱 .

(ii) Let T be the linear transformation T : V → V deﬁned by

T (w1 ) = w1 + 2w2 _ w3 ,

T (w2 ) = w1 _ w2 ,

T (w3 ) = w1 + w3 .

Using your answer from (i) ﬁnd the matrix [T] with respect to the basis 岱.

Unseen, combines change of basis and linear transformation, concept seen in general in lecture notes..