MATH 0047: Advanced Linear Algebra
University College London 2021-2022
MATH 0047 final assessment - 2020-2021: a set of solutions
1. (a) The system of equations may be written as the following matrix equation:
1 −2 −3 3
−2 4 5 −2
4 −8 −8 −4
−1 2 2 1


x1
x2
x3
x4
 =

2
−4
8
−2

The augmented matrix for this system of equations is
1 −2 −3 3 | 2
−2 4 5 −2 | −4
4 −8 −8 −4 | 8
−1 2 2 1 | −2

1
Let us first use elementary row operations to find the reduced row echelon form of this matrix:
1 −2 −3 3 | 2
−2 4 5 −2 | −4
4 −8 −8 −4 | 8
−1 2 2 1 | −2
 R2→R2+2R1−−−−−−−→

1 −2 −3 3 | 2
0 0 −1 4 | 0
4 −8 −8 −4 | 8
−1 2 2 1 | −2

R3→R3−4R1−−−−−−−→

1 −2 −3 3 | 2
0 0 −1 4 | 0
0 0 4 −16 | 0
−1 2 2 1 | −2

R4→R4+R1−−−−−−−→

1 −2 −3 3 | 2
0 0 −1 4 | 0
0 0 4 −16 | 0
0 0 −1 4 | 0

R2→−R2−−−−−→

1 −2 −3 3 | 2
0 0 1 −4 | 0
0 0 4 −16 | 0
0 0 −1 4 | 0

R1→R1+3R2−−−−−−−→

1 −2 0 −9 | 2
0 0 1 −4 | 0
0 0 4 −16 | 0
0 0 −1 4 | 0

R3→R3−4R2−−−−−−−→

1 −2 0 −9 | 2
0 0 1 −4 | 0
0 0 0 0 | 0
0 0 −1 4 | 0

R4→R4+R2−−−−−−−→

1 −2 0 −9 | 2
0 0 1 −4 | 0
0 0 0 0 | 0
0 0 0 0 | 0

So, the original system of real linear equations is equivalent to:
x1 − 2x2 − 9x4 = 2
x3 − 4x4 = 0
We may rearrange each of these equations, to obtain:
x1 = 2 + 2x2 + 9x4
x3 = 4x4
2
Thus, the general solution to this system of real linear equations is given by:
x1 = 2 + 2λ+ 9µ
x2 = λ
x3 = 4µ
x4 = µ
for λ, µ ∈ R.
The given system of real linear equations has infinitely many solutions.
(b) For all 1 ≤ i ≤ n, 1 ≤ j ≤ n:
((A+B)(2C))ij =
n∑
k=1
(A+B)ik(2C)kj by definition of matrix multiplication
=
n∑
k=1
(Aik +Bik)(2C)kj by definition of matrix addition
=
n∑
k=1
(Aik +Bik) · 2(Ckj) by definition of scalar multiplication
=
n∑
k=1
(2AikCkj + 2BikCkj) since we are dealing with numbers
=
n∑
k=1
2AikCkj +
n∑
k=1
2BikCkj since we are dealing with numbers
=2
(
n∑
k=1
AikCkj +
n∑
k=1
BikCkj
)
since we are dealing with numbers
=2((AC)ij + (BC)ij) by definition of matrix multiplication
=2((AC +BC)ij) by definition of matrix addition
=(2(AC +BC))ij by definition of scalar multiplication
So, since for all i, j: ((A+B)(2C))ij = (2(AC +BC))ij , we deduce that
(A+B)(2C) = 2(AC +BC)
as required.
3
2. (a) We row reduce the augmented matrix (M | I3 ), where I3 is the 3× 3 identity matrix:0 4 10 | 1 0 00 1 2 | 0 1 0
1 −2 −4 | 0 0 1
 R1↔R3−−−−→
1 −2 −4 | 0 0 10 1 2 | 0 1 0
0 4 10 | 1 0 0

R1→R1+2R2−−−−−−−→
1 0 0 | 0 2 10 1 2 | 0 1 0
0 4 10 | 1 0 0

R3→R3−4R2−−−−−−−→
1 0 0 | 0 2 10 1 2 | 0 1 0
0 0 2 | 1 −4 0

R3→ 12R3−−−−−→
1 0 0 | 0 2 10 1 2 | 0 1 0
0 0 1 | 1
2
−2 0

R2→R2−2R3−−−−−−−→
1 0 0 | 0 2 10 1 0 | −1 5 0
0 0 1 | 1
2
−2 0

At the end of the row reduction, we obtain the identity matrix on the “left half” of the augmented
matrix (i.e. the reduced row echelon form of M is the identity matrix). So, the inverse of M is the
matrix on the “right half” of the final augmented matrix:
M−1 =
 0 2 1−1 5 0
1
2
−2 0

4
(b) (i) The matrix C is a matrix that contains a zero column (the third column of C is a zero column),
so we may use the result that states that, if a square matrix has a zero row or column, then its
determinant is equal to 0; using this, we may deduce that
det(C) = 0
The matrix D is upper triangular, so we may use the result that states that the determinant of an
upper triangular matrix is the product of its diagonal entries; using this, we may deduce that
det(D) = D11D22D33D44 = (1)(9)(
1
12
)(4)
i.e.
det(D) = 3
(ii) We may use the result that states that, for any square n× n matrices A and B:
det(AB) = det(A)det(B)
Then: det(D)det(D−1) = det(DD−1) = det(I) = 1. As a result, we obtain:
det(D−1) =
1
det(D)
= 1
3
Similarly, det(D−4) = det(D−2D−2) = det(D−2)det(D−2) = det(D−1D−1)det(D−1D−1),
so that
det(D−4) = det(D−1)det(D−1)det(D−1)det(D−1)
We obtain
det(D−4) = 1
3
× 1
3
× 1
3
× 1
3
= 1
81
i.e. det(D−4) = 1
81
(iii) We may use the result, mentioned in the solution to part (b)(ii), which states that, for any square
n× n matrices A and B, det(AB) = det(A)det(B); here, we obtain
det(CD10) = det(C)det(D10) = 0× det(D10)
i.e. we obtain det(CD10) = 0.
We may then use the result that states that a square matrix is invertible if and only if it has
nonzero determinant, in order to conclude that, since the determinant of the matrix CD10 is
equal to 0, the relevant matrix, CD10, is not invertible.
5
3. (a) Let us verify that the relevant defining conditions for a real subspace hold for the given subset of R3:
(1) The zero vector is of the required form, i.e. ifx1x2
x3
 =
00
0

then the condition of being in S is satisfied:
7x1 − 5x2 + x3 = 7(0)− 5(0) + (0) = 0
so that 0 ∈ S.
(2) Suppose that we start with vectors x and y in S, say
x =
x1x2
x3
 , y =
y1y2
y3

Then, since x and y are in S, we may assume that the following equations hold:
7x1 − 5x2 + x3 = 0
7y1 − 5y2 + y3 = 0
Consider the sum of the vectors x and y:
x+ y =
x1x2
x3
+
y1y2
y3
 =
x1 + y1x2 + y2
x3 + y3

Let us verify that x+ y also satisfies the defining equation of S:
7(x1 + y1)− 5(x2 + y2) + (x3 + y3) = 7x1 + 7y1 − 5x2 − 5y2 + x3 + y3
= (7x1 − 5x2 + x3) + (7y1 − 5y2 + y3)
= 0 + 0 using the fact that x and y are in S
= 0
So, as required, x+ y ∈ S.
6
(3) Suppose that we are given any real number λ and a vector x in S, say
x =
x1x2
x3

Then, since x is in S, we may assume that the following equation holds:
7x1 − 5x2 + x3 = 0
Consider the scalar product λx:
λx =
λx1λx2
λx3

Let us verify that λx also satisfies the defining equation of S:
7(λx1)− 5(λx2) + (λx3) = 7λx1 − 5λx2 + λx3
= λ(7x1 − 5x2 + x3)
= λ× 0 using the fact that x is in S
= 0
So, as required, λx ∈ S.
Having verified that the relevant defining conditions hold, we may deduce that S is, indeed, a real
subspace of R3, as required.
(b) (i) The given set S is not a real subspace of R2.
For example, the zero vector does not satisfy the equation x1 − 4x2 = 7.
Explicitly, if we set (
x1
x2
)
=
(
0
0
)
then
x1 − 4x2 = (0)− 4(0) = 0 ̸= 7
So, since x1− 4x2 ̸= 7 for the zero vector, we deduce that the zero vector does not belong to S.
Therefore, one of the conditions in the relevant definition fails to hold; S is not a real subspace
of R2.
7
(ii) The given set S is not a real subspace of R2.
It fails to satisfy one of the conditions in the definition of a real subspace.
The following two vectors are in S:
v =
(
7
1
)
, w =
(−7
1
)
Let us consider the sum of these vectors:
x = v + w =
(
0
2
)
This sum is not in S, since in this case
x21 − 49x22 = (0)2 − 49(2)2 = −196 ̸= 0
Therefore, we have found vectors v, w, which are in S, but such that their sum, v +w, is not in
S. This shows that one of the conditions in the relevant definition fails to hold, and so allows us
to deduce that S is not a real subspace of R2.
(iii) The given set S is a real subspace of R2.
The only possible (real) solution of the equation x21 + 49x
2
2 = 0 is the zero solution, i.e. the
solution for which x1 = 0 and x2 = 0. So, S contains only one vector, the zero vector:
S =
{(
0
0
)}
We may now verify that the relevant defining conditions for a real subspace hold:
(1) The zero vector is contained in the set S: 0 ∈ S.
(2) Suppose that x and y are vectors in S. Since S consists only of the vector 0, the only
possibilities for x and y are x = 0 and y = 0. In such a case:
x+ y = 0 + 0 = 0
so that x+ y = 0 is once again in S, as required.
(3) Suppose that we are given any real number λ and a vector x in S, i.e. the vector x = 0.
Then, the product λx = λ× 0 = 0 is also in S, as required.
Since we have shown that the relevant defining conditions hold, we may conclude that S is a
real subspace of R2.
8
(c) We suppose that the vectors u1, u2, u3 span a subspace U of Rn, so that the vectors u1, u2, u3 are
vectors in U and each vector in U can be written as a linear combination of the vectors u1, u2, u3; as
such, given any vector u in U , there are real numbers λ1, λ2, λ3 such that
u = λ1u1 + λ2u2 + λ3u3
By rearranging u1− 4u2 +8u3 = 0, we obtain u1 = 4u2− 8u3, and, by substituting u1 = 4u2− 8u3
into the above equation, we obtain
u = λ1(4u2 − 8u3) + λ2u2 + λ3u3
= 4λ1u2 − 8λ1u3 + λ2u2 + λ3u3
= (4λ1 + λ2)u2 + (−8λ1 + λ3)u3
As a result, we obtain
u = (4λ1 + λ2)u2 + (−8λ1 + λ3)u3
We may therefore conclude that every vector u in U can be written as a linear combination of the
vectors u2, u3 (which are, themselves, vectors in U ). This shows that the vectors u2, u3 span U , as
required.
9
4. (a) (i) We shall try to make use of the following equality, relevant in the setting of the standard real
inner product on Rn, which, for θ being the angle between two non-zero vectors x, y in Rn,
relates cos θ to ||x||, ||y|| and ⟨x, y⟩:
cos θ =
⟨x, y⟩
||x|| ||y||
We shall also try to make use of the following general results, in the setting of the standard real
inner product on Rn, for vectors a, b, c in Rn and a real number λ:
◦ ⟨a, b⟩ = ⟨b, a⟩
◦ ⟨a+ c, b⟩ = ⟨a, b⟩+ ⟨c, b⟩ , ⟨a, b+ c⟩ = ⟨a, b⟩+ ⟨a, c⟩
◦ ⟨λa, b⟩ = λ ⟨a, b⟩ , ⟨a, λb⟩ = λ ⟨a, b⟩
Here, we are given that ||x|| = 2, ||y|| = 1, and that the angle between the two vectors is 120◦.
In terms of inner products, this information allows us to calculate ⟨x, y⟩, using the equality
mentioned above :
cos θ =
⟨x, y⟩
||x|| ||y||
Substituting ||x|| = 2, ||y|| = 1 and θ = 120◦ above, we obtain:
cos 120◦ =
⟨x, y⟩
2× 1
i.e.
⟨x, y⟩ = 2× cos 120◦ = 2×−1
2
= −1
We may now compute the norm of the vector 4x+ 3y. By definition of norm:
||4x+ 3y||2 = ⟨4x+ 3y, 4x+ 3y⟩
We may compute the inner product on the right, by using, in the setting of the standard real
inner product on Rn, the assumed general results mentioned above, involving relevant inner
products, to “break the product down” (below, we also make use of the definition of norm):
⟨4x+ 3y, 4x+ 3y⟩ = ⟨4x, 4x+ 3y⟩+ ⟨3y, 4x+ 3y⟩
= ⟨4x, 4x⟩+ ⟨4x, 3y⟩+ ⟨3y, 4x⟩+ ⟨3y, 3y⟩
= 4 ⟨x, 4x⟩+ 4 ⟨x, 3y⟩+ 3 ⟨y, 4x⟩+ 3 ⟨y, 3y⟩
= 16 ⟨x, x⟩+ 12 ⟨x, y⟩+ 12 ⟨y, x⟩+ 9 ⟨y, y⟩
= 16||x||2 + 24 ⟨x, y⟩+ 9||y||2
10
By substituting the values of ||x||, ||y||, ⟨x, y⟩ above, we obtain:
⟨4x+ 3y, 4x+ 3y⟩ = 16||x||2 + 24 ⟨x, y⟩+ 9||y||2
= 16 · (2)2 + 24 · (−1) + 9 · (1)2
= 64− 24 + 9 = 49
Finally, using the definition of the norm, we obtain:
||4x+ 3y|| =
√
⟨4x+ 3y, 4x+ 3y⟩ =
√
49 = 7
So, the norm of the vector 4x+ 3y is 7.
(ii) It might be helpful to note that, in general, vectors a and b, inRn, are orthogonal, with respect to
the standard real inner product on Rn, if a and b are nonzero vectors in Rn satisfying ⟨a, b⟩ = 0
i.e. satisfying
⟨a, b⟩ = a1b1 + · · ·+ anbn = 0
Here, we consider the vectors 4x + 3y and x; we may compute the inner product ⟨4x+ 3y, x⟩
by using the assumed general results, involving relevant inner products, in the setting of the
standard real inner product on Rn, mentioned in the solution to part (a)(i), in order to “break the
product down” (below, we also make use of the definition of norm):
⟨4x+ 3y, x⟩ = ⟨4x, x⟩+ ⟨3y, x⟩
= 4 ⟨x, x⟩+ 3 ⟨y, x⟩
= 4||x||2 + 3 ⟨x, y⟩
By substituting the values of ||x|| and ⟨x, y⟩, e.g as described in the solution to part (a)(i), we
obtain:
⟨4x+ 3y, x⟩ = 4||x||2 + 3 ⟨x, y⟩
= 4 · (2)2 + 3 · (−1)
= 16− 3
= 13
Then, since the relevant inner product is not equal to zero, i.e. since ⟨4x+ 3y, x⟩ ≠ 0, we may
use the relevant definition of orthogonality described above in order to conclude that the vector
4x+ 3y is not orthogonal to the vector x, with respect to the standard real inner product on Rn.
11
(b) (i) To find the kernel of T , we may solve the equation T (x) = 0, or, equivalently: 1 4 12 9 3
−2 −3 3
x1x2
x3
 =
00
0

This is a (homogeneous) matrix equation that we can solve by row reducing 1 4 1 | 02 9 3 | 0
−2 −3 3 | 0

 1 4 1 | 02 9 3 | 0
−2 −3 3 | 0
 R2→R2−2R1−−−−−−−→
 1 4 1 | 00 1 1 | 0
−2 −3 3 | 0

R3→R3+2R1−−−−−−−→
1 4 1 | 00 1 1 | 0
0 5 5 | 0

R1→R1−4R2−−−−−−−→
1 0 −3 | 00 1 1 | 0
0 5 5 | 0

R3→R3−5R2−−−−−−−→
1 0 −3 | 00 1 1 | 0
0 0 0 | 0

So, the original matrix equation reduces to the following equation:1 0 −30 1 1
0 0 0
x1x2
x3
 =
00
0

Therefore, we deduce that
x1 − 3x3 = 0 , x2 + x3 = 0 , 0 = 0
Since columns 1 and 2 contain ‘leading ones’, the above row reduction suggests that we can
obtain expressions for x1 and x2.
We obtain
x1 = 3x3 , x2 = −x3
Substituting these into the starting vector, composed of x1, x2, x3, allows us to deduce that a
general element satisfying the above equation (i.e. a general element in the kernel of the map)
12
has the form: 3x3−x3
x3
 for x3 in R
This means that we have essentially described the kernel of T :
Ker(T ) =

3x3−x3
x3
 : x3 ∈ R

In order to give a basis for the kernel of T , we may break down the general element of the kernel
into its constituent parts:3x3−x3
x3
 = x3
 3−1
1
 for x3 in R
So, a basis of the kernel of T is, for instance, the following set:
 3−1
1

13
(ii) There does exist a basis for the kernel of T that is orthonormal, with respect to the standard real
inner product on R3.
As indicated above, the following is a possible basis of/for the kernel of T :
 3−1
1

We may obtain an orthonormal basis, from the given basis, by, for instance, exchanging the
relevant vector for a scalar multiple, of the vector, that has norm equal to 1, with respect to the
standard real inner product on R3.
We may achieve this by computing the norm of the relevant vector, and then dividing the vector
by its norm; this amounts to applying the Gram-Schmidt (orthonormalisation) process, with
respect to the standard real inner product on R3.
For instance, if we refer to as v the vector involved in the basis given above:
v =
 3−1
1

then we obtain:
||v|| =
√
⟨v, v⟩ =
√√√√√〈
 3−1
1
 ,
 3−1
1
〉 = √11
As such, the following is a vector that has norm equal to 1, with respect to the standard real
inner product on R3:
e1 =
v
||v|| =
1√
11
 3−1
1
 =

3√
11−1√
11
1√
11

Hence, the following is, for instance, a basis for the kernel of T that is orthonormal, with respect
to the standard real inner product on R3: 

3√
11−1√
11
1√
11


14
5. (a) We will first determine the eigenvalues of A by solving the characteristic equation of A:
det(A− λI) = 0 ⇔
∣∣∣∣8− λ 4i−4i 2− λ
∣∣∣∣ = 0
⇔ (8− λ)(2− λ)− (4i)(−4i) = 0
⇔ λ2 − 10λ+ 16− 16 = 0
⇔ λ2 − 10λ = 0
⇔ λ(λ− 10) = 0
So, the matrix A has two eigenvalues:
λ = 0 with algebraic multiplicity 1
λ = 10 with algebraic multiplicity 1
Let us next find the eigenvectors corresponding to each eigenvalue, by substituting each of the
eigenvalues of A into the equation
(A− λI)x = 0
and then solving the resulting matrix equation.
15
For the eigenvalue λ = 0, we will solve the equation (A− 0I)x = 0, i.e. Ax = 0:(
8 4i
−4i 2
)(
x1
x2
)
=
(
0
0
)
Each row of the matrix A is a multiple of the other row (e.g. if we multiply row 2 through by 2i,
we obtain row 1), so we essentially have a single equation: −4ix1 + 2x2 = 0, or, equivalently,
−2ix1 + x2 = 0.
So, setting, for example, x2 = 2ix1 above, we see that a general solution of the above equation is of
the form: (
x1
2ix1
)
In other words, a general eigenvector of A corresponding to the eigenvalue 0 is of the form:(
x1
2ix1
)
= x1
(
1
2i
)
forx1 ∈ C , x1 ̸= 0
For the eigenvalue λ = 10, we will solve the equation (A− 10I)x = 0:(−2 4i
−4i −8
)(
x1
x2
)
=
(
0
0
)
Each row of the matrix A− 10I is a multiple of the other row (e.g. if we multiply row 1 through by
2i, we obtain row 2), so we essentially have a single equation: −2x1 + 4ix2 = 0, or, equivalently,
x1 − 2ix2 = 0.
So, setting x1 = 2ix2 above, we see that a general solution of the above equation is of the form:(
2ix2
x2
)
In other words, a general eigenvector of A corresponding to the eigenvalue 10 is of the form:(
2ix2
x2
)
= x2
(
2i
1
)
forx2 ∈ C , x2 ̸= 0
16
(b) Using the solution to part (a), we may find two linearly independent eigenvectors for the matrix A
(one corresponding to each eigenvalue). Therefore, we deduce that the matrix A is diagonalisable.
For example, we may choose
P =
(
1 2i
2i 1
)
In that case, since we have chosen to place an eigenvector corresponding to the eigenvalue 0 in
column 1 of P and an eigenvector corresponding to the eigenvalue 10 in column 2 of P , the product
P−1AP (which is the required diagonal matrix) will be equal to
D =
(
0 0
0 10
)
To find a formula for An, we shall use a/the result that states that, for a square matrix A, if there
exists an invertible matrix P such that P−1AP = D, then, for any positive integer n:
An = PDnP−1
If we consider the particular case of the matrix A considered here, then, starting from the particular
matrix D determined above, Dn is simple to compute (since D is a diagonal matrix):
D =
(
0 0
0 10
)
⇒ Dn =
(
0 0
0 10n
)
Also, we can use P to find P−1, for example by row reducing the augmented matrix (P | I ):
(
1 2i | 1 0
2i 1 | 0 1
)
R2→R2−2iR1−−−−−−−−→
(
1 2i | 1 0
0 5 | −2i 1
)
R2→15R2−−−−−→
(
1 2i | 1 0
0 1 | −2i
5
1
5
)
R1→R1−2iR2−−−−−−−−→
(
1 0 | 1
5
−2i
5
0 1 | −2i
5
1
5
)
The identity matrix now forms the “left half” of the row reduced matrix, so the “right half” must be
the inverse of P :
P−1 =
(
1
5
−2i
5−2i
5
1
5
)
i.e. P−1 = 1
5
(
1 −2i
−2i 1
)
17
We may now determine the required formula, for An, using the result from part (b):
An = PDnP−1
=
(
1 2i
2i 1
)(
0 0
0 10n
)(
1
5
−2i
5−2i
5
1
5
)
=
(
0 2i(10n)
0 10n
)(
1
5
−2i
5−2i
5
1
5
)
=
(
4(10n)
5
2i(10n)
5−2i(10n)
5
10n
5
)
Therefore:
An =
(
4(10n)
5
2i(10n)
5−2i(10n)
5
10n
5
)
i.e. An = 1
5
(
4(10n) 2i(10n)
−2i(10n) 10n
)
(c) In order to find suitable matrices Q and E, we may start by noting that, for the matrices P and D
determined in the solution to part (b):
D5 =
(
P−1AP
)5 by using D = P−1AP
=
(
P−1AP
) (
P−1AP
) (
P−1AP
) (
P−1AP
) (
P−1AP
)
= P−1A
(
PP−1
)
A
(
PP−1
)
A
(
PP−1
)
A
(
PP−1
)
AP
= P−1AIAIAIAIAP by using PP−1 = I
= P−1A5P
As a result, we obtain, P−1A5P = D5, where
D5 =
(
0 0
0 10
)5
=
(
0 0
0 105
)
So, we may conclude that, for the matrix P described above, the product P−1A5P will be equal to
the diagonal matrix
E =
(
0 0
0 105
)
As such, overall, we may determine an invertible matrix Q and a diagonal matrix E such that
Q−1A5Q = E, by, for instance, setting Q = P and E = D5, i.e. by, for instance, setting:
Q =
(
1 2i
2i 1
)
, E =
(
0 0
0 105
)
=
(
0 0
0 100000
)
18