UPSC Civil Services Main 1992 - Mathematics

Linear Algebra
Sunder Lal
Retired Professor of Mathematics
Panjab University
Chandigarh

June 14, 2007

Question 1(a) Let U and V be vector spaces over a field K and let V be of finite dimension.
Let T : V −→ U be a linear transformation, prove that dim V = dim T(V) + dim nullity T.

Solution. See question 3(a), year 1998.

Question 1(b) Let S = {(x, y, z) | x + y + z = 0, x, y, z ∈ R}. Prove that S is a subspace

of R3 . Find a basis of S.

Solution. S = 6 ∅ because (0, 0, 0) ∈ S. If (x1 , y1 , z1 ), (x2 , y2 , z2 ) ∈ S then α1 (x1 , y1 , z1 ) +

α2 (x2 , y2 , z2 ) ∈ S because (α1 x1 + α2 x2 ) + (α1 y1 + α2 y2 ) + (α1 z1 + α2 z2 ) = α1 (x1 + y1 + z1 ) +
α2 (x2 + y2 + z2 ) = 0. Thus S is a subspace of R3 .
Clearly (1, 0, −1), (1, −1, 0) ∈ S and are linearly independent. Thus dim S ≥ 2. However
(1, 1, 1) 6∈ S, so S = 6 R3 . Thus dim S = 2 and {(1, 0, −1), (1, −1, 0)} is a basis for S.

Question 1(c) Which of the following are linear transformations?

1. T : R −→ R2 defined by T(x) = (2x, −x).

2. T : R2 −→ R3 defined by T(x, y) = (xy, y, x).

3. T : R2 −→ R3 defined by T(x, y) = (x + y, y, x).

4. T : R −→ R2 defined by T(x) = (1, −1).

Solution.

1
 1.

T(αx + βy) = (2αx + 2βy, −αx − βy)

= (2αx, −αx) + (2βy, −βy)
= αT(x) + βT(y)

Thus T is a linear transformation.

2. T(2(1, 1)) = T(2, 2) = (4, 2, 2) 6= 2T(1, 1) = 2(1, 1, 1) Thus T is not a linear transfor-
mation.

3.

T(α(x1 , y1 ) + β(x2 + y2 )) = T(αx1 + βx2 , αy1 + βy2 )

= (αx1 + βx2 + αy1 + βy2 , αy1 + βy2 , αx1 + βx2 )
= α(x1 + y1 , y1 , x1 ) + β(x2 + y2 , y2 , x2 )
= αT(x1 , y1 ) + βT(x2 , y2 )

Thus T is a linear transformation.

4. T(2(0, 0)) = T(0, 0) = (1, −1) 6= 2T(0, 0) Thus T is not a linear transformation.

Question 2(a) Let T : M2,1 −→ M2,3 be a linear transformation defined by (with the usual
notation)        
1 2 1 3 1 6 1 0
T = ,T =
0 4 1 5 1 0 0 2
 
x
Find T .
y

Solution.
       
x 1 1 1
= x −y +y
y 0 0 1
       
x 2 1 3 6 1 0 2x + 4y x 3x − 3y
T = (x − y) +y =
y 4 1 5 0 0 2 4x − 4y x − y 5x − 3y

2
Question 2(b) For what values of η do the following equations

x+y+z = 1
x + 2y + 4z = η
x + 4y + 10z = η 2

have a solution? Solve them in each case.

1 1 1

Solution. Since the determinant of the coefficient matrix 1 2 4 is 0, the system has to
1 4 10
be consistent to be solvable.
Clearly x + 4y + 10z = 3(x + 2y + 4z) − 2(x + y + z). Thus for the system to be consistent
we must have η 2 = 3η − 2, or η = 1, 2.
If η = 1, then x + y + z = 1, x + 2y + 4z = 1 so y + 3z = 0, or y = −3z, x = 1 + 2z. Thus
the space of solutions is {(1 + 2z, −3z, z) | z ∈ R}. Note that the rank of the coefficient
matrix is 2, and consequently the space of solutions is one dimensional.
If η = 2, then x + y + z = 1, x + 2y + 4z = 2, so y + 3z = 1 or y = 1 − 3z, hence x = 2z.
Consequently, the space of solutions is {(2z, 1 − 3z, z) | z ∈ R}.

Question 2(c) Prove that a necessary and sufficient condition of a real quadratic form
x0 Ax to be positive definite is that the leading principal minors of A are all positive.

Solution. Let all the principal minors be positive. We have to prove that the quadratic
form is positive definite. We prove the result by induction.
If n = 1, then a11 x2 > 0 ⇔ a11 > 0. Suppose as induction hypothesis the result is true
B B1

for n = m. Let S = B01 k be a matrix of a quadratic form in m + 1 variables, where
B is m × m, B1 is m × 1 and k is a single element. Since all principle minors of B are
leading principal minors of S, and are hence positive, the induction hypothesis gives that B
is positive definite. This means that there exists a non-singular m × m matrix P such that
P0 BP = Im (We shall prove this presently). Let C be an m-rowed column to be determined
soon. Then
 0 
P0 BP P0 BC + P0 B1
   
P 0 B B1 P C
=
C0 1 B0 1 k 0 1 C0 B0 P + B01 P C0 BC + C0 B1 + B0 1 C + k

Let C be so chosen that BC + B1 = 0, or C = −B−1 B1 . Then

 0     0 
P 0 B B1 P C P BP 0
=
C0 1 B0 1 k 0 1 0 B0 1 C + k

Taking determinants, we get |P0 ||S||P| = B0 1 C + k, because P0 BP = Im , and B0 1 C + k is

0 0 2
a single element. Since |S|
 > 0,itfollows that
 B 1 C + k > 0, so let B 1 C + k = α . Then
P C Im 0
Q0 SQ = Im+1 with Q = . Thus the quadratic forms of S and Im+1 take
0 1 0 α−1
the same values. Hence S is positive definite, so the condition is sufficient.

3
 The condition is necessary - Since x0 Ax is positive definite, there is a non-singular matrix
P such that P0 AP = I ⇒ |A||P|2 = 1 ⇒ |A| > 0.
Let 1 ≤ r < n. Let xr+1 = . . . = xn = 0, then we obtain a quadratic form in r variables
which is positive definite. Clearly the determinant of this quadratic form is the r×r principal
minor of A which shows the result.
Proof of the result used: Let A be positive definite, then there exists a non-singular P
such that P0 AP = I.
We will prove this by induction. If n = 1, then the form corresponding to A is a11 x2 and
√
a11 > 0, so that P = ( a11 ).
Take  
1 −a−1 11 a12 0 ... 0
0 
P1 =  ..
 

. (n − 1) × (n − 1) 
0
then  
a11 0 a13 ... a1n
 0 
 
P01 AP1 =  a13
 

 .. 
 . (n − 1) × (n − 1) 
a1n
Repeating this process, we get a non-singular Q such that
 
a11 0 ... 0
Q0 AQ =  ...
 
(n − 1) × (n − 1) 
0

Given the (n − 1) × (n − 1) matrix on the lower right, we get by induction P∗ s.t. P∗ 0 ((n −
1) × (n − 1) matrix)P∗ is diagonal. Thus ∃P, |P| 6= 0, P0 AP = [α1 , . . . , αn ] say. Take R =
√ √
diagonal[ α1 , . . . , αn ], then R0 P0 APR = In .

2 1


Question 3(a) State the Cayley-Hamilton theorem and use it to find the inverse of 4 3 .

Solution. Let A be an n × n matrix. If |λI − A| = λn + a1 λn−1 + . . . + an = 0 is the

characteristic equation of A, then the Cayley-Hamilton theorem says that An + a1 An−1 +
. . . + an I = 0 i.e. a matrix satisfies its characteristic

equation.
The characteristic equation of A = 4 3 is2 1

2−λ 1
= λ2 − 5λ + 2 = 0
4 3−λ

4
By the Cayley-Hamilton theorem, A2 − 5A + 2I = 0, so A(A − 5I) = −2I, thus A−1 =
− 12 (A − 5I). Thus
  3
− 21
   
−1 1 2 1 5 0
A =− − = 2
2 4 3 0 5 −2 1

Question 3(b) Transform the following into diagonal form

x2 + 2xy, 8x2 − 4xy + 5y 2
and give the transformation employed.
8 −2



Solution. Let A = 11 10 , B = −2 5

1 − 8λ 1 + 2λ
Let 0 = |A − λB| = = −5λ + 40λ2 − 4λ2 − 4λ − 1
1 + 2λ −5λ
√
Thus 36λ2 − 9λ − 1 = 0, so λ = 9± 81+144 72
= 13 , − 12
1
.

x1 1 5 5
Let (x1 , x2 ) be the vector such that (A − λB) x2 = 0 with λ = 3 . Thus − 3 x1 + 3 x2 =
0 ⇒ x1 = x2 . We take x1 = 11 so
that (A − λB)x1 = 0 with λ = 31 . Similarly, if (x1 , x2 ) is

the vector such that
(A − λB) xx12 = 0 with λ = − 12 1

, then 53 x1 + 56 x2 = 0, so 2x1 + x2 = 0.
1
We take x2 = −2 .
Now
x01 Ax1 = ( 1 1 ) 11 10 11 = ( 1 1 ) 21 = 3

x02 Ax2 = ( 1 −2 ) 11 10 −2 = ( 1 −2 ) −1

1


1 = −3
x01 Ax2 = ( 1 1 ) 11 10 −2 = ( 1 1 ) −1

1



1 =0
If P = (x1 x2 ), then P0 AP = 30 −3



0
, thus x2 + 2xy ≈ 3X 2 − 3Y 2 by P = 11 −2 1
.
Similarly
x01 Bx1 = ( 1 1 ) −2 8 −2

1
6


5 1 = (1 1) 3 = 9
x02 Bx2 = ( 1 −2 ) −2 8 −2

1
12


= ( 1 −2 ) −12 = 36
0 8 −2
5

1−2
12


x1 Bx2 = ( 1 1 ) −2 5 −2 = ( 1 1 ) −12 = 0

Thus P0 BP = 90 36 0 , so 8x2 − 4xy + 5y 2 is transformed to 9X 2 + 36Y 2 by X = P x

Y y

Question 3(c) Prove that the characteristic roots of a Hermitian matrix are all real, and
the characteristic roots of a skew Hermitian matrix are all zero or pure imaginary.
Solution. For Hermitian matrices, see question 2(c), year 1995.
0 0
If H is skew-Hermitian, then iH is Hermitian, because (iH) = iH = −iH = iH as
0
H = −H . Thus the eigenvalues of iH are real. Therefore the eigenvalues of H are −ix
where x ∈ R. So they must be 0 (if x = 0) or pure imaginary.