Test: Linear Algebra - 7 - Mathematics MCQ

Let T be a linear operator on a finite dimensional space V and c is any scalar. Then
we need to find the valid condition for which c is characteristic value of T.
Let c be a characteristic value of 7, then there exist a non-zero vector x ∈ V such that
T(x) = cx.
or equivalently (T - cl)x = 0.
Since x ≠ 0, Therefore, T - cl = 0
Taking the determinant on both sides, we get det(T- cl) = 0.
which shows that T - cl is singular.
Thus, if c is a characteristic value of T. Then T- cl is singular.

Let T be a linear transformation on the vector space V₂(F) defined by
T(a, b) = (a, 0)
We need to find the matrix of T relative to the ordered basis {(1,0), (0,1)} of V₂(F).
Now,T(1, 0) = (1, 0) = 1(1,0) + 0(0,1) and T (0, 1) = (0, 0) = 0(1,0) + 0(0,1), Therefore the matrix of T relative to the ordered basis {(1, 0), (0, 1)}
of V₂(F) is 

We are given that the matrices

say A₁ ,A₂ and A₃
A₁ is a diagonal matrix which has eigen values 1 and 0
A₂ is upper triangular matrix which has eigen values 0, 0 and A₃ is a lower triangular matrix which has eigen values -1 , -1.
Hence,only one matrix which has eigen value 1.

We are given that the matrix A =  has an
eigen value 3. We need to find the other two eigen values;
We have, trace A = sum of eigen values
2 - 1 + 0 = λ₁ + λ₂ + λ₃
1 = 3 + λ₂ + λ₃
or λ₂ + λ₃ = -2
Hence, option (b) satisfies this condition.

We are given that the matrix A =  has the eigen values 4 and 8. We need to find the values of x and y.
We have, trace A = sum of eigen values
or 2+y= 4 + 8
y= 12 -2 = 10
and, the determinant of the matrix A = product of eigen values.
2y - 3x = 4 x 8
or 20 - 3x = 32
or 3x = -12
or x = -4
and y= 10

We need to find the eigen values of the matrix A = 
The characteristic equations of the given matrix, 

(3 - λ) [ (- 1 - λ)² + 9] = 0
(3 - λ) (λ² + 2λ + 10) = 0
λ = 3 ,-1 + 3i,- 1- 3i.

We are given that the characteristic equation of a 3 x 3 matrix ,A is
λ³ + λ² + 2λ + 1 = 0
By Cayley Hamilton theorem, this characteristic equations may be written in the matrix form as,
A³ + A²+ 2A + I=0
or A³ + A² + 2A = - I
or A(A² + A + 2I) = - I
or A^-1 = - (A² + A + 2I)

Let T : R³ -> R² be a linear transformation defined by
T(x,y,z) = (x + y ,x - z ) .
We need to find the dimension of the null space of T.
Let (x, y, z) ∈ ker T. Then
T(x,y,z) = ( 0,0)
using the definition of T, we get
(x + y , x - z) = (0, 0)
or equivalently
x + y = 0 , x - z = 0
Implies y = -x, z = x.
Therefore,ker T = {(x, - x , x ) : x is arbitrary}.
Hence,dim ker T = 1

We are given that T₁ : Rⁿ —> R^m  and T₂ : R^m —> Rⁿ be linear transformation such that T₁T₂ is bijective, where n ≠ m
Thus, T₁T₂ ---> R^m ---> Rⁿ is a linear transformation given by
(T₁T₂) (n) = T₁(T₂(n))
Since T₁T₂ is bijective. Therefore, T₁T₂ is non-singular and hence
Rank (T₁T₂) = m
But Rank (T₁T₂) < min Rank {T₁T₂} Therefore,m ≤ min Rank {T₁T2}
Case I : If m ≤ n, Then rank of both T₁ and T₂ will be m.
Thus, we get
Rank (T₁) = Rank (T₂) = m
Case II : If m ≥ n. Then since T₁ is n x m matrix. Therefore, its rank cannot exceed n. Similarly, Rank T₂ cannot exceed n. Thus, we get m ≤ n. Therefore m = n.

Let T : R³ -> R³ be the linear transformation whose matrix with respect to the standard basis of R³ is  where a, b, c are real number not all zero. The determinant = -abc + bac = 0
Thus, Rank of T is 2. Using Rank nullity theorem, Nullity T = 3 - 2 = 1.
Hence, T is not one-one. Also T is not onto because range has two vectors, so it will not generate R³.
Next, we know that the equation of line passing through origin is 
Let (l, m, n) be the point on the line. Then (l, m, n) = l(1, 0, 0) + m(0, 1, 0) + n(0 , 0 , 1)
Taking the image under T, we get
T(1,m,n) = T(l(1, 0,0) + m(0,1,0) + n(0,0,1))
= IT(1, 0, 0) + mT(0,1, 0) + nT(0, 0,1)
= l(0, -a, -b) + m(a, 0, -c) + n(b, c, 0)
= (ma + nb, -al + nc, - bl - mc)
But the point (ma + nb, -a l + nc, - bl - mc) does not lie on the line

Let T: R² —-> R² be a linear transformation such that
T(1, 2) = (2,3) and T(0,1) = (1, 4)
We need to find the image of (5, 6) under T.
Let there exist scalars α and β
such that (5, 6) = α (1 , 2) + β(0, 1)
or equivalently (5,6) = (α , 2α + β) Comparing the components of the coordinates,
we get α = 5, 2α + β = 6
Solving for α and β, we get
α= 5 ,β = - t Therefore,(5, 6) = 5(1, 2) - 4(0, 1)
Taking the image under T, we get
7(5, 6) = 7[5(1, 2) - 4(0, 1)]
= 5T(1,2) -4T(0,1)
= 5(2, 3 ) - 4 (1,4) = (6 ,-1)

Let S = {T : R³---> R³ such that T is a linear transformation with
T(1, 0, 1) = (1, 2, 3), T(1, 2, 3) = (1,0,1)}.
Then we can define the linear transformation for the third independent element in any way. Therefore, we can get uncountably many linear transformation.

Null space is also a subspace. We known that the dimension of Null space consisting of n × n symmetric matrices is

For particular n = 6, dim of Null space =

We are given the matrix A = 
The characteristic equation of the given matrix is |A-λI| = 0
or 
or λ² + 3λ + 2 = 0
By Cayley Hamilton theorem, this characteristic equation can be written in the matrix form as,


 

Let T : R³ —» R³ be a linear transformation and I be the identity transformation of R³. If there exist a scalar c and a non zero vector x ∈ R³, such that T(x) = cx. Then
(T - c l) (x) = T(x) - (cl) (x)
= cx - cx = 0.
Thus, for non zero x, (T - cl) (x) is zero. Hence, nullity of ( T - cl) cannot be 0. Using Rank nullity theorem.
Rank (T - cI) = 3 - Nullity of (T - cl) Therefore, Rank of (T- cl) can not be 3.

For a positive integer n, let P_n denote the space of all polynomials P(x) with coefficients in M such that degree P(x) ≤ n and let Bn denote the standard basis of Pn given by B_n = {1, x, x²,....., xⁿ}
If T : P₃---->P₄ is a linear transformation defined by
T(P(x)) = x²P'(x) +
and A = (a_ij) is the 5 x 4 matrix of T with respect to the standard bases B₃ and B₄


Hence, the matrix of T related to the basis
Hence a₃₂ = 3/2 and a₃₃ = 0