Linear Transformation and Diagonalization: 2 Marks Important Questions with Answer

LINEAR TRANSFORMATION AND DIAGONALIZATION

2 Marks Important Questions with Answer

1. If T is linear transformation, then prove that T (0) = 0.

Proof:

We know that T(x+y)=T(x)+T (y) and

T(0) = T (0+0) = T (0) + T (0)

= 0 and

T(0)=T(0x)=0T (x)=0.

2. If T is linear, then prove that

 T (cx+y)=cT (x)+T (y) for all x,y ∈ V and c ∈ F.

Proof:

Assume T is linear

T (cx+y) = T (cx) + T (y)

 = cT (x)+T (y)

Conversely,

Assume T (cx + y) = cT (x) + T (y)

put c=1

T (x + y) = T (x) + T (y).

 y ∈ V; y = 0 ∈ V

y(cx+0) = cT (x)+T (0)

= cT(x)+0 = cT(x)

 T is linear

3. If T is linear then prove that T(x−y) = T(x) − T (y) for all x, y ∈ V.

Proof

To prove T (ax+by) = aT(x)+bT (y) for x, y ∈ V and a, b ∈ F.

The scalars are considered as 1,1 ∈ F,  we assume

 a = 1 and b = −1

T(x − y) = (1)T (x) + (− 1) T(y) = T(x) − T(y)

 T(x−y)=T(x)−T(v)

4. Show that T is not linear when T: R² → R² which is defined by T (a₁, a₂) = (sin a₁, 0).

Solution:

It is given that T (a₁, a₂) = (sin a₁, 0)

Let T(2π/2, 0) = Т (π, 0) = (sin π, 0)

= (0, 0)                   ...(1)

Now let T(2π/2, 0) = 2T(2π/2, 0)

= 2(sin(π/2), 0) =

 = 2 (1, 0)

 = (2,0)                                 ...(2)

From (1) and (2) we have (0, 0) ≠ (2, 0). Hence T is not linear.

5. Let T:P_n(R) → P_n−1(R). Define T[f(x)]=f '(x) where f '(x) denotes the derivatives of ƒ(x). Show that T is linear.

Solution:

 Let g(x), h(x) ∈ P_n(R), c ∈ F,

T (cg(x) + h(x)) = d/dx (cg (x) + h (x))

 = cg'(x) + h'(x)

 = cT(g(x)) + T(h (x))

T is linear.

6. State the dimension theorem in vector space

Proof:

Suppose that dim (V) = n.

 'N (T) is a subspace of V.

dim (N (T))   n

Let dim N (T) = k ≤ n, and

 { v1, v2, …… vk} is a basis of N(T).

We know that if W is a subspace of a finite dimensional vector space V then any basis for W can be extended to a basis of V.

  we may extend { v1, v2, v3,..., vk } to a basis

 B = { v1, v2, v3,..., vk, vk+1, …… vn } for V.

Now we claim that,

 s = {T (vk+1), T (vk+2), ... T(vn)} is a basis for R (T).

First we prove that S generates R (T).

We know that if B={v1, v2, ..., vn} is a basis for V, then R (T) = span (T (B)).

Since { v1, v2, ..., vk } is a basis for N (T)

⇒ vi ∈ N(T)

⇒ T(vi) = 0

for all i = 1 to k.

we have

R(T) = span ({T (V1), T (V2), ... T(vn) }) then R (T) = span (T (B))

Since { v1, v2, ..., vk} is a basis for N(T)

      vi ∈ N(T) ⇒ T (vi) = 0 for all i = 1 to k.

R(T) = span {T (vk+1), T (vk+2), ... T(vn)}

 = span (S)

Next we prove that S is linearly independent.

Let bk+1 T (vk+1) + bk+2 T (vk+2) + ... + bn T (vn) = 0

where     bk + 1, bk +2,... bn ∈ F.

Using the fact that T is linear, we have

Since { v1, v2, ..., vk} is a basis for N(T)

Since { v1, v2, v3,..., vk, vk+1, …… vn } is a basis for V.

 (ie) { v1, v2, v3,..., vn } is linearly independent.

−ci=0 for all i = 1 to k and

 bi=0 for all i=k+1 to n.

In particular

 bk+1 = bk+2 = ... = bn = 0

S = { T (vk+1), T (Vk + 2), ... T (vn) } is Linearly independent.

  S is a basis for R (D)

 dim (R (T)) = n−k

 rank (T) = dim V − dim (N (T))

 Nullity (T) + rank (T) = dim (V)

7. Is there a linear transformation T: R³ → R² such that T (1, 0, 3) = (1, 1) and T (−2, 0, − 6) = (2, 1).

Solution:

T (− 2, 0, − 6) = T (− 2 (1, 0, 3))

= −2T (1, 0, 3)

= −2 (1, 1)

= (−2,−2)

(2, 1) ≠ (−2,−2)

It is not a linear transformation.

8. Let T: R² → R³. Define T(a₁, a₂) = (a₁+a₂, 0, 2a₁−a₂). Verify whether T is one−to−one or on−to

Solution:

We know that if T: V→W is a linear transformation and N (T), R (T) are nullspace and range of T then,

(i) T is 1−1 if and only if N (T) = 0 and

(ii) T is on to iff R (T) = W.

Here N (T) = 0, So T is 1−1 and T is not on to because T never maps all the values of R³.

  T is 1−1 but not on−to.

9. Verify that T: R³ → R and T(u) = || u || is a linear transformation or not.

Solution:

T(u + v) = || u+v || ≤ || u || + || u ||

Here T (u+v) ≠ T (u) +T (v) for all u, v ∈ R³.

Hence it is not linear transformation.

10. Prove that similar matrices have the same characteristic polynomial.

Solution:

Let A and B be two similar matrices and p(k) denote the k^th degree polynomial.

Now we have to prove that p(A) and p(B) are similar matrices.

Since A and B are similar matrices, B = P^‒1AP for some matrix P.

Now let

 |B| = |P^‒1AP| = |P^‒1| |A| |P|

= |A| |P^‒1| | P |

= |A| |P^‒1P|

= |A| |I|

= |A|

|B| = |A|

Now

P(B) = | B−λI |

  (B ‒ λI) = P^‒1 AP − λI = P^‒1 AP − λ(P^‒1IP)

= P^‒1(A−λI)P

Since |B−λI| is same as A−λI

⇒ P(A) and P (B) are similar matrices.

 The similar matrices have the same characteristic polynomial.

11. Let be a square matrix. Prove that A and A^T have the same eigen values.

Solution:

To prove A and A^T have the same eigenvalues let λ be the eigenvalue of A.

Then its characteristic equation is |A−λI|.

 | A − λI | = | (A − λI)^T | = | (A^T – λI^T) |

  |A− λI| = |A^T − λI |

Since the characteristic equations of A and A^T are identical in any field, the eigenvalues are same.

Thus A and A^T have the same characteristic equations and have the same eigenvalues.

12. Prove that similar matrices have the same trace.

Solution:

Trace of [A]_n×n Sum of its diagonal elements.

Let A & B be two matrices of same order n.

Trace (A) = ⁿΣ_i=1 A_ii

Let A and B be two similar matrices.

  Trace (AB) = Trace (BA)

Then there exists an invertible matrix P such that B = PAP⁻¹.

Trace (B) = Trace (P A P^‒1) = Trace (P^‒1AP)

= Trace (P⁻¹ PA)

= Trace (IA)

= Trace (A)

  Trace (B) = Trace (A)

Hence, similar matrices have the same trace.

13. Find the eigenvalues of the matrix A =  ∈ M_2×2(R).

Solution:

Let A =  then the characteristic equation is

 f(t) = |A − λI| = 0

 = (1 – λ)2 ‒ 4 = 0

 1+ λ2−2λ−4 = 0

 λ2−2λ−3=0

(λ − 3) (λ + 1) = 0

The eigenvalues are 3,− 1.

14. Let T be the linear operator on P₂(R) defined by T[f(x)] = f(x) + (x + 1)ƒ'(x). Let B be the standard ordered basis for P₂(R). Compute characteristic polynomial and eigenvalue of T.

Solution:

Let us consider β = { 1, x, x² }

T(1)=1+0 (1+1)=1=1+0x+0x²

T(x)=x+1(1+x)=1+2x=1+2x+0x²

 T(x²) = x² + (x + 1)2x = 3x²+2x=0+2x+3x²

f(t) = |A−λ| = = − λ3 + 6λ2 − 11λ + 6

= − λ³ + 6λ² − 11λ + 6

= −(λ−1) (λ−2) (λ−3)

⇒ λ=1, 2, 3

The eigenvalues are 1, 2, 3.

15. Test the matrix A =  ∈ M_2×2(R) for diagonalizable.

Solution:

 f(t) =|A ‒ λI_n| = 

 = λ² − 2λ = λ(2−2)

λ=0, λ =2.

⇒ λ=0,2 the eigenvalues are distinct.

They are diagonalizable.

16. Test the matrix A =  ∈ M_2×2(R) for diagonalizable.

Solution:

Given that 

Then |A− λI| = (1−λ)²−16=1+λ²−2λ−16

= λ²−2λ−15 = (λ − 5) (λ+3)

(ie) λ = −3 and λ = 5.

 λ=−3, 5. Since the two eigenvalues are distinct, it is diagonalizable.

17. Find the algebraic multiplicity of all eigenvalues of

Solution:

⇒ (3 − t) [(3 − t) (4−t) − 0] − 1[00] + [0+0] = 0

⇒ (3−t)² (4−t) = 0

 f(t) = (3 − t)² (4 − t)

⇒ t = 3,3 and t=4

The algebraic multiplicity of λ=3 is 2

The algebraic multiplicity of λ=4 is 10