Solution
David answered on
Dec 29 2021
Ans1: True
Its given that T=T*, and T is defined for real nos, trace(T²)=0
So above conditions will be satisfied only if T will be symmetric matrix.
Let’s take an ex. g a b
T = a e c
b c f
Trace (T²) = a²+b²+g²+a²+e²+c²+b²+c²+f²=0
since a, b, c, g, e, f all are real, their squares are also positive, so summation of all these positive
numbers will be zero only if they all have value 0.
This implies all the elements of matrix T =0.
Ans2: False
Here it is given trace (T) =0
In above example
Trace (T) = g + e + f =0
So g, e, f can take any values satisfying above condition, it is not necessary that they must be zero here,
so T=0 is not the only condition.
Ans3: True
Columns of matrix A are linearly independent if and only if AX=0 has only trivial solution,
Means det (A) ≠ 0
For columns of A² to be independent det (A²) ≠ 0
Det (A²) = det (A)*det (A)
So multiplication of two non-zero terms is also nonzero
This implies det (A²) ≠ 0, and it also has only trivial solution.
Ans4: True
Columns of matrix A2 are linearly independent if and only if AX=0 has only trivial solution,
For columns of A to be independent det (A) ≠ 0
Det (A²) = det (A)*det (A)
Means det (A2) ≠ 0
So multiplication of two non-zero terms is also nonzero
This implies det (A) ≠ 0, and it also has only trivial solution.
Ans5: True
(v₁, v₂……) are Eigen vectors of T, and λ₁, λ₂……. are eigenvalues of T
(v₁, v₂, …….vᵣ) are independent in V
a₁ v₁+a₂ v₂+………aᵣ vᵣ=0
Apply T to both sides, we get,
a₁ T v₁ + a₂ T v₂ +………aᵣ T vᵣ=0
a₁ λ ₁v₁+a₂ λ₂ v₂+……….aᵣ λᵣ vᵣ=0
v₁, v₂…….and λ₁, λ₂……..all are non zero, so a₁, a₂…….are all zero……this means {T (v₁), T (v₂)……T (vᵣ)}
is linearly independent in W.
Ans6: True
Same reason as above.
Ans7: True
Pick an a
itrary v ε V . Since the vectors v1; v2; : : : ; vk span V, there exist scalars c1;...