Inner Product Spaces, Orthonormal Basis MCQ Quiz in বাংলা - Objective Question with Answer for Inner Product Spaces, Orthonormal Basis - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:

Gram-Schmidt Method: Given an arbitrary basis {u₁, u₂, …, u_n} for an n-dimensional inner product space V, the Gram-Schmidt algorithm constructs an orthogonal basis {v₁, v₂,…,v_n} for V such that

v1 = u1, 
v2 = }{}v_1\)

v₃ = }{}v_1\) - }{}v_2\)

v₄ = u₄ - }{}v_1\) - }{}v_2\) - }{}v_3\)

Explanation:

Here  ....(i)

Given basis {1, t, t2, t3}

Let u₁ = 1, u2 = t, u3 = t2, u4 = t3

Result: ......(ii)

Using Gram-Schmidt Method

v1 = 1

v₂ = t - }{}1\)

=  = 0 (using (i) and (ii))

So v2 = t - 0 = t

v3 =  - }{}1\) - }{}t\)

Now, 2, 1> = 

= 

2, t> =  = 0

So v3 =  -  which can be written as v3 = 3 - 1

v4 =  - }{}1\) - }{}t\) - }{}(3t^2-1)\)

Now, 3, 1> =  = 0

= 

3, t> = 

= 

Using same method \) = 0

So v4 =  -  which can be written as v3 = 5 - 3

 Hence {1, t, 3t2 - 1, 5t3 - 3t} is orthogonal basis

According To the Official Answer key 

No option provided is correct 

We will Update the Question and Solution 

Once the Result will comes out.

Given - Any orthogonal set of non-zero vectors in an inner product space V is ?

Concept - Any orthogonal set of non-zero vectors in an inner product space V is Linearly independent.

Explanation - 

Using the Concept, we directly get the result.

Hence the option (ii) is correct.

For example - Orthogonal sets are automatically linearly independent.

To see this result, suppose that   are in this orthogonal set, and there are constants  such that For any j between 1 and k, take the dot product of  with both sides of this equation. We obtain  and since  is not(otherwise the set could not be orthogonal), this forces  Thus the only linear combinations of vectors in the set which equal the 0 vector are those in which all of the coefficients are zero, which means that the set is linearly independent.

Explanation:

We have, ⟨f, g⟩ = \(\int_0^1\) f(t) (g(t))2dt

for all f, g ∈ C[0, 1] then, 

= \int_0^1 f(t) (\alpha g(t))^2 dt\)

= \alpha^2 \int_0^1 f(t) ( g(t))^2 dt \)

= \alpha^2 \)

It is not a bilinear form so it is not an inner product.

If we choose f(t) =  we get,

= \int_0^1 f(t)(f(t))^2 dt \)

Therefore, options 1, 2 and 4 are wrong.

The correct answer is option (3). 

Explanation:

Recall: The inner product on Rⁿ satisfies the following properties (u, v, ω ∈ [Rⁿ, a, b ∈ R

(1) Linearity : = a + b ;

(2) Symmetric property : =

(3) Positive Definite :  ≥ 0 and = 0 Iff u = 0

(1) Let u = (x₁, x₂) then =  Take, (x₁, x₂) = (-1, 1) ⇒ 1 + 1 - 4 = -3  0

∴ It is not an inner product on R²

Similarly, (3), =  and for (x₁, x₂) = (-1, 1),   0

option (1) & (3) discorded.

(2) Linearity Property : Take u = (x₁, x₂), v = (y₁, y₂), then = 1, x2) + b(y1, y2), (z1, z2)>

= 1 + by₁, ax₂ + by₂), (z₁, z₂)>

= (ax₁ + by₁) z₁ + (ax1 + by1) z₂ + (ax₂ + by₂) z₁ +  2(ax2 + by2) z₂ 

= 1, x₂)(bz₁, z₂)> + 1, y₂), (z₁, z₂)>

= +

Symmetric : 1, x₂), (y₁, y₂)> = x₁y₁ + x₂y₂ + x₂y₁ + 2x₂y₂

= y₁x₁ + y₁x₂ + x₁y₂ + 2x₂y₂

= 1, y₂), (x₁, x₂)>

Positive Definite: 1, x₂), (x₁, x₂)> = 

= 0\)

And (x₁, ) +  = 0 ⇒ (x₁, x₂) = (0, 0) (∴ sum of two positive term can't be zero)

∴ =x_1y_1+x_1y_2+x_2y_1+2x_2y_2\)

is an inner product on ℝ²

Similarly, =x_1y_1-\frac{1}{2}x_1y_2-\frac{1}{2}x_2y_1+x_2y_2\)

is an inner product on ℝ²

option (2), (4) are true

Concept:

• Let v = (v₁, . . . , v_n) and w = (w₁, . . . , w_n) ∈ Rⁿ .
• We define the inner product (or dot product or scalar product) of v and w by the following formula:  = v₁w₁ + · · · + v_nw_n.
• Define the length or norm of v by the formula || v || = √ =√ { v₁² 1 + · · · + v_n² } .

We have the following properties for the inner product:

1. (Bilinearity) For all v, u, w ∈ R ,  = + and  = + . For all v, w ∈ Rn and t ∈ R, = = t.

2. (Symmetry) For all v, w ∈ Rⁿ , = .

3. (Positive definiteness) For all v ∈ Rⁿ , = || v ||² ≥ 0, and   = 0  if and only if v = 0.

Inner product spaces over the field of complex numbers are sometimes referred to as unitary spaces.

​Hence, the correct answer is option 4)

Explanation:

We have a set  of unit vectors.

For each pair of distinct vectors  and  (with i  j ), the inner product  is a non-positive integer,

which means it could be 0 or any negative integer.

We are asked to maximize N() , defined as the number of pairs (r, s) with  such that  .

Since there are 5 vectors, the total number of unique pairs (r, s) with  is



To maximize N() , we want as many pairs as possible to have . 

1. Non-positive Integer Condition: Each inner product  (for i  j ) is a non-positive integer, so it can be 0 or a negative integer.

2. Unit Vector Condition: Each vector  is a unit vector, meaning  = 1 .

Since we are maximizing the count of non-zero inner products, we want as many pairs as possible

to have a negative inner product, while satisfying the given conditions.

as for atleast one choice of i and j,  =0

Hence required maximum value of  N() = 9.

Hence option 1) is correct.

Concept:

A square matrix A is called an orthogonal matrix if AT = A-1 

Explanation:

(v1, v2, ..., vn) be n column vectors forming an orthonormal basis of ℝn.

A is the n × n matrix formed by the column vectors v1, ... vn. 

i.e., A is formed by the orthonormal vectors.

Hence A is orthogonal.

AT = A-1

Option (3) is true

Concept -

(i) We know that the formula for the orthogonal complement -

dim(M) + dim(M^T) = dim (ℝ4 )

(ii) dim (M) = dim (ℝ4 ) - number of restriction

Explanation -

we have 𝑀 = {(𝑥1, 𝑥2, 𝑥3, 𝑥4 ) ∈ ℝ4 ∶ 𝑥1 = 𝑥3 } 

now use the formula -

dim (M) = dim (ℝ4 ) - number of restriction

dim(M) = 4 -1 = 3

Now use the another formula -

dim(M) + dim(MT) = dim (ℝ4 )

⇒ dim(MT) = 4 - 3 = 1

Hence the correct answer is 1.