Inner Product Spaces, Orthonormal Basis MCQ Quiz in தமிழ் - Objective Question with Answer for Inner Product Spaces, Orthonormal Basis - இலவச PDF ஐப் பதிவிறக்கவும்

For a, b ∈ ℝ, let

p(x, y) = a2x1y1 + abx2y1 + abx1y2 + b2x2y2,

x = (x1, x2), y = (y1, y2) ∈ ℝ2.

Explanation - 

p(x, y) = a2x1y1 + abx2y1 + abx1y2 + b2x2y2,

p(x, x) = a²x₁²+ abx2x1 + abx1x2 + b²x₂2

= (ax₁+bx₂)²   where x = (x₁,x₂)

Now if a, b ∈ ℝ, then for \(x_1=\frac{1}{a} \ \ and \ \ x_2=\frac{-1}{b}\)

\((a(\frac{1}{a})+b(\frac{-1}{b}))=1-1=0\)
So it can not be an inner product.

If a = 0 and \(b \neq 0\), then for \(x_1 =1 \ \ and \ \ x_2 = 0\)
we get p(x, x) = 0 so it can not be an inner
product.

If \(a \neq 0\) and b = 0, then but but \(x_1 =0 \ \ and \ \ x_2 = 1\)

we get p(x, x)=0 so it can not be an inner product.

If a=0, b=0, then for every \(x \neq 0\)

p(x, x)=0 which is not possible in an inner product,

so option (d) is correct option.

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 = \(u_2-\frac{}{}v_1\)

v₃ = \(u_3-\frac{}{}v_1\) - \(\frac{}{}v_2\)

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

Explanation:

Here \(\rm  ....(i)

Given basis {1, t, t2, t3}

Let u₁ = 1, u2 = t, u3 = t2, u4 = t3

Result: \(\int_{-a}^af(x)dx = \begin{cases}2\int_0^af(x)dx, \text{if f(x) is even}\\0, \text{if f(x) is odd}\end{cases}\)......(ii)

Using Gram-Schmidt Method

v1 = 1

v₂ = t - \(\frac{}{<1, 1>}1\)

= \(\int_{-1}^1td t\) = 0 (using (i) and (ii))

So v2 = t - 0 = t

v3 = \(t^2\) - \(\frac{}{<1, 1>}1\) - \(\frac{}{}t\)

Now, 2, 1> = \(\int_{-1}^1t^2d t=2\int_0^1t^2dt=\frac23\)

<1, 1> = \(\int_{-1}^1d t=2\int_0^1dt=2\)

2, t> = \(\int_{-1}^1t^3d t\) = 0

So v3 = \(t^2\) - \(\frac13\) which can be written as v3 = 3\(t^2\) - 1

v4 = \(t^3\) - \(\frac{}{<1, 1>}1\) - \(\frac{}{}t\) - \(\frac{}{<(3t^2-1), (3t^2-1)>}(3t^2-1)\)

Now, 3, 1> = \(\int_{-1}^1t^3d t\) = 0

<1, 1> = \(\int_{-1}^1d t=2\int_0^1dt=2\)

3, t> = \(\int_{-1}^1t^4d t=2\int_0^1t^4dt=\frac25\)

= \(\int_{-1}^1t^2d t=2\int_0^1t^2dt=\frac23\)

Using same method \(\) = 0

So v4 = \(t^3\) - \(\frac35\) which can be written as v3 = 5\(t^3\) - 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  \(v_1, . . ., v_k \) are in this orthogonal set, and there are constants \(c_1, . . ., c_k\) such that \( c_1 v_1 + · · · + c_k v_k = 0. \)For any j between 1 and k, take the dot product of \(v_j\) with both sides of this equation. We obtain \(c_j ||v_j ||^2 = 0,\) and since \(v_j\) is not(otherwise the set could not be orthogonal), this forces \(c_j = 0.\) 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, 

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

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

\( = \alpha^2 < f, \alpha g>\)

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

If we choose f(t) = \(\frac{1}{2}\) we get,

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

 \(= \int_0^1 (f(t))^3 dt = \int_0^1 (\frac{-1}{2})^3 dt\) 

\(\frac{-1}{8} < 0\)

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 = \(\rm x_1^2+x_2^2+4x_1x_2\) Take, (x₁, x₂) = (-1, 1) ⇒ 1 + 1 - 4 = -3 \(\ngtr\) 0

∴ It is not an inner product on R²

Similarly, (3), <4, 4> = \(\rm x_1^2+2x_1x_2+x_2^2\) and for (x₁, x₂) = (-1, 1), <4, 4> \(\ngtr\) 0

option (1) & (3) discorded.

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

= <(ax₁ + 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 : <(x₁, x₂), (y₁, y₂)> = x₁y₁ + x₂y₂ + x₂y₁ + 2x₂y₂

= y₁x₁ + y₁x₂ + x₁y₂ + 2x₂y₂

= <(y₁, y₂), (x₁, x₂)>

Positive Definite: <(x₁, x₂), (x₁, x₂)> = \(\rm x_1^2+x_1x_2+x_2x_1+2x_2^2\)

= \(\rm x_1^2+2x_2^2+2x_1x_2=(x_1+x_2)^2+x_2^2>0\)

And (x₁, \(\rm x_2^2\)) + \(\rm x_2^2\) = 0 ⇒ (x₁, x₂) = (0, 0) (∴ sum of two positive term can't be zero)

∴ \(\left<\left|\begin{matrix}x_1\\\ x_2\end{matrix}\right),\left(\begin{matrix}y_1\\\ y_2 \end{matrix}\right)\right>=x_1y_1+x_1y_2+x_2y_1+2x_2y_2\)

is an inner product on ℝ²

Similarly, \(\left<\left|\begin{matrix}x_1\\\ x_2\end{matrix}\right),\left(\begin{matrix}y_1\\\ y_2 \end{matrix}\right)\right>=x_1y_1-\frac{1}{2}x_1y_2-\frac{1}{2}x_2y_1+x_2y_2\)

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: < v, w > = v₁w₁ + · · · + v_nw_n.
• Define the length or norm of v by the formula || v || = √< v, v > =√ { v₁² 1 + · · · + v_n² } .

We have the following properties for the inner product:

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

2. (Symmetry) For all v, w ∈ Rⁿ , < v, w > = < w, v >.

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

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.

Concept:

|e^iθ|² = 1

Explanation:

F(x, y) = sup ||eiθx + e-ϕ ||2 where ∀ θ, ϕ ∈ ℝ

            = sup{|eiθ|2 ||x||² + |e-ϕ|2 ||y||² + 2e^iθ-iϕ Re }

            =  sup{||x||2 + ||y||2 + 2ei(θ-ϕ) Re }

            ≤ ||x||2 + ||y||2 + 2 |< x, y >|

(1), (4) are correct