Bilinear Forms,Quadratic Forms MCQ Quiz in मराठी - Objective Question with Answer for Bilinear Forms,Quadratic Forms - मोफत PDF डाउनलोड करा

Given - Define a real valued function B on ℝ2 × ℝ2 as follows. If v = (x1, x2), w = (y1, y2) belong to ℝ2 define B(u, w) = x1y1 – x1y2 – x2y1 + 4x2y2. Let v0 = (1, 0) and let W = {v ∈ ℝ2 : B (v0, v) = 0}.

Explanation -

we have B(u, w) = x1y1 – x1y2 – x2y1 + 4x2y2

v0 = (1, 0) 

W = {v ∈ ℝ2 : B (v0, v) = 0}.

Let v = (x,y)

\(B (v_0, v) =x.1-y.1+0.x-4.0.y\)

\(B (v_0, v) =x-y\)

therefore \(B (v_0, v) =0\)

⇒ x - y = 0

​⇒ x = y

W = {(x,x) | x ∈ ℝ}.

Hence option (iv) is correct.

Given - Let B : ℝ × ℝ → ℝ be the function B(a,b) = ab. 

Explanation -  Here B : ℝ × ℝ → ℝ  and defined as B(a,b) = ab. 

Now B(α (a,b)) = B(αa, αb) = αa.αb = α²ab 

But α.B(a,b) = α.ab

So B is not a linear Transformation.

Hence option (1) is false.

Now for any \(a,a', \alpha_1,\alpha_2, \beta \) ∈ ℝ

\(B((a\alpha_1 +a' \alpha_2,\beta)=(a\alpha_1 +a' \alpha_2)\beta\)

\(= a\alpha_1\beta + a'\alpha_2\beta\)

\(= a B(\alpha_1, \beta)+a'B(\alpha_2,\beta)\)

Also for any \(\alpha, b, b', \beta_1,\beta_2 ∈ R\)

\(B(\alpha, b\beta_1+b'\beta_2)= \alpha(b\beta_1+b'\beta_2)\)

\(= \alpha b\beta_1+ \alpha b'\beta_2\)

\(= b\alpha \beta_1+b' \alpha \beta_2\)

\(= bB(\alpha,\beta_1)+b'B(\alpha, \beta_2)\)

So B is Bilinear form.

Now B(a,b) = ab = ba = B(b,a) ∀ a,b ∈ R

Hence B is symmetric bilinear form.

Also, B(a, a) = a² > 0 for all a ≠ 0

So, B is positive definite.

Thus option (2) is correct 

Explanation:

Options (1):

Q(x, y, z) = xy + z2

(-1, 1, 1) ∈ \(\mathbb{R}\)3 \{(0, 0, 0) and Q(-1, 1, 1) = -1 + 1 = 0

∴ Q(x, y, z) = xy + z2 over \(\mathbb{R}\) represent 0 non trivially

Options (2):

Q(x, y, z) = x2 + 3y2 - 2z2

(-1, 1, √2) ∈ \(\mathbb{R}\)3 \{(0, 0, 0) and Q(-1, 1, √ 2) = 1 + 3 - 2 × 2 = 0

∴ Q(x, y,  z) = x2 + 3y2 - 2z2 over \(\mathbb{R}\) represent 0 non trivially

Options (4):

Q(x, y, z) = x2 + xy + z2

(1, -2, 1) ∈ \(\mathbb{R}\)3 \{(0, 0, 0) and Q(1, -2, 1) = 1 - 2 + 1 = 0

∴ Q(x, y,  z) = x2 + xy + z2 over \(\mathbb{R}\) represent 0 non trivially

Options (3):

if possible such (a, b, c) ∈ \(\mathbb{R}\)3 \{(0, 0, 0) exist such that

Q(x, y, z) = x2 - xy + y2 + z2 = 0

⇒ (x-y)² + z² + xy = 0

⇒ (x-y)2 + z2> 0 so xy must less than 0

So if xy < 0 then Q(x, y, z) = x2 - xy + y2 + z2  > 0 for any (a, b, c) ∈ \(\mathbb{R}\)3 \{(0, 0, 0).

Q(x, y, z) = x2 - xy + y2 + z2 over \(\mathbb{R}\) does not represent 0 non-trivially.

Options (1), (2), (4) are correct.

Concept:

Let V be a vector space of dimension n over a field K. A map B : V × V → K  is a symmetric bilinear form on the space if

(i) B(u, v) = B(v, u) ∀ u, v ∈ V

(ii) B(u + v, w) = B(u, w) + B(v, w) ∀ u, v, w ∈ V

(iii) B(cu, v) = cB(u, v) ∀ u, v ∈ V, c ∈ K

It will be skew-symmetric bilinear if B(u, v) = - B(v, u) ∀ u, v ∈ V and is alternating if B(u, u) = 0 ∀ u ∈ V 

Explanation:

 f(A, B) = 2tr(AB) - tr(A)tr(B).

(i)  f(B, A) = 2tr(BA) - tr(B)tr(A) = 2tr(AB) - tr(A)tr(B) = f(A, B) (∵) tr(AB) = tr(BA)) ∀ A, B ∈ \(\mathbb M_{2× 2}\)

(ii) f(A + B, C) = 2tr((A + B)C) - tr(A + B)tr(C) 

                      = 2tr(AC) + 2tr(BC) - (tr(A) + tr(B))tr(C) (∵ tr(A + B) = tr(A) + tr(B))

                     = 2tr(AC) - tr(A)tr(C) + 2tr(BC) - tr(A)tr(C) = f(A, C) + f(B, C) ∀ A, B, C ∈ \(\mathbb M_{2× 2}\)

(iii) f(cA, B) =  2tr(cAB) - tr(cA)tr(B) = 2c tr(AB) - c tr(A)tr(B) = c f(A, B) (∵ tr(kA)=k tr(A)), ∀ A, B ∈ \(\mathbb M_{2× 2}\), c ∈ F

Hence f is symmetric bilinear form

(1) is true and (2)  is false.

f(A, A) = 2tr(A²) - tr(A)tr(A) = 2tr(A2) - [tr(A)]² which is not equal to zero always.

f is not alternating

Option (3) is false.

Explanation:

Given  

b(X, Y) = x1y1 + 2x1у2 - x2y1 + 3x2y2 

u1 = (1, 0) and u2 = (1, 1).

the matrix   B =( \(b_{ij}\)) and   \(b_{ij} = b(u_{i},u_j) \)  where  \(1 \leq i \leq2 \)   and   \(1 \leq j \leq2\) 

X= (x1, x2) and Y = (y1, y2)

Now,   \(b_{11}\) = b((1, 0); (1, 0)) = 1 + 0 - 0 + 0 = 1 

         \(b_{12}\) = b((1, 0); (1, 1)) = 1 + 2 - 0 + 0 = 3 

        \(b_{21}\) = b((1,1), (0,1)) = 0 + 2 - 0 + 3 = 5

        \(b_{22}\) = b((1,1), (1,1)) = 1 + 2 - 1 + 3 = 5 

Hence required matrix  is 

B = \(\left[\begin{array}{cc}1 & 3 \\ 5 & 5\end{array}\right]\)

Therefore the correct option is option (2)   

Concept:

A quadratic form Q = x^TAx is called positive definite if Q > 0 for all x ≠ 0 or if all the eigenvalues of A or positive or if A is symmetric and has positive leading principal minors

Explanation:

 Given quadratic form is Q(X1, X2, X3) = xt Ax where x = (X1, X2, X3)t. where A = (ai, j) be a real symmetric 3 × 3 matrix.

Let A = \(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end {bmatrix}\) 

Here ai, j = 0 for all i ≠ j but Q(x1, x2, x3) is positive definite as all the eigenvalues of A ire positive.

(1) is false

If ai, j > 0 for all i and A is symmetric then leading principal minors of A are positive,

Hence Q is positive definite.

(2) is true

Let A = \(\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end {bmatrix}\)

Here ai, j > 0 for all i ≠ j but Q(x1, x2, x3) is not positive definite.

(3) is false

For same example (4) is also false

Given - We have four statement in options.

Concept - 

Theorem -(i) - Two quadratic forms each in n variables are isomorphic over R if and only they have same rank and same index or their same rank and the same signature.

Theorem -(ii) - Two quadratic forms each in n variables are isomorphic over C if and only they have the same rank.

Explanation -

Option (i) is false.

Option (ii) is true.

Option (iii) is false.

For option (iv) -  Two quadratic forms are isomorphic if they become identical after a linear invertible change of variables. 

Hence the option (iv) is false.

Explanation:

Given data  

b(X; Y) = x1y1 - 2x1у2 + x2y1 + 3x2y2   ,   u1 = (0, 1) and u2 = (1, 1).

the matrix   B =( \(b_{ij}\)) and   \(b_{ij} = b(u_{i},u_j) \)  where  \(1 \leq i \leq2 \)   and   \(1 \leq j \leq2\) 

now   \(b_{11}=b((0,1);(0,1))=3\)

         \(b_{12}=b((0,1),(1,1))=4\) 

        \(b_{21}=b((1,1),(0,1))=2\) 

        \(b_{22}=((1,1),(1,1))=3\) 

now then our required matrix  is 

        B \(=\left[\begin{array}{cc}3 & 4 \\ 2 & 3\end{array}\right]\)

Therefore the correct option is option (4)   

Concept:

1. Two quadratic forms over R are isomorphic if they have same rank and some signature.

2. Two quadratic forms over ℂ are isomorphic if they have same rank.

Explanation:

(1) Option (1) is false using result 1, because with same rank they can have a different signature.

(2) Option (2) is true using result 2.

(3) Option (3) quadratic form in 3 variables (non-zero) can have rank 1, 2 or 3, only and being over ℂ, with same rank they are isomorphic. Thus there are 3 quadratic forms upto isomorphic over ℂ and (c) is true.

(4) Option (4) it is false as over ℂ with rank 2, there is only one quadratic form (result 2).

The correct options are (2) and (3).