Matrix Decomposition: 2 Marks Important Questions with Answer

MATRIX DECOMPOSITION

2 Marks Important Questions with Answer

1. Find <x, y> if 

Solution:

 <x, y> = 1(4)+2(−5)+3(6) = 4−10+18 = 12

2. Find <x,y> if x = [20, −4, 30, 10] and y = [10, −5, −8, −6].

Solution:

 <x, y> = 20 (10) + (− 4) (− 5) + (30) (−8) + 10 (− 6)

 = 200+20−240−60

 <x, y> = −80

3. Let x = . Verify whether the given vectors are orthogonal or not.

Solution:

Let x = 

 <x, y> = 1 (1) + 1 (1) + 1 (− 2) = 1+1−2 = 0

 <y, z> = 1 (1) + 1 (− 1) + (− 2) (0) = 1−1+0 = 0

 <z, x> = 1 (1) + (1) (−1) + 1 (0) = 1−1+0 = 0

Since <x, y> = <y, z> = <z, x> = 0 is an orthogonal set because each vector is orthogonal to every other vector.

4. If x=  then find (i) ||x||₂ (ii) ||x||₁ (iii) ||x||_∞ and ||x||₅.

Solution:

 (i) ||x||₂ = √(x•x) = √(1.1+2.2 +3.3) = √14

 (ii) ||x||₁ = |1| + |2 | + |3| = 6

 (iii) ||x||_∞ =Max [ |1|, |2|, |3| ] = 3

 (iv) ||x||₅ = [ |1|⁵ + |2|⁵ + |3|⁵ ]^1/5 = (276)^1/5 = 3.077.

5. Prove that the eigenvalues of A^HA are non−negative.

Solution:

If λ is an eigenvalue of A^HA, then there must exist a non−zero eigenvector X associated with λ satisfying the equality A^HAX = λX.

 0 ≤ <AX, AX> = <A* AX, X> = <A^HAX, X> = <λX, X> = λ <X, X>.

Since X is an eigenvector, it is non−zero and <X, X> is positive. Dividing by <X, X>, it gives λ≥0.

6. Prove that the eigenvalues of a Hermitian matrix are real.

Solution:

Let λ denote an eigenvalue of a Hermitian matrix A and let X denote a corresponding eigenvector. Then under the Euclidean inner product,

  λ<X,X> = <λX,X> = <AX, X> = <X, A*X>

  = < X, A^HX > = < X, λ X > =  < X, X >.

Since X is an eigenvector, it is non−zero and <X, X> is also non−zero. Dividing by <X, X> then the above relation gives λ= which implies that λ is real.

7. Prove that the diagonal element of a positive definite matrix must be positive.

Solution:

If A has order n×n, define X to be an n−dimensional vector having one of its components, say the k^th equal to unity and all the other components equal to zero. For this vector 0< <AX, X> = <AX> .  = a_kk.

8. Prove that the eigenvalues of a positive definite matrix are positive.

Solution:

Let A be positive definite with eigenvalue λ and corresponding eigenvector X. Then for this X,

 0 < <AX, X> = <λX, X> = λ<X,X>

                          ……….(1)

Since X is an eigenvector, it is not zero and <X, X> is positive. Dividing (1) by <X, X> we obtain λ >0.

9. Show that the determinant of a positive matrix is positive.

Solution:

The determinant of a matrix is the product of its eigenvalues, and each eigenvalue of a positive definite matrix is positive.

10. Show that a positive definite matrix is invertible.

Solution:

The determinant of a positive definite matrix is positive and so non−zero and therefore, that matrix must have an inverse.