Matrix MCQ - 4 - Mathematics MCQ

Characteristic equation is given by 


According to Caylay Hamilton theorem, Characteristic equation is satisfied by A.
- A ³ + 2A² + A + 1 = 0 
multiplying the whole equation by A^-1
 A ² + 2A + I + A ^-1 = 0
A^-1 = A² - 2A - I

The sum o f eigen values = trace o f the matrix = 1 2 
The eigen values are given by |A — λI| = 0

As A is upper triangular. So, A⁴ + 3A — 2I also, and hence eigen values are diagonal elements. 
Diagonal elements of A⁴ are 1, 16, 81 
Diagonal elements of A⁴ + 3A — 2I are 1 + 3 — 2, 16 + 6 — 2, 81 + 9 — 2 ie. 2 , 20 , 88 .

A^T = A,B^T = B 
(AB)^T = B^TA^T
= BA
Hence, AB is not a symmetric matrix 
option (a)discard 
Now, if AB = BA 
(AB)^t = (BA)^t 
= A^tB^t
= AB
Hence, AB is symmetric
Here, only one choice is correct. So, option (c) true.

Let  is eigenvector corresponding to λ = i

Let  s eigenvector correspinding to λ = - 1

P raid to the power 102 is zero, means P is nilpotent matrix and nilpotent matrix raise to the power its order is always null matrix. Hence, (a) option is correct

If P is non singular, there exist P^-1 
P² = P
p ^-lp ² = p^-l p
P = I
So, P is identity matrix.
Alternate: As P² = P 
PP = P
(pp)^T = p^T 
p^Tp^T = p^T
( p^t)² = p^t
P^T is idempotent 
option (a) is discard 
P² = P
as matrix P satisfies characteristic polynomial, by Caylay Hamilton theorem 

option (b) discard
take  Then P² = P
option (c) discard.

eigenvalues are given by |P — λI| = 0 
λ = 0,0 not distinct 
=>P is not diagonalizable 

=> singular (also, |P| = 0, product of eigenvalues) 
eigenvector corresponding to λ = 0 is given by

X₁ + ix₂ = 0 after apply R₂ → R₂ — iR₁

P has only one eigenvector
P is diagonalizable if algebraic multiplicity and geometric multiplicity of eigenvalues are equal. λ = 0 has algebraic multiplicity 2 whereas its geometric multiplicity is 1.
Hence, P is not diagonalisable.

Eigen values of A are given by |A — λI| = 0


λ = 0 , 2 , 3 all distinct 
So, D is equal to 

Let N be a nilpotent matrix of order 4.
⇒ N⁴ = 0; N satisfied its characteristic equation 
i.e λ⁴ = 0
λ = 0 eigenvalue of N
Since, eigenvalue of N⁴ is λ⁴ where λ is eigenvalue of N. 
zero is the only eigenvalue as eigenvalue of N is all zero.

Since, all the eigenvalues are distinct, the matrix will be similar to diagonal matrix and hence diagonalizable. Since, none of the eigenvalue is zero. So, P is invertible.

Let other two eigenvalues are λ₁and λ₂. 
sum of eigenvalues = trace of matrix.

now, product of eigenvalues = detA

So, eigenvalues of A are 1,1 + i, 1 — i 
eigenvalues of A² = l², ( l + i )², ( l — i )² 
= 1 , 2i, —2i
now, (A² - 2I)X = A²X - 2IX 

eigenvalue of (A² — 2I) is λ² — 2 where A are eigenvalues of A.
So, eigenvalues are — 1,2i — 2, — 2i — 2

matrix is upper triangular. So, eigenvalues are 1,1. Let us find eigenvector corresponding to 1 given by 
[A — λI]x = 0
for λ = 1  
[A - I]x = 0

The characteristic polynomial of a S x 5 matrix will be in degree 5, and we know that complex root (eigenvalue) comes in pair.
Hence, at least one real eigenvalue because we cannot arrange 5 in pairs (2,2,1 (unpaired)).

Characteristic equation is given by 

according to Caylay Hamilton theorem, every matrix satisfies its characteristic equation. 
So, A³ - 2A² - 5A + 61 = 0 
Multiplying whole equation with A^-1 
A² - 2A - 51 + 6A^-1 = 0

If one from the distinct eigenvalues is zero. Then it will be singular and hence, not invertible but T is necessarily diagonalizable.

The eigenvalue of unitary matrix is of unit modulus and eigenvalue of Hermitian matrix is always real.

set n = 2
and let characteristic polynomial is x² + (sum of eigenvalues)x + product of eigenvalues.
In general, det A = product of eigenvalues = c₀

If we think T in matrix form whose rank is n — 2 means two rows are zeroes, that is, zero will come in diagonal element Hence, 0 must be an eigenvalue of T. 1 may or may not be eigenvalue.
Therefore, only (b) option is correct.

Eigenvalues of (I + A¹⁰⁰) is 1 + λ¹⁰⁰
i e . l + 1 , 1 + 1,0 + 1
i.e.2,2,1
So, determinant of I + A¹⁰⁰ is product of eigenvalues i. e. 4

According to Caylay Hamilton theorem, every matrix satisfies its characteristic equation.
So, M² - M — I = 0
multiplying by M^-1, M — I — M^-1 = 0
M - I = M^-1

M is Hermitian matrix and Hermitian matrix has always real eigen values.

Matrix is lower triangular matrix. So, eigenvalues are 1,1,2, and 2 
∴ minimal o r characteristic polynomial ( x — l )2( x — 2 ) ².

Let us find the eigenvalues of A 

Since, rank of M is 1. So, its determinant will be zero (because M has a zero row). So, one of the eigenvalue must be zero. If other eigenvalue is zero then it is nilpotent and if other eigenvalue is non zero then it will be diagonalizable.

M is upper triangular matrix. So, 1,4,9 are eigenvalue. Hence, diagonalizable.

eigenvalues of M² are distinct. So, diagonalizable.

Therefore   x₃ = - x₁  i.e.  x₁ and  x₃ are dependent on each other.

Let x₁ = 1 then x₃ = -1 

and x1 + x₂ - x₃ = 0 

⇒ x₂ = -2

∴ eigenspace corresponding to the eigen value λ₃ = 3   is (1,-2,-1) .

So, 4 is dimension of above.
Statement for linked Answer Question 
Let T :  be defined by T ( x₁, x₂, x₃) = (X₁ + x₂ + x₃, — x₁ — x₂, —X₁ — x₃) and M be its matrix with respect to the standard ordered basis.

Matrix multiplication exists only when column of first matrix is same as rows of second i.e b = c.

Properties of eigenvalues-

Eigenvalues of unitary and orthogonal matrices are of unit modulus |λ| = 1

• If λ1, λ2…….λn are the eigenvalues of A, then kλ1, kλ2…….kλn are eigenvalues of kA
• If λ1, λ2…….λn are the eigenvalues of A, then 1/λ1, 1/λ2…….1/λn are eigenvalues of A-1
• If λ1, λ2…….λn are the eigenvalues of A, then λ1k, λ2k…….λnk are eigenvalues of Ak

Eigenvalues of A = Eigen Values of AT (Transpose)

Sum of Eigen Values = Trace of A (Sum of diagonal elements of A)

Product of Eigen Values = |A|

Maximum number of distinct eigenvalues of A = Size of A

Calculation:

The given eigenvalues are 5 and −2.

Q = M3 - 4M2 - 2M

Eigen values of M³ are 125 and -8,4M²  are 100 and 16, 2M are 10 and -4.

SO eigen values of Q, λ1= 125-100-10 = 15,λ2 = -8-16-(-4)=-20

So the eigen values are 15 and -20