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Let A be a $4 \times 4$ matrix with eigen values -5,-2,1,4. Which of the following is an eigen value of  the matrix$\begin{bmatrix} A & I \\ I & A \end{bmatrix}$, where $I$ is the $4 \times 4$ identity matrix?

  1. $-5$
  2. $-7$
  3. $2$
  4. $1$
asked in Linear Algebra by (47 points)
edited by | 3k views
0
some one explain it in better way .
0

By Cayley Hamilton theorem:- Every square matrix satisfy it's own characteristic equation.

3 Answers

+34 votes
Best answer

Ans is (C) 2

$Ax = λx$, where $λ$ is the eigen value of $A$ . Hence $(A−λI)x = 0$ or $|A−λI| = 0$
So, for our given matrix, we have

$\begin{bmatrix} A-\lambda I &I \\ I& A - \lambda I \end{bmatrix} = 0$

This is a $ 2 \times 2$ block matrix where the first and last and the second and third elements are the same. So, applying the formula for determinant of a block matrix as given here

When A = D and B = C, the blocks are square matrices of the same order and the following formula holds (even if A and B do not commute)

$det \begin{pmatrix} A & B\\ C& D \end{pmatrix} = det (A-B) \ det(A+B)$

(second last case) https://en.wikipedia.org/wiki/Determinant#Block_matrices

we get,

$|A−\lambda I− I| \times |A-\lambda I + I|= 0 $

$\implies |A-(\lambda+1)I| \times |A - (\lambda -1)I| = 0$

Each of the eigen value of $A$ is the solution of the equation $|A- \alpha I |= 0$ ($\alpha$ being the eigen value of $A$). So, we can equate $\lambda +1$ and $\lambda -1$ to any of the eigen value of $A$, and that will get our value of $\lambda$. If we take $\alpha = 1$, we get $\lambda = 2$, and that is one of the choice. For no other choice, this equation holds. So, (c) 2 is the answer.

answered by Loyal (5.8k points)
edited by
0
{A−(λ+1)I}{A−(λ−1)I}=0

From here, how we get λ=−1   or  λ=1?
0
Yes, that's a mistake. In hurry I did it. Thanks.
0
I have edited the answer, i think this makes sense to me, is there anything i am missing?
0
Is there any easy way to solve this question ? can anyone tell me please ?
0
Please update the second line.-->"λ is the eigen value of A", rather it should be -"

λ is the eigen value of the matrix A I | I  A
+51 votes

We find the eigenvalues of given matrix by solving its characteristic equation :

2007-25
|(A – x I)2 – I2| = 0
|(A – (x – 1) I) * (A – (x + 1) I)| = 0
|(A – (x – 1) I)| * |(A – (x + 1) I)| = 0
So, |(A – (x – 1) I)| = 0 or |(A – (x + 1) I)| = 0

Let y be eigenvalues of A, then |(A – y I)| = 0 .

So, by comparing equations we get,
either x – 1 = y or x + 1 = y

Therefore , x = y + 1 or x = y – 1

y = -5, -4, 1, 4 (given)

So, x = −4, −1, 2, 5, −6, −3, 0, 3

 
Thus, option (C) is the answer.

answered by Active (4.7k points)
0
Is this method of expantion or soln of det is aplicable for all types of Block matrices or only for this?
0
Which of the following is an eigen value of  the matrix  [A  I,       where I is the 4×4 identity matrix?

                                                                                       I  A  ]

what does that mean ? inside this 4 * 4 matrix there is I which is again a 4 *4 matrix ?

might be very silly

but this line is confusing me
0
@warrior

NO , not for every matrix

Can be used only for matrices  where the first and last and the second and third elements are the same
0
A is 4x4.

I is 4X4.

And  the representation that is given would be of 8 x8.
0
@gari

why then a 2*2 matrix is taken and solved
0

@Ai$H

in the selected solution the second method is used

in this solution, first method is used 

see the text from wiki

0
the second methd is to be used
+2 votes
Characteristics equation of given matrix is:

$(A- \lambda)^2 - I^2 = 0$

$(A-\lambda + I)(A- \lambda -I)=0$

$\lambda = A+I, A-I$

Given eigen values of $A=-5,-2,1,4$

Then $A+I$ and $A-I$ eigen values of given matrix are: (according to property of Cayley Hamilton theorem):

$A+I=(-4,-1,2,5)$

$(A-I)=(-6,-3,0,3)$
answered by Boss (14.6k points)
Answer:

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