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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 (61 points) | 1.4k views
some one explain it in better way .

2 Answers

+25 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

(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 Boss (6.3k points)
edited by
{A−(λ+1)I}{A−(λ−1)I}=0

From here, how we get λ=−1   or  λ=1?
Yes, that's a mistake. In hurry I did it. Thanks.
I have edited the answer, i think this makes sense to me, is there anything i am missing?
Is there any easy way to solve this question ? can anyone tell me please ?
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
+15 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 Boss (8.2k points)
Is this method of expantion or soln of det is aplicable for all types of Block matrices or only for this?
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
@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
A is 4x4.

I is 4X4.

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

why then a 2*2 matrix is taken and solved

@Ai$H

in the selected solution the second method is used

in this solution, first method is used 

see the text from wiki

 

the second methd is to be used


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