edited by
16,522 views
79 votes
79 votes

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. $1$
edited by

5 Answers

Best answer
58 votes
58 votes

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.

edited by
114 votes
114 votes
Characteristics equation of given matrix is:

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

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

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

$ \implies \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):

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

$(A-I)=(-6,-3,0,3)$
edited by
83 votes
83 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.

8 votes
8 votes

 

So above given matrix has 8 eigen values and we know the eigen value of identity matrix is always 1. if identity matrix size 4x4 then four eigen value={1,1,1,1}.

Answer:

Related questions

33 votes
33 votes
4 answers
1
61 votes
61 votes
8 answers
3
Ishrat Jahan asked Oct 29, 2014
16,029 views
Let $A$ be the matrix $\begin{bmatrix}3 &1 \\ 1&2\end{bmatrix}$. What is the maximum value of $x^TAx$ where the maximum is taken over all $x$ that are the unit eigenvect...