in Linear Algebra
4,029 views
3 votes
3 votes
Let the eigenvalues of 2 x 2 matrix A be 1, -2 with eigenvectors x1 and x2 respectively. Then the eigenvalues and eigenvectors of the matrix A^2 - 3A+4I would respectively, be

(a) 2,14; x1,x2

(b) 2,14; x1+x2:x1-x2

(c) 2,0; x1, x2

(d) 2,0; x1+x2,x1-x2
in Linear Algebra
4.0k views

2 Answers

9 votes
9 votes

Let  λ  be an eigenvalue of A and X be the eigenvector of A.

Then according to the definition of eigenvalues and eigenvector we get

AX =  λ X ........................(1)

Now, multiplying both sides of equation 1 with A we get

 A2X = A λ X =  λ AX =  λ  λ X =  λ2X ............... (2)

So, the matrix A2 has the same eigenvector as in A but the eigenvalue is square of the eigenvalue of A.

Now, multiplying both sides of equation 1 with -3 we get

-3 AX = -3  λX ..........................(3)

Now, adding equation (2) and (3) we get

(A2 - 3A)X =  ( λ2 - 3 λ )X ........................(4)

Now, we add 4$I$ X to the both sides of equation 4

(A2 - 3A) X + 4$I$ X = ( λ2 - 3 λ )X + 4$I$X 

=> (A2 - 3A + 4$I$)X = ( λ2 - 3 λ +4 )X  [ since $I$X = X]

So,  the eigenvalue of the matrix (A2 - 3A +4$I$) is (  λ2 - 3 λ + 4 ) and the eigenvector remains same as A.

Now, the value of  λ  is -2 and 1

So, putting those values in the expression of the eigenvalue we get the eigenvalues of the given matrix which are 2 and 14

So, Option A is the correct answer.

0 votes
0 votes
$\rightarrow$Matrix A

$\rightarrow$Eigen values 1 , -2

$\rightarrow$So, Matrix  A$^{2}$ - 3A + 4

                   $\Rightarrow$ 1-3(1)+4 and (-2)$^{2}$ -3(-2) +4

                   $\Rightarrow$ Eigen values are 2 and 14 respectively.

$\therefore$ A & P(A) = a$_{0}$ I + a$_{1}$ A + a$_{2}$A have same eigen vectors.

So, option A is the correct answer.

2 Comments

My doubt is:

Say 'e' is eigen value of A , then eigen value of A^2 is e^2. Also the eigen value of pA will be p*e.

But this rules doesn't apply on + and -  then how can we say that eigen value of matrix Matrix  A^2 - 3A + 4I will be

eigen of (A^2) - eigen of(3A) +4
0
0
edited by
If λ$_{i}$ is any eigenvalue of A and φ$_{i}$ is an eigenvector corresponding to λ$_{i}$ ,then for any polynomial p(λ) we have  p(A).φ$_{i}$ = p(λ$_{i}$).φ$_{i}$.

Better u should read the properties of eigen value and eigen vector, u will get it.
0
0