how to solve the following equation to get lambda values from it?

modulus([2 -1; -1 2]-λ[1 0; 0 1])=0

please help me. thanks.

[Rafael Hernandez-Walls]

if is a problem of eigenvalue, then using:

A=[2 -1; -1 2];
eig(A)

[Venkata Rama Krishna Gona]

Thanks for the help.

Answer 1

[Walter Roberson]

modulus is a matrix operator; on a 2 x 2 input, it would give a 2 x 2 output. You ask that it be equal to 0: are you asking that it equal [0 0; 0 0]?

If so then because the modulus of a matrix A is X such that X*X = A'A and X would be [0 0; 0 0] then we can see that A'A would have to equal 0.

syms lambda
A = [2 -1; -1 2]-lambda * [1 0; 0 1]
B = simplify(A'*A)

B =
 
[ 1 - ((2*lambda - abs(lambda)^2)*(lambda - 2))/lambda,                                   2*real(lambda) - 4]
[                                   2*real(lambda) - 4, 1 - ((2*lambda - abs(lambda)^2)*(lambda - 2))/lambda]

B(2,1) and B(1,2) are 2*real(lambda)-4 and that has to equal 0, which requires that lambda = 2 + 1i*LI where LI is real. Substituting that in,

syms LI real
D = subs(B,lambda,2+1i*LI)

The diagonal becomes the same, 1 - (LI*(- LI^2 + LI*2i)*1i)/(2 + LI*1i) . Normalize it,

>> simplify(D(1)*(2+LI*1i)) 
ans =
LI^3*1i + 2*LI^2 + LI*1i + 2

and that has to equal 0, and by construction LI is real valued. Looking at the real part, 2*LI^2 + 2 = 0... which is only possible for imaginary LI, but by construction LI is real. LI^3 = -LI would also have to be true to take care of the imaginary components, but that is not possible for real-valued LI.

We have arrived at a contradiction, which can potentially be resolved in one of two ways:

1. That no such modulus exists; or
2. That the normalization (2+LI*1i) was invalid because (2+LI*1i) = 0. But for (2+LI*1i) = 0, LI would have to be imaginary, which again leads to a contradiction.

So, there is no solution.

[Venkata Rama Krishna Gona]

Thanks for the Analysis..helped a lot.