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

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Venkata Rama Krishna Gona
编辑： Walter Roberson 2020-8-5
modulus([2 -1; -1 2]-λ[1 0; 0 1])=0
please help me. thanks.

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回答（1 个）

Walter Roberson
Walter Roberson 2020-8-4
编辑：Walter Roberson 2020-8-5
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.

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