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Where $\lambda_{1}, \lambda_{2}, \lambda_{3}$ are the eigenvalues to work out. Now, let's say you feel lucky and want to assume that all the eigenvalues are integer. Then, from equation $(3)$ you know the largest one could be only $3$ or $4$ in absolute value, in which case the second largest would have to be $\pm 2$ and then $\pm 1$.
This is a soft question without objective answers, but I can list some cases where finding the eigenvalues (or at least one eigenvalue) is easy: Singular matrices with obvious kernels Diagonal or triangular matrices
In fact, you can see both equations are essentially the same (the one below is the upper multiplied by two). So we have. −v1 − 2v2 = 0 − v 1 − 2 v 2 = 0. That leads to. v1 = −2v2 v 1 = − 2 v 2. And the vectors in the eigenspace for λ = 9 λ = 9 will be of the form. (−2v2 v2) (− 2 v 2 v 2)
the eigenvalues are real: our instruments tend to give real numbers are results :-) As a more concrete and super important example, we can take the explicit solution of the Schrodinger equation for the hydrogen atom. In that case, the eigenvalues of the energy operator are proportional to spherical harmonics:
1. For a 3 × 3 matrix, the coefficients of the characteristic polynomial are. 1, − tr(X), tr2(X) − tr(X2) 2, − det(X) which could be easier to compute. In many exercises, a solution can be found by means of the rational root theorem. In the case of three equal values on the main diagonal, you might as well have solved for λ − 1.
13. It is not a good idea to label your eigenvalues λ1 λ 1, λ2 λ 2, λ3 λ 3; there are not three eigenvalues, there are only two; namely λ1 = −2 λ 1 = − 2 and λ2 = 1 λ 2 = 1. Now for the eigenvalue λ1 λ 1, there are infinitely many eigenvectors. If you throw the zero vector into the set of all eigenvectors for λ1 λ 1, then you ...
$\begingroup$ @LGezelis The restriction that an eigenvector need not be 0 is not necessary with the way I defined the terms, and, I want $0$ to be an eigenvector, so I can define the eigenspace as the set of all eigenvectors and it will be a subspace.
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16. If you are looking at a single eigenvector v v only, with eigenvalue λ λ, then A A just acts as the scalar λ λ, and any reasonable expression in A A acts on v v as the same expression in λ λ. This works for expressions I − A I − A (really 1 − A 1 − A, so it acts as 1 − λ 1 − λ), its inverse (I − A)−1 (I − A) − 1 ...
It's expanded form is k (λ) = a λ + a − 1λ − 1 + ⋯ + a1λ + a0, noting that the degree of the polynomial is n because the determinant is a sum of products of n matrix entries (each entry having at most one λ). Observe that k (0) = det (A), so a0 = det (A). Notice that the only way to get the power λ as a product of n matrix entries ...