Linear Algebra 17d: Easy Eigenvalues - Nontrivial Null Space

[satyanathbhat]

Dear Professor,

One additional idea. Since we know in the particular example discussed, the column space is just linear combination of (1, 2, 3)^T. In particular, if we apply the transformation to (1, 2, 3) we should get some scalar multiplication of (1, 2, 3) itself. So (1, 2, 3) is a eigen vector and then we can use the trace trick to get the eigen value as 35.

P.S: Really enjoy your lectures :)

[MerrillHutchison]

Just to check, when A is singular, then N(A) = E_0 = N(A-λI), or in other words,   the nullspace of A is also the nullspace of characteristic equation when λ=0. This happens because N(A--λI) becomes N(A).

E_0 is the nullspace of A is the eigenspace created by the eigenvectors when the eigenvalue = 0.

[trickyabb]

how did we get the eigen vectors? here 2, -1, 0 anD 10, 0, -1, I understood the part the columns are factors c2 = 2*C1 and C3 = 10*C1, therefore Eigen Value = 0, but how we calculated the Eigen Vector in this Case?

[debendragurung]

following this lecture, an attempt to solve eigenvalue for the diapad matrix [1 2 3 ; 4 5 6 ; 7 8 9], Since its Nullspace is (1 -2 1) which is thus one of its eigenvector corresponding to 0 eigenvalue. using characteristic polynomial also eigen values are 0 and (15±√297)∕2. we know the eigenvector corresponding to 0. the other two impossible to construct. Its a terrible construct....