My textbook states
Theorem 1 The eigenvalues of a triangular matrix are the entries on its main diagonal.
After which it shows that the matrix
A=
[ 3 6 -8]
[ 0 0 6]
[ 0 0 2]
Has an eigenvalues of $\{3,0,2\}$
The book also states that a non-invertible matrix has an eigenvalue of $0$.
However matrix $A$ is non invertible due to the $0$ in its diagonal.
Does theorem 1 override the invertible matrix theorem?
Also the book states that the matrix $B$, which is equivalent to the matrix $C$ due to linear dependence. Has an eigenvalue of $0$ because it is not invertible.
All non zero elements in $C$ are above the diagonal, therefore it is triangular. So the eigenvalues of $C$ are $\{1,0,0\}$ because it is triangular?
B= C=
[ 1 2 3] [ 1 2 3]
[ 1 2 3] = [ 0 0 0]
[ 1 2 3] [ 0 0 0]
It seems that I interpreted the problem in the book incorrectly.
The question was to find one eiganvalue, I had understood it as finding the eiganvalue.
The eiganvalue 0 is part of the set of eiganvalues of matrices A, B, C, it is however not the only eigan value.
eA = {3,0,2}
eB = {6,0,0}
eC = {1,0,0}