Eigenvalues of non-invertible triangular matrices

Question

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?

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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}

Answer 1

Theorem 1 states that the eigenvalues of a triangular matrix are the entries on the main diagonal. So the matrix $A$ has eigenvalues $\{3,0,2\}$.

The invertible matrix theorem states that a matrix is non-invertible if, and only if, $0$ is an eigenvalue. Since $A$ has $0$ as an eigenvalue, $A$ is not invertible by the invertible matrix theorem. There is no inconsistency here. An important consequence is any triangular matrix with a $0$ on the diagonal is not invertible.

A very important fact is that row reductions do not preserve eigenvalues! Just because $B$ and $C$ are row equivalent does not mean they share the same eigenvalues. So $C$'s eigenvalues are $\{1,0,0\}$ but you can't say anything about $B$'s eigenvalues. In fact, the eigenvalues of $B$ are $\{6,0,0\}$.