Quick question about eigenvalues of $3\times3$ matrix

Question

I'm looking over solutions from past exams. For this problem, the solution states “By inspection, we see that $2$ is an eigenvalue”. Given arbitrary $3\times3$ matrix below$$\begin{bmatrix}2&a&b \\0&c&d\\0&e&f\end{bmatrix}$$

How is it obvious that $2$ is an eigenvalue? other constants are arbitrary so I can't assume it's an upper triangular matrix.

Answer 1

It is obvious since, if $M$ is your matrix, then$$M.(1,0,0)=(2,0,0)=2(1,0,0).$$

Answer 2

Consider $det (A-\lambda I)$ and expand the determinant through first column.

Answer 3

This is a block upper triangular matrix of the form of $$M=\begin{bmatrix} A & B \\ O & D\end{bmatrix}$$, the eigenvalues of $M$ are eigenvalues of $A$ and $D$ since

$$det(M-\lambda I)=det(A-\lambda I_1) \det(D-\lambda I_2)$$

where $I_2$ and $I_2$ are identity matrices of appropriate sizes.

Answer 4

It is obvious since, if you subtract $2I$, the matrix has a zero column. Thus if $M$ is your matrix, $$ \det(M-2I)=0 $$

Answer 5

The columns of the matrix are the images of the basis vectors. The first column is a scalar multiple of $(1,0,0)^T$, therefore...