matrix similarity upper triangular matrix How to show:
Any matrix A with real or complex entries is similar to an upper triangular matrix M whose diagonal entries are the eigenvalue of A.
Thank you!
 A: In $M_n(\mathbb R)$ this would be false.  There are matrices without real eigenvalues.
In $M_n(\mathbb C)$ this is true.  Hogben's Handbook of linear algebra contains an algorithm that produces a unitary matrix to conjugate a given matrix to a triangular matrix.
A: I'll assume you're working in the field of complex numbers, but I believe it holds for any algebraically closed field(?)
Let $(\lambda, v)$ be an eigenvalue-eigenvector pair of an $n$-by-$n$ complex matrix $A$. (This is possible because we're working in an algebraically closed field.) Find $u_2, \ldots, u_n$ such that $\{v, u_2, \ldots, u_n\}$ forms a basis of $\mathbb C^n$, i.e., the matrix
$$
B =
\begin{bmatrix}
| & | & \ldots & |\\
v & u_2 & \ldots & u_n \\
| & | & \ldots & |
\end{bmatrix}
$$
is non-singular, and so
$$
B^{-1}AB =
\begin{bmatrix}
\lambda & * & \ldots & * \\
0 & * & \ldots & * \\
\vdots & \vdots & \ddots & \vdots \\
0 & * & \ldots & *
\end{bmatrix}.
$$
Repeat the process with the bottom-right $(n-1)$-by-$(n-1)$ submatrix.
$B$ can even be made orthogonal. This is called the Schur decomposition.
