matrix similarity upper triangular matrix [closed]

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!

Solution 1:

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.

Solution 2:

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.