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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!

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Matrices with complex entries? – Jonas Meyer Jan 19 '13 at 7:06
@JonasMeyer So can the statement be wrong or complex matrices? Is the statement certainly true for all real matrices? Why? Why does it not hold in complex case? Thank you – Salih Ucan Jan 19 '13 at 7:12
@Yobo Yes it can. – Tunococ Jan 19 '13 at 7:12
@Yobo: The statement is true if you are working with complex matrices. I'm trying to ask, what are you asking about? Are you asking about matrices with complex entries, or something else? Please clarify your question; if you want to know about, say, the real case too, you could mention that. You could also indicate what you have tried to do so far to answer your question. – Jonas Meyer Jan 19 '13 at 7:14
@JonasMeyer I am asking how to show it in matrices with complex entries? Why did you make a remark then whether it is complex or real if it does not matter? Thank you.. – Salih Ucan Jan 19 '13 at 7:17
up vote 4 down vote accepted

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.

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When you repeat the process how do you change B, by keeping the first row and column untouched?? – Salih Ucan Jan 19 '13 at 7:40
@Yobo Yes. So when you find the next $B$, call it $\tilde B$, you assume that $\tilde B$ has an identity in its upper-left corner, and work only with the lower-right submatrix. – Tunococ Jan 19 '13 at 7:47

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.

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