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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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  • $\begingroup$ Matrices with complex entries? $\endgroup$
    – Jonas Meyer
    Jan 19, 2013 at 7:06
  • $\begingroup$ @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 $\endgroup$
    – Salih Ucan
    Jan 19, 2013 at 7:12
  • $\begingroup$ @Yobo Yes it can. $\endgroup$
    – Tunococ
    Jan 19, 2013 at 7:12
  • $\begingroup$ @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. $\endgroup$
    – Jonas Meyer
    Jan 19, 2013 at 7:14
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    $\begingroup$ @Yobo: It does matter. Will you please edit your question to clarify what you are asking about (so that one doesn't have to read the comments to guess the context of what you are asking)? In what context did you encounter the problem, and what have you tried? $\endgroup$
    – Jonas Meyer
    Jan 19, 2013 at 7:19

2 Answers 2

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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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  • $\begingroup$ When you repeat the process how do you change B, by keeping the first row and column untouched?? $\endgroup$
    – Salih Ucan
    Jan 19, 2013 at 7:40
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    $\begingroup$ @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. $\endgroup$
    – Tunococ
    Jan 19, 2013 at 7:47
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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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    $\begingroup$ In $M_n(\mathbb{R})$, if all (complex) eigenvalues of a matrix are real, then will it be similar to an upper triangular matrix? $\endgroup$
    – Janson A.J
    Jun 26, 2020 at 4:47
  • $\begingroup$ @JansonA.J: I think yes but don't have a reference or proof in mind. If searching doesn't reveal an answer and you want, it could be asked as a new question. $\endgroup$
    – Jonas Meyer
    Jun 28, 2020 at 17:19

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