I read the article from wiki about triangular matrix, it says "A matrix which is conjugate to a triangular matrix is called triangularizable." I do not quite understand: isn't any triangularizable matrix still a triangular matrix, since the conjugate of any triangular matrix is still triangular?

This "conjugate" means "similar to" or something else, or it does not say anything non-trivial here?

Thanks.

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I don't think triangularizable matricies have to be triangular. For instance, diaginal matricies are triangular so any diagonalizable matrix is triangularizable. – user3180 Apr 22 '11 at 23:11
"Conjugate" means "similar to" in this context. en.wikipedia.org/wiki/Similar_matrix – Jonas Meyer Apr 22 '11 at 23:13
@Jonas, ah! This "conjugate" terminology means always "transposition"-related to me. It is quite confusing here. :) – Qiang Li Apr 22 '11 at 23:16
The link is really confusing! – Qiang Li Apr 22 '11 at 23:23

Here, conjugate is understood in the Algebraic sense: two matrices A and B are conjugates (also said to be similar to each other) if there exists an invertible matrix C s.t. $A = C^{-1} B C$.

Any triangular matrix is triangularizable, but not the other way around. Any square matrix will full rank is triangularizable, for example, though not necessarily triangular. This comes from theorems about diagonalizeable matrices (the concept of a matrix being triangularizable was used in the first proof of the spectral theorem I saw, but I don't remember how it was used anymore).

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I am confused by the sentence "Any square matrix will full rank is triangularizable, for example." Triangularizability depends on the field. E.g., $\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ has full rank but is not triangularizable in $M_2(\mathbb R)$. Every matrix, regardless of rank, is triangularizable in $M_n(\mathbb C)$. – Jonas Meyer Apr 23 '11 at 1:55