According to this question, strictly upper triangular matrices are nilpotent (over any field).

Does it also hold tha nilpotent matrices $N$ are strictly upper triangularizable? (i.e. they are conjugate that an strictly upper triangular matrix $N = JUJ^{-1}$ with $J$ and $U$ both complex).

In the complex case, it seems trivial to me that the Jordan canonical form has zeros on its diagonal and is thus strictly upper triangular.

However, I have not seen this characterization of complex nilpotent matrices so I might be missing something. This would be related to the Jordan decomposition of matrices/endomorphisms into a semisimple and nilpotent part.

For the complex case, semisimple is the same as diagonalizable, and, if my intuition is correct, the nilpotent part is the same as strictly upper triangularizable. This perfectly relates to the Jordan canonical form of complex matrices, which implies that all complex matrices are triangularizable.

Also, what about the real case? Does my statement about nilpotent matrices being strictly upper triangularizable still hold? The Jordan canonical form should be the same (with given signature), although they might not be conjugate via a real matrix, so I'm not sure whether it is still true.

For the real case, the semisimple part is conjugate to a block decomposition where you might also have some two-dimensional blocks which are non-scalar/not proportional to the identity (as per this question), so "kind of" diagonalizable. Does something similar hold for real nilpotent matrices - "kind of" strictly upper triangularizable?

Hope my questions were clear, thanks!

  • $\begingroup$ Over the reals as well? $\endgroup$ Commented Sep 29, 2023 at 16:39
  • $\begingroup$ Great, thanks! If you want to write an answer I can accept it. $\endgroup$ Commented Sep 29, 2023 at 16:42

3 Answers 3


This is true over any field as others have said, but it's potentially circular to cite the Jordan normal form theorem here, because some proofs of the Jordan normal form theorem reduce to the nilpotent case, e.g. this one from Terence Tao.

Actually it's straightforward to prove the desired statement directly. The following more general argument suffices. People often cite the Jordan normal form theorem in situations where this easier theorem would suffice.

Upper triangularization theorem: Let $T : V \to V$ be a linear transformation of a finite-dimensional vector space over a field $k$ such that there exists a polynomial $f(t) \in k[t]$ which splits over $k$ such that $f(T) = 0$. Then there exists a basis of $V$ in which $T$ is upper triangular with diagonal entries given by some of the roots of $f$ (with some multiplicities).

The desired result for nilpotent matrices follows by setting $f(t) = t^k$. We don't need anything about algebraic closures or the existence of eigenvalues in general, and we don't even need the concept of minimal or characteristic polynomial.

Proof. Let $v \in V$ be any nonzero vector and write $f(t) = \prod_{i=1}^k (t - \alpha_i)$ for some $\alpha_i \in k$. Then

$$f(T) v = \left( \prod_{i=1}^k (T - \alpha_i) \right) v = 0$$

so one of the vectors $v, (T - \alpha_1) v, (T - \alpha_1)(T - \alpha_2) v, \dots $ is the first one to equal zero. This produces a nonzero vector in $\text{ker}(T - \alpha_i)$ for some $i$ and hence an eigenvector of $T$. For ease of notation WLOG this eigenvector $v_1$ has eigenvalue $\alpha_1$.

We now induct on the dimension of $V$. The result is clearly true if $\dim V = 1$. Now consider the induced action of $T$ on $V/v_1$, which still satisfies the same polynomial $f$. By the induction hypothesis there is some basis of $V/v_1$ with respect to which $T$ is upper triangular with diagonal entries given by some of the roots of $f$; choose any lift of this basis to vectors $v_2, \dots v_n \in V$. The upper triangular condition is equivalent to the condition that

$$T \left( \text{span}(v_2, \dots v_i) \right) \subseteq \text{span}(v_2, \dots v_i) \bmod v_1$$

which gives $T \left( \text{span}(v_1, \dots v_i) \right) \subseteq \text{span}(v_1, \dots v_i)$. So with respect to this basis, $T$ is upper triangular with diagonal entries the diagonal entries of the induced map on $V/v_1$ together with $\alpha_1$, which is still a root of $f$. $\Box$

In other words, what this argument shows more abstractly is that a linear transformation is upper triangular with respect to some basis iff it fixes some complete flag iff it satisfies a polynomial which splits over the base field. Note that not requiring $f$ to be the minimal or characteristic polynomial of $T$ makes the induction a little more flexible.


Over any algebraically closed field, a nilpotent matrix is strictly upper triangularizable by Jordan normal form. However, if two matrices become similar over a field extension, then they were actually already similar over the original field (this follows from the existence of rational canonical form which does not change when you extend the field; see here). So if $A$ is a nilpotent matrix over a field $k$, it is similar to a strictly upper triangular matrix with only $0$s and $1$s as entries over the algebraic closure $\overline{k}$ by Jordan normal form, and hence also over $k$.

(More generally, this means that the Jordan normal form exists over a non-algebraically closed field for matrices whose eigenvalues are all in your field.)


This answer would fit better as a comment to Eric Wofsey but it does give a very elementary proof to the original question by showing that if two real matrices are conjugates over complex matrices, then they also are over real matrices. This is much weaker then showing the result over any field extension, but it's a lot simpler.

Indeed suppose that $N = PUP^{-1}$ with $P = R + iJ$ and $R,J$ both real matrices.

Consider $P_x = R + xJ$, since $NP=PU$, by separating real and imaginary parts you see that $NP_x=P_xU$ for any complex $x$, and by picking a real $x$ such that $P_x$ is invertible, we would have a real $P_x$ such that $N=P_x U P_x^{-1}$.

How do we know such an $x$ exists? This is the nicest part of the proof, consider the polynomial $Q(x)=det(P_x)$, we need to show that $Q$ is non zero over $\mathbb{R}$ (or equivalently over $\mathbb{C}$), but we simply know $Q(i)=det(P)\neq0$.


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