Let $A$ be an $n \times n$ real matrix with the following property:

All the conjugates of $A$ have only zeros on the diagonal. Does $A=0$?

(By conjugates, I mean all the matrices similar to it, over $\mathbb{R}$, that is I require the conjugating matrix to be real).

Of course, if $A$ is diagonalizable, then clearly it must be zero.

The only idea I have is to use the Jordan form for real matrices, but after some thought I am not sure this is a good approach.

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Asaf Shachar
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  • What are the conjugates. The submatrices of some size? – AnyAD May 29 '18 at 11:55
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    @AnyAD: Conjugate in this context is another term for [similar](https://en.wikipedia.org/wiki/Matrix_similarity). – joriki May 29 '18 at 11:58
  • @AnyAD Yes, I meant similar (over $\mathbb{R}$). – Asaf Shachar May 29 '18 at 11:59
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    You can triangulate the matrix over the reals, which shows $A$ has only the zero eigenvalue. Therefore it is nilpotent. Take a nonzero nilpotent matrix; does it have the property? – egreg May 29 '18 at 12:13
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    @AnyAD You are right of course. However, the requirement here is much stronger than having zero trace. So far, I could not come up with a non-zero example. – Asaf Shachar May 29 '18 at 12:14
  • @egreg Surely not all of them. I checked that the nilpotent matrix $\begin{pmatrix} 0 & 1 \\\ 0 & 0 \end{pmatrix}$ is similar to a matrix with non-zero diagonal. – Asaf Shachar May 29 '18 at 12:16
  • And every nilpotent matrix of size $2 \times 2$ is similar to that one. So at least the property is true for $n=2$. – nicomezi May 29 '18 at 12:19

3 Answers3


I think it is true: Suppose $A$ is nonzero. Then we find nonzero $v,w\in \mathbb R^n$ with $Av=w$. If $w$ and $v$ are linearly dependent, extend $v$ to a basis, then $A$ written in that basis will have a nonzero entry in the diagonal. If they are independent, then so are $v$ and $v+w$. Extending $\{v,v+w\}$ to a basis will then yield a nonzero diagonal element in $A$ written in this basis.

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An alternative, coordinate-based answer: Conjugate with the mutually inverse matrices $S^\pm_{ij}=\delta_{ij}\pm\delta_{i\alpha}\delta_{j\beta}$ for arbitrary $\alpha\ne\beta$ (that is, the matrices where $\pm1$ is added to the unit matrix in some off-diagonal entry) and consider the $\alpha$-th diagonal element of the result:

$$ \left.\sum_{jk}\left(\delta_{ij}+\delta_{i\alpha}\delta_{j\beta}\right)A_{jk}\left(\delta_{kl}-\delta_{k\alpha}\delta_{l\beta}\right)\right|_{i=l=\alpha}=\left.\sum_{jk}\left(\delta_{ij}+\delta_{j\beta}\right)A_{jk}\delta_{kl}\right|_{i=l=\alpha}=A_{\alpha\alpha}+A_{\beta\alpha}\;. $$

Clearly $A_{\alpha\alpha}=0$ (from conjugating with the unit matrix), and since $\alpha\ne\beta$ were arbitrary, it follows that all entries of $A$ are zero.

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The field is unimportant. The given condition implies that $\operatorname{tr}(APDP^{-1}) = \operatorname{tr}(P^{-1}APD) = 0$ for every invertible matrix $P$ and every diagonal matrix $D$. Since the set of all diagonalisable matrices spans the whole matrix space (there is actually an even better result, namely, every square matrix is the sum of at most three diagonalisable matrices), the previous observation in turn implies that $\operatorname{tr}(AB)=0$ for every matrix $B$. Hence $A=0$.

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