I can't seem to find an answer to this online, basically what I am asking is

$$\mathrm{Tr} ((4)_{1\times 1})=4?$$ or in general is $\mathrm{Tr} ((k)_{1\times 1})=k?$ Where $k$ is a constant.

I think the answer is yes, since a $1\times 1$ matrix is simply a number (or constant) so the sum of the diagonal elements is just that element.

But I would like to see some confirmation of this if possible.

  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Mar 17 at 18:05
  • $\begingroup$ The answer is indeed yes $\endgroup$ Mar 17 at 18:49

1 Answer 1


I think this is more appropriate to be a comment rather than answer. But it is too long for a comment.

If I have guessed correctly what you mean, I would suggest that you should write $$\mathrm{Tr} ((4)_{1\times 1})$$ which causes less confusion. As the answer for your question, the trace of a $1\times 1$ matrix $(a)$ is indeed $a$. What do you think is ambiguous? If we define the trace of a $1\times 1$ matrix $(a)$ as $a$, then the properties of traces for $n\times n$ $(n\ge 2)$ matrices still holds for $1\times 1$ matrix. For example $a$ is the sum of all of its eigenvalues (since $0=\det (\lambda (1)-(a))=\det (\lambda -a)\Longleftrightarrow \lambda=a$, i.e. $a$ is the unique eigenvalue of $(a)_{1\times 1}$.)


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .