# Is the trace of a constant simply that constant?

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.

• 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.
– Community Bot
Mar 17 at 18:05
• The answer is indeed yes Mar 17 at 18:49

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 $$(a)_{1\times 1}$$.)