Consider the trace map $M_n (\mathbb{R}) \to \mathbb{R}$. What is its kernel?

Question

The map is the trace map. I.e, it takes any $n$ by $n$ matrix and associates to that matrix, a number of the form $\mathrm{Tr}(A) = \sum_{i=1}^n a_{ii}$, where $A \in M_n (\mathbb{R})$.

I need to find the kernel of this map, give a basis and its dimension (which is easy once I have the basis.

I also am asked if the trace is surjective (and to prove this), and to describe the elements in $M_n(\mathbb{R}) / Ker \,(\mathrm{Tr})$.

Thanks in advance for any help I recieve. I love this community :) I will probably post many questions in regards to linear algebra this semester.

EDIT. I know that for any real number, we can associate to it, a matrix whose trace is that number. (all we have to do is let the other elements on the diagonal be $0$). So I suppose the surjectivity isn't an issue here.

However, finding the kernel and basis for the kernel seems tough for me. There are an awful lot of matrices whose trace is precisely $0$.

Answer 1

The trace of a matrix $M$ is $0$ if and only if the sum of the elements on the (main) diagonal of $M$ is $0$. Since the dimension of all $n\times n$ matrices is $n^2$ and the dimension of its image $\mathbb R$ is $1$ (see below), we know that the dimension of the kernel of $\mathrm{tr}$ is $n^2-1$. (That follows from the fact that the dimension of the image plus the dimension of the kernel is the dimension of the domain.) We know that the numbers of $M$ that are not on the diagonal are not important for the trace. Let $M_{ij}$ be the matrix with only zeros apart from a $1$ at index $(i,j)$. We know that $\mathrm{tr}(M_{ij})=0$ iff $i\neq j$. Also, we have that $\mathrm{tr}(M_{ii}-M_{nn})=0$ (if $i\neq n$, otherwise $M=0$.). Now, we have found $n^2-1$ matrices with trace $0$ that form a basis of the kernel.

As your said yourself, it is not hard to prove that the image is indeed all of $\mathbb R$, its codomain, and thus $\mathrm{tr}$ is surjective. For example, you can take $M=\alpha M_{11}$. Then $\mathrm{tr}(M)=\mathrm{tr}(\alpha M_{11})=\alpha$. Thus, for any $\alpha\in \mathbb R$ we can find a matrix $M$ such that $\mathrm{tr}(M)=\alpha$.

Answer 2

The trace of a matrix $A\in\mathbb R^{n\times n}$ is nothing but the sum of its eigenvalues.

So Ker(Trace) is the set of matrices with vanishing sum of eigenvalues, and its dimension is $n^2-1$.

The quotient space ${\mathcal M}_n/$Ker(Trace) is a subspace of dimension $1$ isomorphic to $\mathbb R$.

The elements of this quotient space are classes of matrices. All the elements of one class have the same trace, and we can thus conveniently represent each trace by a suitable multiple of the unit matrix.

This class corresponding to matrices of traces $s$ is represented by the matrix $(s/n)I$.