Uniqueness of identity matrix for one matrix

Question

I know that there is only one matrix $I$ such that for all matrices $M$, $M = I M = M I$. But in general, suppose I have a particular matrix $M$, and the matrix equation $$ M = M T $$ Can I conclude from here that $T = I$? I know that if $M$ is invertible we can conclude that, but what about in general?

Answer 1

Consider $M=0\in\mathbb{K}^{n\times n}$. Like you already conjectured, this only has to be true when $M\in GL_n(\mathbb{K})$

Answer 2

Suppose $M$ is not invertible. Let $K$ be its (nonzero) kernel, and $W$ a complement to $K$ in the underlying vector space $V$. Then a vector $v \in V$ can be written uniquely as $v = k + w$, for $k \in K$ and $w \in W$. Clearly $M v = M w$.

Define $T$ to be the identity on $W$, and an arbitrary linear transformation $S$ on $K$. (For instance $S = 0$ will do.)

Then $M T v = M T (k + w) = M (S k + w) = M S k + M w = M w$, as $S k \in K$.

So $M = M T$ for all these $T$.

Answer 3

If you think about the matrices as linear transformations you can easily see, how to get the answer.

If $M$ is invertible, every vector, that you multply on the right hand side, must be, as if only multiplied by $M$, so $T$ must be the identity.

If $M$ is not invertible, there are some things you can do. Since $M$ does not have a full rank, there is a non-trivial nullspace. As long as $T$ does only transform in this nullspace, there is nothing, $M$ can detect about the transformations made by $T$.