The question states:
Which are the following are subspaces of M (nxn):
And the main problem I'm having trouble understanding is with this question:
The set of all n x n matrices A for which Ax=0 has only the trivial solution.
How would I go about proving if this is a subspace?
If we know $V$ is a vector space a priori, then a space $W$ contained in $V$ is a subspace of $V$ if there is closure under scalar multiplicaton and addition.
What does it mean for a square matrix $A$ such that $A\vec{x}=\vec{0}$ to have only the trivial solution?
This means that $A$ is invertible.
Can you find two invertible matrices such that when you add them together you do not get an invertible matrix?
Yes! Let $A=\begin{pmatrix} 2&1\\ 1&1\\ \end{pmatrix}$ and $B=\begin{pmatrix} -2&-1\\ -1&-1\\ \end{pmatrix}$
Thus $A+B=\begin{pmatrix} 0&0\\ 0&0\\ \end{pmatrix}$ which is not an invertible matrix since $\det(A+B)=0$.
Since we have demonstrated that adding two matrices together in our supposed subspace yields a matrix not in our subspace, the set of all $n\times n$ matrices $A$ for which $A\vec{x}=\vec{0}$ has only the trivial solution is not a subspace of $M_{nn}$ matrices.