Let n>=2. Which of the conditions defining a subspace are satisfied for the following subsets of the vector space Mnxn(R) of real (nxn)-matrices? (Proofs of counterexamples needed).
U={A is an element of Mnxn(R) such that rank(A)<=1}
V={A is an element of Mnxn(R) such that det(A)=0}
W={A is an element of Mnxn(R) such that trace(A)=0}
For this I started by stating the 3 conditions for a subspace V to hold:
i. the subspace is non-empty
ii. for all u and v in V, u+v is also in V
iii. for all u in V and t in R, tu is also in V
For U
For i I acknowledged that if rank(A)<=1, then it is linearly dependant and is non empty as there is an c1x1+c2x2+...+cnxn=d in the matrix.
For ii, I let u=(c1x1,c2x2,...,cnxn) and this is where I started to get stuck. As rank(A)<=1, can I take v=(0x1,0x2,...,0xn) so u+v=((c1+0)x1,...,((c1+0)xn))? I'd assume this is just the same as u so this property doesn't hold.
For iii, as it is linearly dependant there exists rows that are (0x1,0x2,...,0xn) so that multiplication by the scalar holds, but would it still apply that tu=u which is an element of V?
For V
On this question I acknowledged that if det(A)=0, then the matrix is linearly dependant and also has a row of zero vectors (at least).
For i can you state that the matrix Mnxn can contain any element and still be non empoty? i.e. isn't this just trivial?
For ii, can you do something similar to U part ii where you just take your row of zero vectors and add them on?
For both $U$ and $V$, consider
$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \\ \end{bmatrix} +\begin{bmatrix} 0 & 0 \\ 0 & 1 \\ \end{bmatrix}$$
For $W$, consider what adding and scalar multiplication does to the trace. You will find that this is indeed a subspace. So all three conditions are satisfied for $W$.
As for condition 1), the n-by-n zero matrix is a member of $U$, $V$, and $W$, so that is satisfied for all three.
As for condition 2), the counterexample I gave shows that is not satisfied in $U$ or $V$. It is satisfied for $W$. I gave you a hint on proving that.
As for condition 3), that is satisfied for all of $U$, $V$, and $W$.
For $U$, multiplying a matrix by a non-zero scalar does not change its rank. You can show that any linear combination of rows in a matrix such as $au+bv$ is equal to a linear combination in the multiplied matrix: $\frac at(tu)+\frac bt(tv)$. Multiplying a matrix by the zero scalar reduces its rank to zero, which means it remains in $U$.
For $V$, use the theorem that $\operatorname{det}(tA)=t^n\cdot \operatorname{det}(A)$. So if $\operatorname{det}(A)=0$, we also have $\operatorname{det}(tA)=t^n\cdot \operatorname{det}(A)=t^n\cdot 0=0$.
I gave you a hint in proving condition 3) for $W$.