$U = \{A | A_{11} = A_{22}=A_{33}, A \in M_{3}(\Bbb{R})\}$ where $A_{11},A_{22},A_{33}$ are the diagonal elements of the matrix $A$.
I think $U$ will be a subspace of $M_{3 \times 3}(\Bbb{R})$, as any linear combination of matrices will still have the same property of equal diagonal elements.
Next I am thinking of subset of $M_{3 \times 2}(\Bbb{R})$
$V = \{A \in \Bbb{R}_{3 \times 2} | ||A||_{F}^2 \leq 1 \}$
where $||A||_{F} = \sum_{i=1}^{3} \sum_{j=1}^{2} A_{ij}^2$ is the Frobenius norm
I dont think it will form a subspace of $M_{3 \times 2}(\Bbb{R})$ as if we consider $cA$ for some real number $c$, then the frobenius norm becomes $c^2 ||A||_{F} \leq c^2$, and we can choose $c$ such that the new matrix's Frobenius norm squared need not be less than 1.
Any ideas?
Your reasoning for $U$ is correct, but the Frobenius norm is $||A||_{F} = (\sum_{i=1}^{3} \sum_{j=1}^{2} A_{ij}^2)^{1/2}$.
Then we have $||cA||_F=|c| ||A||_F$.
If $A \in V$ and $||A||_F=1$, then $||2A||_F^2=4>1$, hence $2A \notin V$.
$V$ is not a subspace of $M_{3 \times 2}(\Bbb{R})$.