Which of the following subsets of $\Bbb R^{3×3}$ are subspaces of $\Bbb R^{3×3}$?
A. The $3×3$ matrices whose entries are all integers
B. The $3×3$ matrices with determinant $0$
C. The $3×3$ matrices with trace $0$
D. The diagonal $3×3$ matrices
E. The $3×3$ matrices $A$ such that the vector $\left( \begin{matrix} 2 \\ 5 \\ 0 \end{matrix} \right)$ is in the kernel of $A$.
F. The invertible $3×3$ matrices
How would I prove these? I am pretty bad at proofs, and I am trying to practice. I mainly only do counterexample proofs. I know to see if something is a subspace of $\Bbb R^n$, you would need to perform these 3 checks:
A subspace of $\Bbb R^n$ is any set $H$ in $\Bbb R^n$ that has three properties:
a. The zero vector is in $H$.
b. For each $u$ and $v$ in $H$, the sum $u+v$ is in $H$.
c. For each $u$ in $H$ and each scalar $c$, the vector $cu$ is in $H$.
For part A, I assume you can simply do:
$$ 0.5 \begin{bmatrix} 1 && 1 && 1 \\ 1 && 1 && 1 \\ 1 && 1 && 1 \end{bmatrix} = \begin{bmatrix} 0.5 && 0.5 && 0.5 \\ 0.5 && 0.5 && 0.5 \\ 0.5 && 0.5 && 0.5 \end{bmatrix} $$ As I assume you can use any real number scalar.
For part B:
$$ \begin{bmatrix} 1 && 0 && 0 \\ 0 && 0 && 0 \\ 0 && 0 && 0 \end{bmatrix}
+ \begin{bmatrix} 0 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 1 \end{bmatrix}
= \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 1 \end{bmatrix} $$
I have no idea how to prove part C, D, and E as they're not counter example proofs.
Part D:
$$ \begin{bmatrix} a && 0 && 0 \\ 0 && b && 0 \\ 0 && 0 && c \end{bmatrix} + \begin{bmatrix} a' && 0 && 0 \\ 0 && b' && 0 \\ 0 && 0 && c' \end{bmatrix} = \begin{bmatrix} (a+a') && 0 && 0 \\ 0 && (b+b') && 0 \\ 0 && 0 && (c+c') \end{bmatrix} $$
$$ k \begin{bmatrix} a && 0 && 0 \\ 0 && b && 0 \\ 0 && 0 && c \end{bmatrix} = \begin{bmatrix} ka && 0 && 0 \\ 0 && kb && 0 \\ 0 && 0 && kc \end{bmatrix}, k \in \Bbb R, $$
Haven't attempted F.
Best Answer
For $A$, good job.
For $B$, good job.
For $C$, first note that $0$ is an element of this set. Then let $A$, $B$ $\in \Bbb R^{3\times 3}$ and $k\in\Bbb R$ such that $trace(A) = 0 = trace(B)$ . From the properties of trace we know that $trace(A+kB) = trace(A)+trace(kB) = trace(A)+k\cdot trace(B) = 0$. Thus $A+kB$ is also traceless. So $A,B$ in the set $\implies A+kB$ in the set for any $k\in \Bbb R$.
For $D$, you just need to note that the zero matrix is a diagonal matrix. Then good job.
For $E$, first note that $0$ is an element this set. Then let $A$, $B$ $\in \Bbb R^{3\times 3}$ and $k\in \Bbb R$ such that $null(A)\ni (2,5,0)^T \in null(B)$. Then $A(2,5,0)^T = 0 = B(2,5,0)^T$. Consider $$(A+kB)(2,5,0)^T = A(2,5,0)^T+kB(2,5,0)^T$$ due to the properties of matrix multiplication. But then $(A+kB)(2,5,0)^T=0$. So $A,B$ in the set $\implies A+kB$ in the set for any $k\in \Bbb R$.
For $F$, consider $I+(-I)$.