I was asked this question:
Let $U_1,U_2 \subseteq M_2(\mathbb{F})$, $\mathbb{F} = \mathbb{R}$ define such as:
$U_1 = \{ A \in M_2(\mathbb{F}) | AB=BA, B=\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \}$.
$U_2 = \{ \begin{pmatrix} 2a-3b+c & 2b+c-3a \\ 0 & 2a-3b+c \end{pmatrix} | a,b,c \in \mathbb{F} \}$
Prove that $U_1$ is a subspace of $M_2(\mathbb{F})$ and find a basis and a dimension for $U_1$. Also prove that $U_1 = U_2$ and check if over every field $U_1=U_2$ exists, if so, prove it, if not, then find a field where $U_1 \neq U_2$.
Here is my try:
- We'll prove that $U_1$ is a subspace of $M_2(\mathbb{F})$. We'll check if $U_1 = \emptyset$.
$A=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \in U_1$, and $AB=BA=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$, therefore $U_1 \neq \emptyset$.
Now, we'll check if its closed under multiplication and addition. Let $ \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \begin{pmatrix} e & f \\ g & h \end{pmatrix} \in U_1$, and let $\lambda_1, \lambda_2 \in \mathbb{R}$: $\lambda_1\left(\begin{matrix}a&b\\c&d\\\end{matrix}\right)+\lambda_2\left(\begin{matrix}e&f\\g&h\\\end{matrix}\right)=\left(\begin{matrix}\lambda_1a&\lambda_1b\\\lambda_1c&\lambda_1d\\\end{matrix}\right)+\left(\begin{matrix}\lambda_2e&\lambda_2f\\\lambda_2g&\lambda_2h\\\end{matrix}\right)=\left(\begin{matrix}\lambda_1a+\lambda_2e&\lambda_1b+\lambda_2f\\\lambda_1c+\lambda_2g&\lambda_1d+\lambda_2h\\\end{matrix}\right)$. (I'm not sure how to continue from here...)
- We'll find a basis for $U_1$. (Was I supposed to come up with an A such that it is a basis for $U_1$? rather than what I've done here?)
We first check if $U_1$ is linearly independent. We can perform gauss elimination on $B$ and get that B can be row-reduced to the identity $I_2$, and $I$ is a linear independent matrix, which means that $AI=IA=A$ therefore it is a a linear independent matrix.
Now we'll check if $U_1$ a spanning set of $\mathbb{F}$, since B is the identity $I_2$ and $I_2$ can be $\mathbb{R}^2$ by scalar multiplication, therefore $I=\mathbb{R}$.
We'll find the dimensions of $U_1$. Since we found in (2) that $I=\mathbb{R}$ then it follows that $dim(U_1) = dim(\mathbb{R})$.
We'll prove that $U_1=U_2$ over $\mathbb{R}$. Since $U_1$ is a basis which means its linear independent and $U_1=I$ we can do gauss elimination on $U_2$ and get to the identity matrix aswell: $\begin{pmatrix} 2a-3b+c & 2b+c-3a \\ 0 & 2a-3b+c \end{pmatrix} \Rightarrow \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ (Assuming non zero determinant), therefore $U_1=U_2$.
We need to check if $U_1=U_2$ when $\mathbb{F} = \mathbb{C}$. (Not sure how it any different than what I did in (4)...)
I got lost in this entire exercise, and got to a point where I'm guessing. I also looked at https://math.stackexchange.com/a/2699746/1166852 of trying to understand more of what's going on, and how to solve it. I'd love for some guidance on how to solve this. Thank you.
