# Thread: finding matrix A

1. ## finding matrix A

Let $A \in M_{2}(\mathbb(R))$.
If $AB=BA$ for all $B \in M_{2} (\mathbb(R))$,
show that $A=\alpha I_{2}$ for any $\alpha \in \mathbb(R)$.

2. Originally Posted by deniselim17
Let $A \in M_{2}(\mathbb(R))$.
If $AB=BA$ for all $B \in M_{2} (\mathbb(R))$,
show that $A=\alpha I_{2}$ for any $\alpha \in \mathbb(R)$.
One way is to write $A=\left[\begin{matrix}a&b\\c&d\end{matrix}\right]$ then select $B=\left[\begin{matrix}1&1\\1&1\end{matrix}\right]$ which gives you a=d and b=c, then select $B=\left[\begin{matrix}1&0\\0&0\end{matrix}\right]$ which gives you c=b=0. (Find AB and BA and set corresponding cells equal.)

3. Originally Posted by undefined
One way is to write $A=\left[\begin{matrix}a&b\\c&d\end{matrix}\right]$ then select $B=\left[\begin{matrix}1&1\\1&1\end{matrix}\right]$ which gives you a=d and b=c, then select $B=\left[\begin{matrix}1&0\\0&0\end{matrix}\right]$ which gives you c=b=0. (Find AB and BA and set corresponding cells equal.)
Since the question asked "for all B", I let $B=\left [\begin{matrix} w & x \\ y & z \end{matrix} \right]$.
But I can't continue.

4. Originally Posted by deniselim17
Since the question asked "for all B", I let $B=\left [\begin{matrix} w & x \\ y & z \end{matrix} \right]$.
But I can't continue.
If it's true for all B, it's certainly true for the particular B's I mentioned in my first post. Your approach might work but it seems a bit of a mess. Note that you're only asked to prove implication in one direction. Also note that the problem statement is a little imprecise in language, I would write "for some alpha" rather than "for any alpha". (Clearly it's an existential rather than universal statement, because the universal statement makes no sense.)

5. You might find this useful.

how to make commutative 2X2 matrices