Image of $\phi: \mathbb{Q}^{2\times 2} \rightarrow \mathbb{Q}^{2\times 2}, \ A \rightarrow A + A^t$

Question

This question is related to the question I previously asked: Kernel.
The following function is given:

$$\phi: \mathbb{Q}^{2\times 2} \rightarrow \mathbb{Q}^{2\times 2}, \ A \rightarrow A + A^t$$

The task is to prove: $im{\phi} = \left\{ A \in \mathbb{Q}^{2\times 2} | A = A^t \right\}$.
I am stuck on this one again even though I feel that it is as easy as the last one was.
I know that for a linear function $f: V \rightarrow W$

$$im(f) = \{w \in W \ | \ \exists v \in V: f(v) = w\}$$.

So I said let $B \in \mathbb{Q}^{2\times 2}$. Thus I have to check for whether there is an $A \in \mathbb{Q}^{2\times 2}$ so that

$$\phi(A) = B \text{ ,right?}$$

But simply saying $A + A^t = B$ is not helping me at all...

Than you very much for your help once again.
FunkyPeanut

Answer 1

Let $T = \{D \in Q^{2 \times 2} | D^t = D\}$

Let $B \in im(\phi)$. Then $B = A + A^t$ so $B^t = A^t + A = B$. So $im(\phi) \subset T$.

Let $A \in T$ so that $A^t = A$. Then write $A = \frac{A + A}{2}= \frac{A}{2} + \frac{A^t}{2}$. Note that $\frac{A}{2}$ is in $Q^{2 \times 2}$. So one can write $A = \phi(\frac{A}{2})$ therefore $A \in im\phi$ so $T\subset im(\phi)$.