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Suppose we have a function $G:M_2(\mathbb R) \to S_2(\mathbb R)$ where $S_2(\mathbb R)$ is a symmetric matrix such that $ S_2(\mathbb R) = \left\{A = \begin{bmatrix} a & b\\ c & d \end{bmatrix} \text{ such that } a,b,c,d \in \mathbb R \text{ and } b = c \right\} $ and

$G$ is defined as:

$ G\left( \begin{bmatrix} a & b\\ c & d \end{bmatrix} \right) = \begin{bmatrix} a & b\\ c & d \end{bmatrix} + \begin{bmatrix} a & c\\ b & d \end{bmatrix} $

Is $f$ surjective? injective?


For injective I got:

Not injective, observe $A = \begin{bmatrix} 1 & 2\\ 1 & 1 \end{bmatrix} $ and $B = \begin{bmatrix} 1 & 1\\ 2 & 1 \end{bmatrix} $

$A \not = B$ but $f(A) = \begin{bmatrix} 2 & 3\\ 3 & 2 \end{bmatrix} = f(B) $

I am not sure how to approach for surjective. I believe that it is surjective... my start of a proof is:

Take arbitrary $D \in S_2(\mathbb R) = \begin{bmatrix} x & y \\ y & z \end{bmatrix}$

Then choose $C \in M_2(\mathbb R)$ such that:

$ \begin{bmatrix} a & b\\ c & d \end{bmatrix} $

where $c + d = y$?

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Your intuition is correct. Hint: $1=\frac{1}{2}+\frac{1}{2}$. –  1015 Dec 7 '13 at 21:35

2 Answers 2

up vote 1 down vote accepted

Indeed, if you take arbitrary $D \in S_2(\mathbb R) = \begin{bmatrix} x & y \\ y & z \end{bmatrix},$

you can choose the following $C \in M_2(\mathbb R)$:

$ C = \begin{bmatrix} \frac{x}{2} & \frac{y}{2}\\ \frac{y}{2} & \frac{z}{2} \end{bmatrix}. $

It is easy to verify that $G(C) = D$, which proves that $G$ is surjective.

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Thanks @Tonci ! –  Stefan Dec 7 '13 at 21:48

If $Α$ is symmetric then $G(A)=2A$.

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