# prove that a linear map is injective - $T(X) = X + 2X^T$

I have the following linear map:

$$T: \operatorname{Mat}_{n\times n}(\mathbb{R}) \to \operatorname{Mat}_{n\times n}(\mathbb{R})\;,$$

$$T(X) = X+2X^T\;.$$

I have to prove that it is injective (one-to-one) and surjective (onto).

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I know that I only need to prove one of them because if I have $T:V\rightarrow W$ and $dimV = dimW$ then T is one-to-one $\Leftrightarrow$ T is onto – Dor Shalom Jan 5 '13 at 14:02

Suppose $X, Y$ are $n\times n$ matrices such that $T(X)=T(Y)$, we want to prove that $X=Y$. If $T(X)=T(Y)$, then by definition we have $$\tag{1}X+2X^T=Y+2Y^T.$$ Taking transponse of $(1)$, we get $$\tag{2}X^T+2X=Y^T+2Y$$ because $(X^T)^T=X$.

Can you combine $(1)$ and $(2)$ to prove that $X=Y$?

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I already have $X^T + 2X = Y^T +2Y$ But it doesn't seem enough to make sure $X=Y$ – Dor Shalom Jan 5 '13 at 14:08
No. What I said is to combine (1) and (2) above to prove that $X=Y$. Did you see how we get (2)? – Paul Jan 5 '13 at 14:11
Oh, okay i got it! Thank you very much! my mind was just stuck :) – Dor Shalom Jan 5 '13 at 14:14
You can accept the answer :) – Paul Jan 5 '13 at 14:25

A linear map is injective if its kernel is zero. Now, if $T(X)=X+2X^T=0$, then by taking transpose, we also have $X^T+2X=0$. So $3X=2(X^T+2X)-(X+2X^T)=0$, or $X=0$. Hence $T$ is injective and in turn it is surjective because $T$ is a linear endomorphism on a finite dimensional vector space.

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