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).
Please help. Thanks
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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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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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