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I have a linear map $T:U \rightarrow V$, with $\dim U = n$ and $\dim V = m$.

What are the conditions of the rank$T$ and ker$T$ for surjectivity and injectivity?

I know $T$ is injective if ker$T = \{0\}$, but still quite unsure about surjective. Is it rank$T$ = $\dim V$?

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Note that $$ \dim U=\dim\ker T+\dim \,\text{im}\, T $$ since $U/\ker T\cong \text{im}\, T$. Hence $T$ is surjective iff $\dim U-\dim\ker T=\dim V$.

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  • $\begingroup$ What does the notation $U/ \ker T$ mean? $\endgroup$ – PhysicsMathsLove May 23 '17 at 17:30
  • $\begingroup$ @PhysicsMathsLove For now you can interpret it as the complementary subspace of $\ker T$ in $U$. $\endgroup$ – Minsheng Liu May 23 '17 at 17:35
  • $\begingroup$ More technically, it means to identify every point in $\ker T$ with $0$. Note that since $\ker T$ is a subspace, you can find a basis for $T$ such that for and only for every point $x \notin \ker T$, the first $\dim \ker T$ entries of its coordinate in the new basis are 0. In this way, ignoring the first $\dim \ker T$ entries gives a $\dim U - \dim \ker T$ vector space. This space is isomorphic to $\operatorname{im} T$. $\endgroup$ – Minsheng Liu May 23 '17 at 17:38
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    $\begingroup$ Injective if and only if the kernel is the zero vector space. You are right about surjective: since the range of $T$ is a subspace of $V$, the only way for surjectivity is for that subspace to be all of $V$. $\endgroup$ – avs May 23 '17 at 17:42
  • $\begingroup$ @avs so the statement works both ways? if the rank $T = \dim V$, then it is surjective? and if it is surjective, rank $T = \dim V$? Is this even when the kernel is non-trivial? $\endgroup$ – PhysicsMathsLove May 23 '17 at 17:46
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yes, consider the linear map T:U→V, with dimU=n and dimV=m. first, wen know that the image of U under the map T is a subset of V (not necessarily V)

surjective map means that for every element v in V, there exists an element u in U such that T(u)=v, i.e imgT=U.

ImgT=U if and only if dimImgT=DimV, ie rankT=DimV. (since imgT is subset of U)

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