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Let $V$ be a finite-dimensional vector space and $T:V\to W$ a linear transformation. Take $\{v_1,\dots,v_r\}$ to be a basis for $\ker T$, and complete this set into a basis $\{v_1,\dots,v_r,u_1,\dots,u_m\}$ of $V$. Taking the spans of $v_1,\dots,v_r$ and $u_1,\dots,u_m$ separately, we obtain $V=U_1\oplus U_2$ where $U_1$ is the kernel of $T$ and $U_2$. It's easy to see that the image of $T$ is isomorphic to $U_2$.

The question: is there a "nice" characterization of $U_2$ which does not involve completing the basis of the kernel? The kernel itself is defined in a very simple way: $\ker T=\{v\in V:T(v)=0\}$. Is there something similar for $U_2$?

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up vote 2 down vote accepted

Your $U_2$ is of course heavily dependent on your choice of bases. You could say $U_2=U_1^\perp$ with respect to the unique inner product such that your v's and u's are orthonormal, but that's just rephrasing exactly the same.

We do have a canonical isomorphism $U_2\cong V/\ker T$ characterizing $U_2$ up to isomorphism, and which is pretty nice.

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There is no such characterization and a reason is that $U_2$ is not unique except in the degenerate cases when $U_1$ is $\{0\}$ or $V$. For example, if $V=\mathbb R^2$ and $U_1=\mathbb R\times\{0\}$, $U_2$ may be $(t,1)\cdot\mathbb R$ for any $t$ in $\mathbb R$.

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