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given $V$ a vector space, let $\dim{V}=n$.
$T\colon V\longrightarrow V$ is a linear transformtion with $\dim(\operatorname{Im}{T})=r$.

Let $X$ be the set of all linear transformtion $F:V\longrightarrow V$ with $T\circ F=0$.

I need to prove that $X$ is a subspace of the space of all linear transformtion $V\longrightarrow V$ and find its dimension.

It's not a problem to prove that $X$ is a subspace of all the linear transformations $V\longrightarrow V$ as in fact $X=KerT\circ F$.

What I'm confused about is $\dim{X}$. Looks like it's $n-r=V-ImT$ but i'm not sure how to exacly explain this, as I don't know what is $KerF$ and don't know how to prove what is the dim of a LT composition.

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  • $\begingroup$ Since $X$ is a set of linear transformation thus we need to define what we mean by $\dim(X)$. $\endgroup$
    – user
    Jun 12, 2018 at 10:03
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    $\begingroup$ $X$ is a subspace of $\operatorname{Hom}(V,V)$. In particular, $X$ is a vector space and has a well-defined dimension. There is no ambiguity on what we mean by $\dim{X}$ $\endgroup$
    – Pedro
    Jun 12, 2018 at 10:07
  • $\begingroup$ @Pedro Ok thanks for the clarification on that point. $\endgroup$
    – user
    Jun 12, 2018 at 10:24
  • $\begingroup$ @Pedro I don't understand why you are considering $n(n-r)$ as dimension and not $(n-r+1)^n$. $\endgroup$
    – user
    Jun 12, 2018 at 10:30
  • $\begingroup$ @gimusi I have given arguments for my answer. It is also a well known result, see for example here. So I think you should instead be asking yourself why are you considering $(n-r+1)^{n}$ as the dimension: this doens't match with the well known result, doesn't match with the 2 examples we have already discussed and most importantly you haven't argued your answer, so we can't really argue much. You are counting some random set of maps. This set of maps isn't a basis of the space we are looking at, hence it doesn't compute the dimension $\endgroup$
    – Pedro
    Jun 12, 2018 at 10:55

1 Answer 1

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1) It doesn't make sense to say $X=\ker{T}\circ F$, because $X$ is the set of such maps $F$. I guess you meant that $X$ is the set of linear transformations $F$ that factor (uniquely) through the kernel of $T$. This is indeed the case by the universal property of the kernel. To show that $X$ is a subspace you must show that the zero map is in $X$ and that $X$ is closed under addition and scalar multiplication.

2) We know that $\dim{\ker{T}}=n-r$. The maps in $X$ are in bijection with the linear maps from $V$ to $\ker{T}$. Hence we want to find the dimension of linear maps from $V\cong \mathbb{K}^{n}$ to $\ker{T}\cong \mathbb{K}^{n-r}$. The dimension of $X$ is therefore $n(n-r)$.

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  • $\begingroup$ I also have some doubts with the interpretation of the OP. Why are you considering only bijection? I thought that the dimension would be $(n-r)^n$, I will check again my conclusion. $\endgroup$
    – user
    Jun 12, 2018 at 10:13
  • $\begingroup$ I don't think there is any doubt about the statement or about what the OP means. Why did you think that is the dimension? You didn't explain your reasoning and the conclusion which you got is wrong. Take $T=0$, then $X$ is the whole $\operatorname{Hom}(V,V)$ which has dimension $n^{2}$, not $n^{n}$ $\endgroup$
    – Pedro
    Jun 12, 2018 at 10:16
  • $\begingroup$ The reasoning is that we can map any basis vector in any basis vector of $ker(T)$ obtaining $(n-r)^n$ different maps. Is it wrong? $\endgroup$
    – user
    Jun 12, 2018 at 10:18
  • $\begingroup$ Yes: suppose the kernel had dimension 1 and V had dimension 2. You are claiming that there is only one map $V\to \ker$. There are infinitely many maps (if the field is infinite), so I assume you mean that $(n-r)^{n}$ is the dimension. But the dimension in this case is 2, not 1. A basis is given by the map sending the first basis vector of V to the only basis vector of the kernel and the other basis vector of V to zero, and the map sending the first basis vector of V to zero and the other basis vector of V to the only basis vector of the kernel $\endgroup$
    – Pedro
    Jun 12, 2018 at 10:23
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    $\begingroup$ @Jneven there aren't two different answers. The other answer was wrong, as I pointed out and explained in the comments. If there is any particular detail you don't unserstand in this answer, I will try to clarify it $\endgroup$
    – Pedro
    Jun 13, 2018 at 0:19

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