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Let $T:X\to Y$ be an $\mathbb{R}$-linear map of $\mathbb{R}$-vector spaces $X$ and $Y$ of finite dimension. Let $W\subseteq Y$ be an $\mathbb{R}$-vector subspace such that $\text{Im} \ T$ and $W$ together span $Y$. Let $Z=T^{-1}(W)$. Show that $\text{dim} \ X+\text{dim} \ W=\text{dim} \ Y+\text{dim} \ Z$.

My approach : Clearly $\text{dim} \ X=\text{rank}\ T+\text{nullity}\ T$. Again since $\text{Im} \ T$ and $W$ together span $Y$, we have $\text{rank}\ T+\text{dim}\ W=\text{dim}\ Y$. Adding these two relations we have $\text{dim} \ X+\text{dim} \ W=\text{nullity}\ T+\text{dim}\ Y$. But I have problem to show that $\text{nullity}\ T=\text{dim}\ Z$. Can anyone help me regarding this issue? Thanks in advance.

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  • $\begingroup$ You don't have $\operatorname{rank}{T}+\dim{W} = \dim{Y}$: $T(X) \cap W$ may be more than $\{0\}$. You do have $\operatorname{rank}{T}+\dim{W} \geq \dim{Y}$ $\endgroup$
    – Chappers
    Apr 12, 2017 at 15:36
  • $\begingroup$ Indeed, the whole point of $Z$ is to account for this. $\endgroup$
    – Chappers
    Apr 12, 2017 at 15:37
  • $\begingroup$ So, how to relate $\text{dim}(\text{Im} T \cap W)$ with $\text{dim}\ Z$? $\endgroup$ Apr 12, 2017 at 15:47

1 Answer 1

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I'm going to write $r_X(T)$ for rank as a map from $X$ and $n_X(T)$ for its nullity.

  1. We have $$ r_X(T) + n_X(T) = \dim{X}, \qquad r_Z(T) + n_Z(T) = \dim{Z} $$ by the rank-nullity theorem.

  2. Now, look at $T$ acting on $Z$. $T(Z) = T(X) \cap W$, so $r_Z(T)=\dim{(T(X) \cap W)}$.

  3. Inclusion-exclusion on an appropriate basis gives $$\dim{Y} = r_X(T) + \dim{W} - \dim{(T(X) \cap W)}.$$

  4. Lastly, $\{0\} \in W$, so $Z$ must include $\ker{T}$, so $$n_Z(T)=n_X(T).$$

Subtracting the two equations from 1. and using 4. gives $$ r_X(T) - r_Z(T) = \dim{X}-\dim{Z}, $$ while substituting into the one in 3. from 2. gives $$ \dim{Y} = r_X(T)-r_Z(T) + \dim{W}. $$ The result follows.

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  • $\begingroup$ Please explain $2$. $\endgroup$ Apr 12, 2017 at 16:07
  • $\begingroup$ Yes, that wasn't quite right. Better? $\endgroup$
    – Chappers
    Apr 12, 2017 at 16:10
  • $\begingroup$ Yes, now it is clear. Thanks. $\endgroup$ Apr 12, 2017 at 16:11

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