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How does one go about proving that $\dim(T(X))\geq \dim(T(V)) - \dim(V) +\dim(X)$ where $X$ is a subspace of $V$, a vector space, and $T$ is a linear transformation? Thanks.

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up vote 2 down vote accepted

Find a basis for $X$ and extend it to a basis for $V$ using $\dim V-\dim X$ additional vectors. Then $T(V)$ is spanned by $T(X)$ and the images of these additional vectors, and the $\dim V-\dim X$ additional vectors can contribute at most another $\dim V-\dim X$ to the dimension, so

$$\dim T(V)\le \dim T(X) + \dim V -\dim X\;.$$

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Here's an approach through the rank-nullity theorem. Let $T : V \to W$ be the linear map, and let $S : X \to W$ be the restriction of $T$ to the subspace $X$. Then clearly $\ker(S)$ is a subspace of $\ker(T)$, and hence: $$ \dim(\ker(S)) \leq \dim(\ker(T)). \tag{1} $$ Using the rank-nullity theorem, $\dim(\ker(T)) = \dim(V) - \dim(T(V))$, and $\dim(\ker(S)) = \dim(X) - \dim(T(X))$. Plugging these in $(1)$ gives the claim.

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