Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

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

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

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.