# $T: V\rightarrow W$ is an injective linear transformation when restricted to subspace $A$ of $V$. Then can we conclude that $\dim(A) = \dim T(A)$

Let $V$ and $W$ be finite dimensional vector spaces and let $A$ be a given subspace of $V$:

Also, we have a linear transformation $T: V\rightarrow W$ such that $T$ is injective on the subspace $A$. Then can we conclude that $\dim(A) = \dim T(A)$?

I think it should hold true. I need little help to prove this. Thanks

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@BrianM.Scott No sir, i am just interested in subspace $A$ and its image. – srijan May 21 '12 at 11:11
Are you sure that you’ve stated this correctly? You have $A$ a subspace of both $V$ and $W$. If you mean that $T$ maps $A$ injectively onto $T[A]$, a subspace of $W$, then the answer to your question is yes. – Brian M. Scott May 21 '12 at 11:11
@BrianM.Scott Sir, i have edited now $A$ is a subspace of $V$ only. – srijan May 21 '12 at 11:13
I think that you need one more small change: ‘is injective onto the subspace $T[A]$’. Then DonAntonio’s answer points you in the right direction. – Brian M. Scott May 21 '12 at 11:15
@srijan: The correct way to say it is that $T$ is injective when restricted to $A$, not that it is "injective onto $A$" or "injective on $A$". – Arturo Magidin May 21 '12 at 16:34

You are talking about the restriction: $T|_A : A \rightarrow T[A]$ is injective and surjective, hence an isomorphism. So $A$ and $T[A]$ have the same dimension.

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Yes. You may want first show that an injective lin. transformation maps linearly independent sets to lin. ind. sets.

BTW, you don't need the finite dimensional thing here.

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If $T\colon V\to W$ is a linear transformation, and $N$ is the nullspace of $T$, then for every subspace $A$ of $V$ we have $$\dim(T(A))+\dim(A\cap N) = \dim (A).$$ This is a simple consequence of the Rank-Nullity theorem.

In particular, since the restriction of $T$ to $A$ is one-to-one if and only if $A\cap N = \{\mathbf{0}\}$, it follows that $\dim(T(A))=\dim(A)$ if and only if the restriction of $T$ to $A$ is one-to-one, if and only if $A\cap N=\{\mathbf{0}\}$.

(This is actually a special case of the homomorphism theorems, applied to linear algebra; the image of $A$ is the same as the image of $A+N$, which is isomorphic to $(A+N)/N \cong A/(A\cap N)$.

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Sincere thanks to you sir. Got new result. – srijan May 21 '12 at 17:04