# The dimension of linear map

I am reading "Introduction to smooth manifolds" by Lee and one place is very unclear for me: Let $P$ and $Q$ be any complementary subspaces of $V$ (which is an $n$-dimensional real vector space) of dimensions $k$ and $n-k$, respectively, so that $V$ decomposes as a direct sum $V=P\bigoplus Q$.(This part is clear for me) The graph of any linear map $X:P\rightarrow Q$ can be identified with a $k$-dimensional subspace $\Gamma(X)\subseteq V$, defined by $$\Gamma(X)=\left \{ v+Xv:v\in P \right \}$$

Earlier in the same book there was a example of a function graph being a manifold. And the graph of the continuous function $f:U \subseteq R^{n} \rightarrow R^{k}$ is the subset of $R^{n}\times R^{k}$ defined by $$\Gamma(X)=\left \{ (x,y)\in R^{n}\times R^{k}:x\in U, y=f(x) \right \}.$$

By this definition the graph of the linear map $X$ would be a subset of $R^{k}\times R^{n-k}$, but the book says it is a $k$ dimensional subspace of $V$. And this is where I am confused. What am I overlooking?

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For any $v\in P$, $(v,Xv)=0$ iff $v=0$. Therefore the nullity of the linear map $f:v\mapsto(v,Xv)$ is zero and by the rank-nullity theorem, $\operatorname{rank}f\equiv\dim\{(v,Xv):v\in P\}=\dim P=k$. Hence the graph of $X$, i.e. $\{(v,Xv):v\in P\}\subset\mathbb{R}^k\times\mathbb{R}^{n-k}$, can be identified with the $k$-dimensional subspace $\{v+Xv:v\in P\}\subset\mathbb{R}^n$.

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This is also a very nice answer. Thanks! – Tomas Jan 18 '13 at 7:19

Since $V$ is finite dimensional you have $V= P \bigoplus Q \simeq P \times Q = R^k \times R^{n-k}$. So it is a subset of $R^k \times R^{n-k}$ and it is a $k$-vector space since

$$P \longleftrightarrow \Gamma(X) = \lbrace v+Xv : v \in P \rbrace: v \longleftrightarrow (v,X(v))$$

is an isomorphism of vector spaces.

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Yes, I know that, but this is not my question. My question is why a graph of linear map $X$ can be identified with $k$-dimensional subspace $\Gamma (X)$? – Tomas Jan 17 '13 at 20:50
What I try to point out is that both definitions are the same, with different notation: $v+Xv =(v, X(v))$ – Z. L. Jan 17 '13 at 20:54
Observe that a linear map is also a continuous function with respect to the topology induced by a norm in that vector space. – Z. L. Jan 17 '13 at 20:55
Since it is parametrized from $v \in P$ you can identify it with $P$ and $P$ is a k-dimensional subspace of $V$. – Z. L. Jan 17 '13 at 20:57
Thanks, now it is clear!! – Tomas Jan 18 '13 at 7:16