Let $Y$ be a finite-dimensional normed space, $X$ a normed space, and $T: X \to Y$ a surjective linear operator. Show that $T$ is an open mapping. 
Let $Y$ be a finite-dimensional normed space, $X$ a normed space, and  $T: X \to Y$ a surjective linear operator. Show that $T$ is an open mapping.

I think if I can show that $T(B_X)$ contains an open ball then I am done where $B_X$ is the unit ball in $X$. But I am unable to show that. Need some help...
 A: Lemma : Let $T$ is an onto linear map from $X$ to $Y$, where $X,Y$ are n.l.s and $Y$ is finite dimensional. Let $U \subseteq X$ is open such that $T(U)$ has only non zero elements.Then $T(U)$ is open.
Proof : Let dimension of $Y$ be a natural no. k. Let $u \in U$ ,then $Tu \ne 0$. Extend $Tu$ to a basis of $Y$ say 
$\{Tu =: u_1,u_2, .... ,u_k\}$. Clearly $u \ne 0$. Choose $v_i \in X$, such that $Tv_i = u_i \forall i = 1, ..., k$ and $v_1 = u$. Thus $\{v_1,v_2, .... ,v_k\}$ is a linearly independent set in $X$. Let  $ X_u := span \{v_1,...., v_k\} \le X$. Let  $T_u$ denote the restriction of $T$ onto $X_u$. Now $T_u$ : $X_u$ $ \rightarrow Y$ is a linear isomorphism (also homeomorphism of topological spaces).Now $U$ $\cap$ $X_u$ is open in $X_u$, therefore $T(U \cap X_u)$ is also open in $Y$. Now $u \in U \cap X_u$ so $Tu \in T(U \cap X_u) \subseteq T(U) \subseteq Y$.
Thus $T(U)$ is open in $Y$.
Answer to the above question.
Enough to show $T(B(0,r))$ is open in $Y$ for all $r \gt 0$ (since translations are homeomorphisms).
We have assumed the dimension of $Y$ to be a natural no., otherwise the result is vacuous.If $T(B(0,r))$ is $Y$ then we are done, else choose $y \in
Y\backslash T(B(0,r))$. Let $t_y$ denote the translation in $Y$ by $y$. Let $x \in X$ such that $Tx = y$. Let $t_x$ denote the translation by $x$ in $X$. (note $y,x$ are non zero vectors) Clearly $T(t_x(B(0,r))=B(x,r)) = t_y(T(B(0,r))) = T(B(0,r)) + y$. We show that the last set do not contain zero :
Let $z \in B(0,r)$, then $Tz + y = 0$ implies $T(-z) = y$, therefore  $y \in T(B(0,r))$, contradiction (since$-z \in B(0,r)$). Thus the last set is open in $Y$ by the lemma. Now since translations are homeomorphisms, we have $T(B(0,r))$ is open. This completes the proof.   
A: *

*If $T$ is continuous, then $\ker(T)$ is closed and the quotient map
$$
\pi : X \to X/\ker(T)
$$
is an open map. Furthermore, $T$ induces an injective map
$$
S : X/\ker(T) \to Y
$$
Since $Y$ is finite dimensional, so is $X/\ker(T)$, and so $S$ (whose range is $Y$) is now a homeomorphism. In particular, $S$ is an open map, so
$$
T = S\circ \pi
$$
is also open.

*If $\dim(Y) = 1$, then it follows from an earlier question that if $U$ is a non-empty open set, then $T(U) = \mathbb{C}$, so it is, in particular, an open map.

*Not sure about the general case (if $T$ is discontinuous and $\dim(Y) > 1$), but perhaps someone else can complete that case (I don't think induction works, but perhaps it could)
A: WLOG, we can take $Y=(\mathbb{C}^{n},\|\cdot\|_{\infty})$(why?). Let $\{e_1,e_2,\dots,e_n\}$ be the standard basis of $\mathbb{C}^{n}$. Since $T$ is surjective, we can find $a_1,a_2,\ldots,a_n$ in $X$ such that $T(a_i)=e_i$ for all $1\leq i\leq n$. Define $\alpha:=\sum_{i=1}^n\|a_i\|$, a positive real number. Now, let $V$ be an open subset of $X$ and consider an element $x\in V$. As $V$ is open, we can find an $r> 0$ such that $B(x;r)\subseteq V$.
Consider $k=(k_1,k_2,\ldots,k_n)\in \mathbb{C}^{n}$ with $\|k\|_{\infty}\leq \frac{r}{\alpha}$. Then $x-\sum_{i=1}^nk_ia_i\in V$, so that $T(x-\sum_{i=1}^nk_ia_i)=T(x)-\sum_{i=1}^nk_ie_i=T(x)-k\in T(V)$. Thus, the set $$\left\{k\in \mathbb{C}^{n}: \|T(x)-k\|_{\infty}\leq \frac{r}{\alpha}\right\}\subseteq T(V),$$showing that $T(V)$ is open in $Y$.
