# Can the range of this operator be closed?

Given an operator $T:H\rightarrow L$, where $H$ is a finite-dimensional Hilbert space and $L$ an infinitedimensional one, is the range of $T$, $T(H)$, a closed set in $L$ ? I know that the image doesn't have to be closed if $H$ were infinite, but I'm not sure in this case.

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$T(H)$ is a finite dimensional subspace and any finite dimensional subspace is closed since it is isomorphic to $\mathbb R^n$ for some $n$ and hence complete.