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Let $U,V$ be finite dimensional vector spaces, and let $N$ be a subspace of $U$. Assuming that $\dim U\leq \dim N+\dim V$, is there an example that there is a linear map $A\in \operatorname{Hom}(U,V)$ with null-space $N(A)=N$? I was thinking about a mapping $A:U\to V$ with two cases, $\dim U\leq \dim V$ and $\dim U>\dim V$, but I do not know what the mapping should be defined as.

Hopeless
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1 Answers1

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Each of the spaces is finite-dimensional so each has a finite basis. Let $$\{v_1, \dots, v_n\}\text{ be a basis for $V$} \\ \{u_1,\dots, u_k\}\text{ be a basis for $N$} \\ \mathcal U = \{u_1,\dots, u_k, u_{k+1},\dots, u_m\}\text{ be a basis for $U$}$$ Now we specify $A$ by its action on $\mathcal U$. First define $Au_i = \begin{cases} 0, & i=1,2,\dots,k \\ v_{i-k}, & i=k+1,\dots, m\end{cases}$. Then extend $A$ by linearity.

Note that, by the rank nullity theorem, $\dim V$ must be at least $\dim U - \dim N$.

Community
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  • Extending $A$ by linearity, you mean $A(x) = 0$ if $1\leq i\leq k$ and $A(x)= \sum_{i=k+1}^m c_i v_{i-k}$ where $c_i$ are scalars? – Hopeless Oct 19 '16 at 16:55
  • I mean that because any vector $u\in U$ can be written as $u = \sum_{i=1}^m c_iu_i$, we define $Au$ for all $u\not\in\mathcal U$ by $$Au = \sum_{i=1}^m c_iAu_i$$ That is, we define $Au$ in terms of $Au_1,\dots, Au_m$. Because of the way we defined $Au_1,\dots, Au_m$ in this case we could also write that as $$Au = \sum_{i=k+1}^m c_iv_{i-k}$$ as you said. –  Oct 19 '16 at 17:30