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I'm trying to solve the following:

Let $V$ be an $n$-dimensional vector space over a field $K$, and let $U$ be a $k$-dimensional subspace of $V.$ Find the dimension of the vector space $$M=\{\varphi:V\to V\mid \varphi \text{ is linear and } \varphi(U)\subseteq U\}.$$

First I noted that $\dim(V^*)=\dim(V) = n$ and $M\subseteq V^*$. Thus $\dim(M)\leq n$. Let $\beta=\{\varphi_1,...,\varphi_n \}$ be a basis for $V^*$. Now if $\varphi \in V^*$, then we can write $\varphi = \lambda_1\varphi_1 + \cdots + \lambda_n\varphi_n$ for some $\lambda_1,...,\lambda_n$. Thus elements of $M$ can also be represented as linear combinations of $\varphi_1,...,\varphi_n$. Also, $\dim(U^*)=\dim(U)=k$, so we must have $\dim(M)\geq k$. However, I don't know how to proceed from here.

I'd appreciate any help.

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    You might find it helpful to take a basis of $U$, extend it to a basis of $V$, and then ask what the matrix of $\varphi$ in this basis needs to look like for it to be an element of $M$. – 2017-01-01
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    @EricWofsey Ok. I think I got it. I extend the basis $u_1,...,u_k$ of $U$ to $u_1,...,u_k,v_1,...,v_{n-k}$ of $V$. Now any $v\in V$ can we written as $v=\lambda_1u_1+\cdots + \lambda_ku_k+\mu_1v_1+\cdots \mu_{n-k}v_{n-k}$. If $v\in U$, we need $\varphi(v)\in U$. In matrix representation we need a $n\times n$ matrix $A$ such that $A(\lambda_1',...,\lambda_k',\mu_1,...,\mu_{n-k}) = (\lambda_1',...,\lambda_k',0,...,0)$. Thus $A = \begin{pmatrix} B & 0 \\ 0 & 0 \end{pmatrix}$, where $B$ is some $k\times k$ matrix. So $\dim(M) = k$. Is that right? – 2017-01-01
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    I believe your matrix $A$ is in error since it maps all vectors in $V$ to $U$. – 2017-01-02
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    @K.Miller Indeed. It should be $A(\lambda_1,...,\lambda_k, \mu_1,...,\mu_{n-k}) = (\lambda_1',...,\lambda_k', \mu_1',...,\mu_{n-k}') $, so $A$ is a block matrix of the form $A=\begin{pmatrix} B & 0 \\ 0&C \end{pmatrix}$. – 2017-01-02
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    I believe it should be an $n\times n$ block upper triangular matrix of the form $A = \begin{bmatrix} A_{11} & A_{12}\\0 & A_{22}\end{bmatrix}$, where $A_{11}$ is $k\times k$. The goal is to map vectors of the form $(\lambda, 0)$ to vectors of the form $(\lambda',0)$, i.e., from $U$ into $U$. – 2017-01-02
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    Oh yes of course. The dimension of $A$ would then be $n$. Relating it back to $M$, would that mean that $\dim(M)=n$? – 2017-01-02
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    The dimension of $M$ is the same as the dimension of the vector space of all block upper triangular matrices over $K$ of the same form as $A$. – 2017-01-02
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    That makes sense. The block $A_{11}$ is generated by the standard basis vectors $E_{ij}$ with $1\leq i,j \leq k$ and similarly for $A_{12}$ and $A_{22}$. So then $\dim(M)= k^2 + k(n-k)+(n-k)^2$. – 2017-01-02

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Let $\{v_1,\ldots,v_k\}$ be a basis of $U$ and extend this basis to $\{v_1,\ldots,v_k,v_{k+1},\ldots,v_n\}$ a basis of $V$. Then for all $v = \alpha_1v_1 + \cdots + \alpha_nv_n \in V$ and $\varphi \in M$ we have

\begin{align} \varphi(v) = \left\{ \begin{array}{ll} \alpha_1\varphi(v_1)+\cdots+\alpha_k\varphi(v_k) & \text{if}~ v\in U\\[1mm] \alpha_1\varphi(v_1)+\cdots+\alpha_n\varphi(v_n) & \text{if}~ v \not\in U \end{array}\right. \tag{1} \end{align}

Now for all $i,j\in\{1,\ldots,n\}$ define the linear maps $e_{ij} : V\to V$ by $$ e_{ij}(v_m) = \left\{ \begin{array}{rl} v_j, & m = i\\ 0, & m \neq i \end{array}\right. $$ and observe that $e_{ij}(v) = \alpha_iv_j$ for all $v\in V$. Since $\varphi(v_i) = \beta_{i1}v_1 + \cdots + \beta_{in}v_n \in V$ for some scalars $\beta_{i1},\ldots,\beta_{in}$, it follows that $$ \varphi(v_i) = \sum_{j=1}^n \beta_{ij}e_{ij}(v_i)~~\forall i = 1,\ldots,n $$ Thus $$ \varphi(v) = \sum_{i=1}^n\alpha_i\sum_{j=1}^n\beta_{ij}e_{ij}(v_i) = \sum_{i=1}^n\sum_{j=1}^n\beta_{ij}e_{ij}(v)~~\forall v\in V \implies \varphi = \sum_{i=1}^n\sum_{j=1}^n\beta_{ij}e_{ij} $$ Considering the restriction $(1)$ placed on all $\varphi\in M$ we find that $\beta_{ij} = 0$ if $1 \leq i \leq k$ and $k < j \leq n$. Otherwise, it would be possible for $\varphi(u) \not\in U$ for some $u\in U$. Thus, $M$ is spanned by the set $E = \{e_{ij}\,|\, 1 \leq i,j \leq k ~\text{or}~ k < i \leq n\}$. Since $E$ is linearly independent (this is easily confirmed) it forms a basis of $M$, which implies that $\dim M = n^2 - k(n-k)$.