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The statement is as follows:

Let the natural numbers $m$ and $n$ be given with $n < m$. Let $V$ be an $n$-dimensional vector space over the field $K$ and $V^*$ be the dual space of $V$. Also, the vectors $v_1,\dots,v_m \in V$ and $\alpha_1,\dots,\alpha_m \in V^*$ are given.

For the matrix $A = (a_{ij}) \in M(m,K)$ ($m\times m$ matrices) with entries $a_{ij} = \alpha_i(v_j)$, show that $\operatorname{rk}(A) \leq n$.

My idea of proof is not very formal. I'm not even sure that it is correct. It looks as follows:

Since $\dim(V) = n$ and the dimension of $M(m,K)$ is strictly bigger than the dimension of $V$, the matrix $A$ has $m-n$ rows consisting of only zeros. Thus we can say that $\operatorname{rk}(A) = n$ at most.

The dual vectors of the vectors $v_1,\dots,v_m \in V$ can be calculated as follows: $$ \alpha_i(v_j) = \delta_{ij} $$ with $\delta_{ij}$ being the Kronecker delta. Not looking at the $m-n$ zero-rows but rather at the remaining $n$ rows of the matrix $A$, the entries of $A$ are given by $ a_{ij} = \delta_{ij} $ meaning that the diagonal entries of $A$ can consist of ones and the rest zeros. This is the previously mentioned case in which $\operatorname{rk}(A) = n$. If any of the $v_1,\dots,v_m$ are linearly dependent we get more zero-rows in $A$, implying that $\operatorname{rk}(A) < n$. So, overall $\operatorname{rk}(A) \leq n$.

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    $\begingroup$ You start out by making a couple of assertions that are just wrong. You write "Since $\dim(V) = n$ and the dimension of $M(m,K)$ is strictly bigger than the dimension of $V$, the matrix $A$ has $m-n$ rows consisting of only zeros." This is wrong. You also write: "The dual vectors of the vectors $v_1,\dots,v_m \in V$ can be calculated as follows: $ \alpha_i(v_j) = \delta_{ij} $." But your given $\alpha$'s and $v$'s do not need to satsify this relation. $\endgroup$ – fredgoodman Apr 4 '18 at 21:30
  • $\begingroup$ @fredgoodman Yes, now I see the my assumptions don't make a lot of sense. I think I got blinded by what I wanted to prove and just assumed anything without questioning it logically. Thanks for the pointers! $\endgroup$ – JtSpKg Apr 5 '18 at 7:38
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There seems to be no assumption that $\alpha_i(v_j)=\delta_{ij}$. By the way this would be impossible, since this would imply $\{v_1,\dots,v_m\}$ is linearly independent: suppose $c_1v_1+\dots+c_mv_m=0$; then $$ 0=\alpha_i(c_1v_1+\dots+c_mv_m)=c_iv_i $$ Also $\alpha_i(v_i)=1$ implies $v_i\ne0$ and therefore $c_i=0$ for $i=1,2,\dots,m$. But no set with $m$ vectors can be linearly independent in an $n$-dimensional space, if $n<m$.

The span of $\{v_1,\dots,v_m\}$ has dimension less than $m$; it is not restrictive to assume that $\{v_1,\dots,v_k\}$ form a basis for the span and that $v_{k+1},\dots,v_m$ are their linear combinations. Indeed a reordering of the columns of the matrix $A$ doesn't change the rank. Moreover $k\le n$ because we're in an $n$-dimensional vector space.

The $j$-th column of $A$ is \begin{bmatrix} \alpha_1(v_j) \\ \alpha_2(v_j) \\ \vdots \\ \alpha_m(v_j) \end{bmatrix} Suppose $k<j\le m$; then $v_j=c_1v_1+\dots+c_kv_k$ and therefore $$ \begin{bmatrix} \alpha_1(v_j) \\ \alpha_2(v_j) \\ \vdots \\ \alpha_m(v_j) \end{bmatrix} = c_1\begin{bmatrix} \alpha_1(v_1) \\ \alpha_2(v_1) \\ \vdots \\ \alpha_m(v_1) \end{bmatrix} + c_2\begin{bmatrix} \alpha_1(v_2) \\ \alpha_2(v_2) \\ \vdots \\ \alpha_m(v_2) \end{bmatrix} +\dots+ c_k\begin{bmatrix} \alpha_1(v_k) \\ \alpha_2(v_k) \\ \vdots \\ \alpha_m(v_k) \end{bmatrix} $$ Hence the columns $k+1,k+2,\dots,m$ of $A$ are linear combination of the columns $1,2,\dots,k$, which implies that the rank of $A$ is at most $k$.

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  • $\begingroup$ Thanks a lot for editing my post and for your proof. I'm facepalming pretty hard right now. $\endgroup$ – JtSpKg Apr 5 '18 at 7:42

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