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An alternate approach is induction on $n=\dim(W)$. The base case $n=0$ is clear, so the hard part is the induction step. For this, it's enough to prove the following result: if $M$ is a closed subspace of $V$ and $x\in V,x\not\in M$, then $M+\mathbb{C}x$ is also a closed subspace. Indeed, by the Hahn-Banach theorem there is a continuous linear functional ...

2

This is in fact never true; suppose you have a matrix $A$ of rank $k$. Then, we can take linearly independent $v_1, \dots, v_k \in \text{Ran}(A)$, so there exist $u_1, \dots, u_k$ such that $Au_i = v_i$ for each $i$. Suppose for a contradiction the statement is true for some $p < k$, so there is a sequence $(B_n)$ of $n \times n$ norm 1 matrices of rank ...

2

The indices are different because the sums are independent from one another. If one intends to multiply the sums, then this distinction is a critical one. Suppose that we were to multiply the summations $\sum_{n=1}^2a_n$ and $\sum_{n=1}^2b_n$ and naively failed to make this distinction. Then, we would have incorrectly \begin{align} ...

1

Orthornormality refers to the basis $e_i$. When a basis is orthonormal it means the inner product between any two elements of the basis $e_i,e_j$ is $\langle e_i, e_j \rangle = \delta_{ij}$ (see kronecker delta). More generally, two vectors $u,v$ are orthogonal if $\langle u, v \rangle = 0$. The normality part comes from elements of the basis having norm ...

1

That's because it is notationally bad and depending on how you read it, it may even give different results. For example, consider simple vectors in $\mathbb{R}^{3}$ given by $v=a_{1}e_{1}+a_{2}e_{2}+a_{3}e_{3}=\sum_{n=1}^{3}a_{n}e_{n}$. Then if we use same index, \begin{aligned} ...

1

Sure, you can do it this way. If $y \neq 0$ is killed by all of the $f_i$, then if $x$ is in a basic weak neighborhood corresponding to the $f_i$, then $x+ty$ is also in this neighborhood for all $t$, since $f_i(x) = f_i(x+ty)$. In particular, the weak neighborhood will be unbounded. Ultimately, I think the point is that the kernel of a linear functional ...

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