Following (not word for word) Theorem 2 page 6 of Kaplansky's $\textit{"Linear algebra and Geometry a second course"}$

Let $V$ be an inner product space with inner product $<-,->$ and let $S$ be a finite dimensional subspace that is non-singular relative to the induced inner product. Then $V$ is the direct sum of $S$ and its orthogonal complement $S'$.

Proof: Since $S$ is nonsingular we have that $S \cap S'=0$. It remains for us to prove that $S$ and $S'$ span $V$. Let $x \in V$; we must express $x$ as a sum of a vector in $S$ and a vector orthogonal to $S$. Pick a basis of $S$, say $u_1,...,u_n$. To solve our problem, we must find scalars $a_1,...,a_n \in K$ an an element $y \in S'$ such that $$x = a_1u_1+...+a_nu_n+y$$.

Suppose $<u_i,u_j>=c_{ij}$. Then we have that $$c_{i1}a_1+c_{i2}a_2+...+c_{in}a_n=<x,u_i> \text{ for each i}$$

We thus have a system of $n$ equations for the $n$ unknowns $a_1,..,a_n$. The matrix of coefficients $(c_{ij})$ is nonsingular since $<,>$ is non-singular. Hence this system of equations has a unique solution and our theorem is proved.

So i'm still trying to see why this proves the theorem. First of all never found the element $y \in S'$ that we were supposedly searching for. Not just that, but I guess I just don't understand the punchline of the proof. I understand everything up to the point where it says "thus the theorem is proved", at which point i go, "huh?".



Many aspects of the proof are strange, either due to your rewording of the original or the choices of the author. First of all, rather than saying "to solve our problem, we must find scalars $a_1,...,a_n \in K$ an an element $y \in S'$ such that $ x = a_1u_1+...+a_nu_n+y $", I think that the intended sentiment is better expressed as

In order to solve our problem (i.e. in order show that $S$ and $S'$ span $V$), we must show that it is possible to find scalars $a_1,...,a_n \in K$ an an element $y \in S'$ such that $x = a_1u_1+...+a_nu_n+y$.

Next, I'm not sure if you understand or if the author makes clear where exactly these equations in terms of the $c_{ij}$ come from. The idea is that we take the equation $x = a_1u_1+...+a_nu_n+y$ and take an inner product of both sides of the equation with each vector $u_i$. That is, each of the $i$ equations in terms of the $c_{ij}$ is an expanded form of the equation $$ \langle x, u_i \rangle = \langle a_1 u_1 + \cdots + a_n u_n + y,u_i\rangle, $$ To rewrite this, we take the expanded form and replace each $\langle u_j,u_i \rangle$ with $c_{ij}$ and note that $\langle u_i, y \rangle = 0$ from the fact that $y \in S'$. Now, the author goes on to explain that the resulting system of equations $$ c_{i1}a_1+c_{i2}a_2+\cdots +c_{in}a_n=\langle x,u_i\rangle \text{ for each } i $$ has a unique solution for the coefficients $a_1,\dots,a_n$. He therefore concludes that our initial system of equations $x = a_1u_1+\cdots +a_nu_n+y$ (where $y$ is specified to be from $S'$) must have a solution.

In fact, I believe that this step is a flawed line of argument. It is clear that if a satisfactory solution to $x = a_1u_1+\cdots +a_nu_n+y$ exists, then the coefficients $a_1,\dots,a_n$ must satisfy the equations in terms of $c_{ij}$. However, it is not clear at this point that the converse holds, i.e. that if such coefficients exist, then the original equation has a solution.

We could show that the converse does hold as follows. Suppose that the coefficients $a_1,\dots,a_n$ solve our equations in terms of $c_i$. Let $y = x - (a_1 u_1 + \cdots + a_n u_n)$. It is clear that we indeed have $$ x = a_1 u_1 + \cdots + a_n u_n + y, $$ but we must show that $y \in S'$. To show that this is the case, it suffices to note that for each $i$, we have $\langle x,u_i \rangle = 0$. Indeed, we see that $$ \langle y,u_i \rangle = \langle x - (a_1u_1 + \cdots + a_n u_n), u_i \rangle \\ = \langle x,u_i \rangle - (c_{i1}a_1 + \cdots + c_{in}a_n) = 0. $$

  • $\begingroup$ Thanks!!!!!!!!!! $\endgroup$
    – Num Toez
    May 10 at 16:09

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