Friedberg book on linear algebra theorem 6.6

Question

in Friedberg book on linear algebra - 4th edition, Theorem 6.6 on page 350

Theorem 6.6: Let $W$ be a finite dimensional subspace of an inner product space $V$, and let $y\in V$. Then there exist unique vectors $u\in W$ and $z\in W^{\perp}$ such that $y=u+z$. Furthermore, if $\{v_1,v_2,...,v_k\}$ is an orthonormal basis for $W$, then

$$u=\sum_{i=1}^{k} \langle y,v_i\rangle v_i$$

Proof: Let $\{v_1,v_2,...,v_k\}$ be an orthonormal basis for $W$, let $u$ be as defined in the preceeding equarion, and let $z=y-u$. Clearly $u\in W$ and $y=u+z$.

To show that $z\in W^{\perp}$, it suffices to show that $z$ is orthogonal to each $v_j$. For any $j$, we have

$$\langle z,v_j\rangle =\langle \left( y-\sum_{i=1}^{k} \langle y,v_i\rangle v_i \right),v_j\rangle=\langle y,v_j\rangle - \sum_{i=1}^{k}\langle y,v_i\rangle\langle v_i,v_j\rangle=\langle y,v_j\rangle-\langle y,v_j\rangle=0$$

....

My question is: In the proof, he used the second part of the theorem ( which is a corollary for the first part) to prove the first part. How can it be a proof ?

Answer 1

Well, no: he only defined a vector by the formula of the second part and showed what he did show. This is not using part of what has to be proved in the proof, which is completely wrong in any logic I can think of.

So he defines $\;u\;$ as he does, and then he shows $\;z=y-u\in W^\perp\;$ , and since trivially $\;y=u+z\;$ we are done except for the uniqueness part, which I pressume is what must be in the part where you wrote periods ...

Nothing wrong here.