Inner Products, Norms & Gram–Schmidt

1. 1
   Step 1:

   $$\overline{\langle v\vert u\rangle}=\overline{\sum_i\overline{v_i}\,u_i}=\sum_i v_i\,\overline{u_i}=\sum_i\overline{u_i}\,v_i=\langle u\vert v\rangle$$

   (i) Conjugation respects sums and products, $\bar{\bar z}=z$, and multiplication commutes.
2. 2
   Step 2:

   $$\langle u\vert av+bw\rangle=\sum_i\overline{u_i}(av_i+bw_i)=a\sum_i\overline{u_i}v_i+b\sum_i\overline{u_i}w_i$$

   (ii) Distribute inside each term and pull the constants $a,b$ out of the sums.
3. 3
   Step 3:

   $$=a\langle u\vert v\rangle+b\langle u\vert w\rangle$$

   Recognise the two sums as inner products.
4. 4
   Step 4:

   $$\langle v\vert v\rangle=\sum_i\overline{v_i}v_i=\sum_i\lvert v_i\rvert^2\ge0$$

   (iii) Each term is a modulus squared, a non-negative real number.
5. 5
   Step 5:

   $$\sum_i\lvert v_i\rvert^2=0\iff v_i=0\ \forall i\iff v=0$$

   A sum of non-negative reals vanishes only when every term does, and $\lvert z\rvert=0$ only for $z=0$.
6. 6
   Step 6:

   $$\langle au+bw\vert v\rangle=\overline{\langle v\vert au+bw\rangle}=\overline{a\langle v\vert u\rangle+b\langle v\vert w\rangle}$$

   Antilinearity: flip the arguments with (i), then use (ii).
7. 7
   Step 7:

   $$=\bar a\,\overline{\langle v\vert u\rangle}+\bar b\,\overline{\langle v\vert w\rangle}=\bar a\langle u\vert v\rangle+\bar b\langle w\vert v\rangle$$

   Conjugation respects sums and products; apply (i) once more.

∎

1. 1
   Step 1:

   $$v=0:\quad \lvert\langle u\vert 0\rangle\rvert=0=\lVert u\rVert\cdot0$$

   If $v=0$ both sides vanish (linearity in the second slot gives $\langle u\vert0\rangle=0$) and $u,0$ are dependent, so assume $v\ne0$ from now on.
2. 2
   Step 2:

   $$\lambda:=\frac{\langle v\vert u\rangle}{\lVert v\rVert^2},\qquad w:=u-\lambda v$$

   Define the component of $u$ along $v$ and the remainder $w$; $\lVert v\rVert^2>0$ by positive-definiteness.
3. 3
   Step 3:

   $$\langle v\vert w\rangle=\langle v\vert u\rangle-\lambda\langle v\vert v\rangle=\langle v\vert u\rangle-\frac{\langle v\vert u\rangle}{\lVert v\rVert^2}\lVert v\rVert^2=0$$

   Linearity in the second slot; the remainder is orthogonal to $v$. Taking conjugates, $\langle w\vert v\rangle=0$ too.
4. 4
   Step 4:

   $$\lVert u\rVert^2=\langle w+\lambda v\vert w+\lambda v\rangle=\langle w\vert w\rangle+\lambda\langle w\vert v\rangle+\bar\lambda\langle v\vert w\rangle+\bar\lambda\lambda\langle v\vert v\rangle$$

   Write $u=w+\lambda v$ and expand: linear in the second slot, antilinear in the first.
5. 5
   Step 5:

   $$=\lVert w\rVert^2+\lvert\lambda\rvert^2\lVert v\rVert^2=\lVert w\rVert^2+\frac{\lvert\langle v\vert u\rangle\rvert^2}{\lVert v\rVert^2}$$

   The cross terms vanish by orthogonality; substitute $\lvert\lambda\rvert^2=\lvert\langle v\vert u\rangle\rvert^2/\lVert v\rVert^4$.
6. 6
   Step 6:

   $$0\le\lVert w\rVert^2=\lVert u\rVert^2-\frac{\lvert\langle v\vert u\rangle\rvert^2}{\lVert v\rVert^2}$$

   Rearrange; a squared norm is non-negative.
7. 7
   Step 7:

   $$\lvert\langle u\vert v\rangle\rvert^2=\lvert\langle v\vert u\rangle\rvert^2\le\lVert u\rVert^2\lVert v\rVert^2$$

   Multiply by $\lVert v\rVert^2>0$; conjugate symmetry gives $\lvert\langle u\vert v\rangle\rvert=\lvert\overline{\langle v\vert u\rangle}\rvert=\lvert\langle v\vert u\rangle\rvert$.
8. 8
   Step 8:

   $$\lvert\langle u\vert v\rangle\rvert\le\lVert u\rVert\,\lVert v\rVert$$

   Both sides are non-negative, so take square roots.
9. 9
   Step 9:

   $$\text{equality}\iff\lVert w\rVert=0\iff u=\lambda v$$

   Equality holds exactly when the discarded term $\lVert w\rVert^2$ is zero, i.e. when $u$ is a multiple of $v$; conversely if $u=\mu v$ then $\lvert\langle u\vert v\rangle\rvert=\lvert\mu\rvert\lVert v\rVert^2=\lVert u\rVert\lVert v\rVert$.

∎

1. 1
   Step 1:

   $$\lVert u+v\rVert^2=\langle u+v\vert u+v\rangle=\langle u\vert u\rangle+\langle u\vert v\rangle+\langle v\vert u\rangle+\langle v\vert v\rangle$$

   Expand using linearity in the second slot and antilinearity in the first (all coefficients are $1$).
2. 2
   Step 2:

   $$=\lVert u\rVert^2+\langle u\vert v\rangle+\overline{\langle u\vert v\rangle}+\lVert v\rVert^2=\lVert u\rVert^2+2\,\mathrm{Re}\langle u\vert v\rangle+\lVert v\rVert^2$$

   Conjugate symmetry, and $z+\bar z=2\,\mathrm{Re}\,z$.
3. 3
   Step 3:

   $$\le\lVert u\rVert^2+2\lvert\langle u\vert v\rangle\rvert+\lVert v\rVert^2$$

   $\mathrm{Re}\,z\le\lvert z\rvert$ for every complex $z$, since $\lvert z\rvert^2=(\mathrm{Re}\,z)^2+(\mathrm{Im}\,z)^2$.
4. 4
   Step 4:

   $$\le\lVert u\rVert^2+2\lVert u\rVert\lVert v\rVert+\lVert v\rVert^2=\bigl(\lVert u\rVert+\lVert v\rVert\bigr)^2$$

   Cauchy–Schwarz, then recognise a perfect square.
5. 5
   Step 5:

   $$\lVert u+v\rVert\le\lVert u\rVert+\lVert v\rVert$$

   Take non-negative square roots.

∎

1. 1
   Step 1:

   $$\lVert u+v\rVert^2=\lVert u\rVert^2+2\,\mathrm{Re}\langle u\vert v\rangle+\lVert v\rVert^2$$

   First two lines of the triangle-inequality proof.
2. 2
   Step 2:

   $$=\lVert u\rVert^2+2\,\mathrm{Re}(0)+\lVert v\rVert^2=\lVert u\rVert^2+\lVert v\rVert^2$$

   Orthogonality kills the cross term.
3. 3
   Step 3:

   $$\Bigl\lVert\sum_{i=1}^{k}v_i\Bigr\rVert^2=\sum_{i,j}\langle v_i\vert v_j\rangle=\sum_{i}\langle v_i\vert v_i\rangle=\sum_i\lVert v_i\rVert^2$$

   Expand the double sum; all terms with $i\ne j$ vanish by pairwise orthogonality.

∎

1. 1
   Step 1:

   $$v=\sum_{j=1}^{n}c_j\,e_j$$

   Expand $v$ in the basis (coordinates exist and are unique).
2. 2
   Step 2:

   $$\langle e_i\vert v\rangle=\sum_{j=1}^{n}c_j\langle e_i\vert e_j\rangle=\sum_{j=1}^{n}c_j\,\delta_{ij}=c_i$$

   Take the inner product with $e_i$, use linearity in the second slot and orthonormality.
3. 3
   Step 3:

   $$\Rightarrow\ v=\sum_{i=1}^{n}\langle e_i\vert v\rangle\,e_i$$

   Substitute $c_i=\langle e_i\vert v\rangle$ back into the expansion.
4. 4
   Step 4:

   $$\lVert v\rVert^2=\sum_{i=1}^{n}\lVert c_ie_i\rVert^2=\sum_{i=1}^{n}\lvert c_i\rvert^2\lVert e_i\rVert^2=\sum_{i=1}^{n}\lvert\langle e_i\vert v\rangle\rvert^2$$

   The vectors $c_ie_i$ are pairwise orthogonal, so Pythagoras applies; $\lVert ce\rVert=\lvert c\rvert\lVert e\rVert$ and $\lVert e_i\rVert=1$.

∎

1. 1
   Step 1:

   $$k=1:\ u_1=v_1\ne0,\ \ e_1=\frac{v_1}{\lVert v_1\rVert},\ \ \lVert e_1\rVert=1,\ \ \operatorname{span}\{e_1\}=\operatorname{span}\{v_1\}$$

   Base case: an independent list contains no zero vector, and a nonzero multiple spans the same line.
2. 2
   Step 2:

   $$\text{assume: } e_1,\dots,e_{k-1}\ \text{orthonormal},\ \ \operatorname{span}\{e_j\}_{j<k}=\operatorname{span}\{v_j\}_{j<k}$$

   Induction hypothesis for $k-1$.
3. 3
   Step 3:

   $$u_k=0\ \Rightarrow\ v_k=\sum_{j<k}\langle e_j\vert v_k\rangle e_j\in\operatorname{span}\{v_1,\dots,v_{k-1}\}$$

   If the remainder vanished, $v_k$ would be a combination of earlier $e_j$, hence (hypothesis) of earlier $v_j$, contradicting independence. So $u_k\ne0$ and $e_k$ is well defined with $\lVert e_k\rVert=1$.
4. 4
   Step 4:

   $$\langle e_i\vert u_k\rangle=\langle e_i\vert v_k\rangle-\sum_{j<k}\langle e_j\vert v_k\rangle\langle e_i\vert e_j\rangle\qquad(i<k)$$

   Take the inner product of $u_k$ with an earlier $e_i$; linearity in the second slot.
5. 5
   Step 5:

   $$=\langle e_i\vert v_k\rangle-\sum_{j<k}\langle e_j\vert v_k\rangle\,\delta_{ij}=\langle e_i\vert v_k\rangle-\langle e_i\vert v_k\rangle=0$$

   Orthonormality of $e_1,\dots,e_{k-1}$ leaves only the $j=i$ term.
6. 6
   Step 6:

   $$\langle e_i\vert e_k\rangle=\frac{\langle e_i\vert u_k\rangle}{\lVert u_k\rVert}=0\quad(i<k)$$

   Scaling by the positive real $1/\lVert u_k\rVert$; hence $e_1,\dots,e_k$ is orthonormal.
7. 7
   Step 7:

   $$e_k=\frac{1}{\lVert u_k\rVert}\Bigl(v_k-\sum_{j<k}\langle e_j\vert v_k\rangle e_j\Bigr)\in\operatorname{span}\{v_1,\dots,v_k\}$$

   Each $e_j$ with $j<k$ lies in $\operatorname{span}\{v_1,\dots,v_{k-1}\}$ by hypothesis, so $\operatorname{span}\{e_1..e_k\}\subseteq\operatorname{span}\{v_1..v_k\}$.
8. 8
   Step 8:

   $$v_k=\lVert u_k\rVert\,e_k+\sum_{j<k}\langle e_j\vert v_k\rangle e_j\in\operatorname{span}\{e_1,\dots,e_k\}$$

   Solve the definition of $u_k$ for $v_k$; together with the hypothesis this gives the reverse inclusion, so the spans are equal. Induction completes the proof.

∎