Orthonormal Basis (Gram–Schmidt)

Linear Algebra, Vectors

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Intro: We build an orthonormal set $u_1,u_2,\dots$ from input vectors $v_1,v_2,\dots$ by subtracting projections and normalizing. Exact fractions and radicals when possible.

Worked example

$v1=(1,1,0), v2=(1,0,1)$

Normalize the first vector:

$$\|v_1\|=\sqrt{1^2+1^2+0^2}=\sqrt{2},\quad u_1=\dfrac{v_1}{\|v_1\|}=\left(\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}},0\right).$$

Project $v_2$ onto $u_1$ and subtract:

$$v_2\cdot u_1=\dfrac{1\cdot1+0\cdot1+1\cdot0}{\sqrt{2}}=\dfrac{1}{\sqrt{2}},\qquad \operatorname{proj}_{u_1}v_2=(v_2\cdot u_1)u_1=\dfrac{1}{2}(1,1,0).$$

$$w_2=v_2-\operatorname{proj}_{u_1}v_2=(1,0,1)-\left(\tfrac{1}{2},\tfrac{1}{2},0\right)=\left(\tfrac{1}{2},-\tfrac{1}{2},1\right).$$

Normalize $w_2$:

$$\|w_2\|=\sqrt{\tfrac{1}{4}+\tfrac{1}{4}+1}=\sqrt{\tfrac{3}{2}}=\tfrac{\sqrt{6}}{2},\quad u_2=\dfrac{w_2}{\|w_2\|}=\left(\tfrac{1}{\sqrt{6}},-\tfrac{1}{\sqrt{6}},\tfrac{2}{\sqrt{6}}\right)=\left(\tfrac{1}{\sqrt{6}},-\tfrac{1}{\sqrt{6}},\tfrac{\sqrt{6}}{3}\right).$$

Orthonormality checks:

$$\|u_1\|=\|u_2\|=1,\qquad u_1\cdot u_2=\tfrac{1}{\sqrt{2}\sqrt{6}}-\tfrac{1}{\sqrt{2}\sqrt{6}}+0=0\;\checkmark$$

Result:

$$\boxed{u_1=\left(\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}},0\right)},\quad \boxed{u_2=\left(\tfrac{1}{\sqrt{6}},-\tfrac{1}{\sqrt{6}},\tfrac{\sqrt{6}}{3}\right)}.$$

$v1=(1,1,0), v2=(1,0,1), v3=(1,1,1)$

From the previous example: $$u_1=\left(\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}},0\right),\quad u_2=\left(\tfrac{1}{\sqrt{6}},-\tfrac{1}{\sqrt{6}},\tfrac{\sqrt{6}}{3}\right).$$

Remove components of $v_3$ along $u_1,u_2$:

$$v_3\cdot u_1=\tfrac{2}{\sqrt{2}}=\sqrt{2}\Rightarrow \operatorname{proj}_{u_1}v_3=(1,1,0).$$

$$v_3\cdot u_2=\tfrac{1-1+2}{\sqrt{6}}=\tfrac{2}{\sqrt{6}}=\tfrac{\sqrt{6}}{3}\Rightarrow \operatorname{proj}_{u_2}v_3=\tfrac{\sqrt{6}}{3}u_2=\left(\tfrac{1}{3},-\tfrac{1}{3},\tfrac{2}{3}\right).$$

$$w_3=v_3-\operatorname{proj}_{u_1}v_3-\operatorname{proj}_{u_2}v_3=(1,1,1)-(1,1,0)-\left(\tfrac{1}{3},-\tfrac{1}{3},\tfrac{2}{3}\right)=\left(-\tfrac{1}{3},\tfrac{1}{3},\tfrac{1}{3}\right).$$

Normalize $w_3$:

$$\|w_3\|=\sqrt{3\cdot(1/9)}=\tfrac{1}{\sqrt{3}},\quad u_3=\dfrac{w_3}{\|w_3\|}=\left(-\tfrac{1}{\sqrt{3}},\tfrac{1}{\sqrt{3}},\tfrac{1}{\sqrt{3}}\right).$$

Final orthonormal basis:

$$\boxed{\{u_1,u_2,u_3\}}=\left\{\left(\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}},0\right),\left(\tfrac{1}{\sqrt{6}},-\tfrac{1}{\sqrt{6}},\tfrac{\sqrt{6}}{3}\right),\left(-\tfrac{1}{\sqrt{3}},\tfrac{1}{\sqrt{3}},\tfrac{1}{\sqrt{3}}\right)\right\}.$$

Orthogonality spot-checks:

$$u_i\cdot u_j=0\;(i\ne j),\;\|u_i\|=1\;\forall i\;\checkmark$$

FAQs

Linear dependence?

If a vector becomes zero after subtracting projections, it is dependent on previous ones and is skipped; the remaining nonzero set is still orthonormal.

Order sensitivity?

Yes. Different orders of the input vectors can produce different orthonormal sets, though all span the same subspace.

Numerical stability?

For floating-point data, Modified Gram–Schmidt improves stability; QR via Householder is even more robust.

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