施密特正交化(Gram-Schmidt method to orthonormalize vectors)

施密特正交化算法

Give an arbitrary basis \begin{Bmatrix} u_1, u_2,\cdots ,u_n \end{Bmatrix} for an n-dimensional inner product space V, we can constructs an orthogonal basis \begin{Bmatrix} v_1, v_2,\cdots ,v_n \end{Bmatrix} for V.

the Gram-Schmidt algorithm is:

Step 1: Let v_1 = u_1

Step 2: Let v_2=u_2-\frac{\left \langle u_2,v_1 \right \rangle}{\left \| v_1 \right \|^2}v_1

Step 3: Let v_3=u_3-\frac{\left \langle u_3,v_1 \right \rangle}{\left \| v_1 \right \|^2}v_1--\frac{\left \langle u_3,v_2 \right \rangle}{\left \| v_2 \right \|^2}v_2

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计算举例

Let V=R^3 with the Euclidean inner product. We will apply the Gram-Schmidt algorithm to orthogonalize the basis \begin{Bmatrix} (1,-1,1), (1,0,1),(1,1,2) \end{Bmatrix}.

Step 1

Let v_1 = u_1, so v_1=(1, -1, 1).

Step 2

Let v_2=u_2-\frac{\left \langle u_2,v_1 \right \rangle}{\left \| v_1 \right \|^2}v_1.

v_2=(1,0,1)-\frac{\left \langle (1,0,1)\cdot (1,-1,1) \right \rangle}{\left \| (1,-1,1) \right \|^2}(1,-1,1)=(1,0,1)-\frac{2}{3}(1,-1,1)=(\frac{1}{3},\frac{2}{3},\frac{1}{3}).

Step 3

Let v_3=u_3-\frac{\left \langle u_3,v_1 \right \rangle}{\left \| v_1 \right \|^2}v_1--\frac{\left \langle u_3,v_2 \right \rangle}{\left \| v_2 \right \|^2}v_2.

v_3=(1,1,2)-\frac{\left \langle (1,1,2)\cdot (1,-1,1) \right \rangle}{\left \| (1,-1,1) \right \|^2}(1,-1,1)--\frac{\left \langle (1,1,2)\cdot (\frac{1}{3},\frac{2}{3},\frac{1}{3}) \right \rangle}{\left \| (\frac{1}{3},\frac{2}{3},\frac{1}{3}) \right \|^2}(\frac{1}{3},\frac{2}{3},\frac{1}{3})=(1,1,2)-\frac{2}{3}(1,-1,1)-\frac{5}{2}(\frac{1}{3},\frac{2}{3},\frac{1}{3})=(-\frac{1}{2},0,\frac{1}{2})

You can verify that \begin{Bmatrix} (1,-1,1),(\frac{1}{3},\frac{2}{3},\frac{1}{3}),(-\frac{1}{2},0,\frac{1}{2}) \end{Bmatrix} forms an orthogonal basis for R^3. Normalizing the vectors in the orthogonal basis, we obtain the orthonormal basis

\begin{Bmatrix} (\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}),(\frac{\sqrt{6}}{6},\frac{\sqrt{6}}{3},\frac{\sqrt{6}}{6}),(-\frac{\sqrt{2}}{2},0,\frac{\sqrt{2}}{2}) \end{Bmatrix}.

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