91影视

The matrix P of the orthogonal projection onto V=span$\left\{u_1{u}_2,.....u_m\right\}$is given by, $\mathbf{P}\mathbf{=}{\mathbf{QQ}}^{\mathbf{T}}\mathbf{}\mathbf{,}\mathbf{}\mathbf{where}\mathbf{}\mathbf{Q}\mathbf{=}\mathbf{\{}{\stackrel{\mathbf{鈬赌}}{\mathbf{u}}}_{\mathbf{1}}\mathbf{}{\stackrel{\mathbf{鈬赌}}{\mathbf{u}}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\stackrel{\mathbf{鈬赌}}{\mathbf{u}}}_{\mathbf{m}}\mathbf{\}}$.

Where the spanning set is orthonormal basis.

Because the given basis is not the orthonormal basis. So, Gram-Schmidt method will be applied here to make them orthogonal.

Consider a basis of a subspace Vof ${\mathit{R}}^{\mathbf{n}}\mathbf{}\mathbf{for}\mathbf{}\mathit{j}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\mathit{m}$for we resolve the vector $\stackrel{\mathbf{鈬赌}}{{\mathbf{v}}_{\mathbf{j}}}$into its components parallel and perpendicular to the span of the preceding vectors $\stackrel{\mathbf{鈬赌}}{{\mathbf{v}}_{\mathbf{1}}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\stackrel{\mathbf{鈬赌}}{{\mathbf{v}}_{\mathbf{j}\mathbf{-}\mathbf{1}}}$.

Then,

localid="1660109869517" $\stackrel{\mathbf{鈬赌}}{{\mathbf{u}}_{\mathbf{1}}}\mathbf{=}\frac{\mathbf{1}}{||{\stackrel{鈬赌}{v}}_1||}{\stackrel{\mathbf{鈬赌}}{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}{\stackrel{\mathbf{鈬赌}}{\mathbf{u}}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{||{{\stackrel{鈬赌}{v}}_1}^{鈯�}||}{{\stackrel{\mathbf{鈬赌}}{\mathbf{v}}}_{\mathbf{2}}}^{\mathbf{鈯�}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{鈬赌}}{\mathbf{u}}}_{\mathbf{j}}\mathbf{=}\frac{\mathbf{1}}{||{{\stackrel{鈬赌}{v}}_j}^{鈯�}||}{{\stackrel{\mathbf{鈬赌}}{\mathbf{v}}}_{\mathbf{j}}}^{\mathbf{鈯�}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{鈬赌}}{\mathbf{u}}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{1}}{||{{\stackrel{鈬赌}{v}}_m}^{鈯�}||}{{\stackrel{\mathbf{鈬赌}}{\mathbf{v}}}_{\mathbf{m}}}^{\mathbf{鈯�}}$

Obtain the value of ${\stackrel{鈬赌}{u}}_1\mathrm{and}{\stackrel{鈬赌}{\mathrm{u}}}_2$ according toGram-Schmidt process. And then put the values in the formula given below.

$$\begin{gathered} {\stackrel{鈬赌}{u}}_1=\frac{{\stackrel{鈬赌}{v}}_1}{||\stackrel{鈬赌}{v}||} \\ {\stackrel{鈬赌}{u}}_2=\frac{{\stackrel{鈬赌}{v}}_2-\left({\stackrel{鈬赌}{u}}_1-{\stackrel{鈬赌}{u}}_2\right){\stackrel{鈬赌}{u}}_1}{||{\stackrel{鈬赌}{v}}_2-\left({\stackrel{鈬赌}{u}}_1-{\stackrel{鈬赌}{u}}_2\right){\stackrel{鈬赌}{u}}_1||} \end{gathered}$$ 鈥︹�� (1)

鈥︹�� (2)

Since, it gives

$$\begin{gathered} ||{\stackrel{鈬赌}{v}}_1||=\sqrt{1^2+1^2+1^2+1^2} \\ =\sqrt{4} \\ =2 \\ {\stackrel{鈬赌}{u}}_1=1/2\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right] \end{gathered}$$

Now, here it needs to find out the values of ${\stackrel{鈬赌}{v}}_2-\left({\stackrel{鈬赌}{u}}_1.{\stackrel{鈬赌}{v}}_2\right){\stackrel{鈬赌}{u}}_1\mathrm{and}||{\stackrel{鈬赌}{\mathrm{v}}}_2-\left({\stackrel{鈬赌}{\mathrm{u}}}_1.{\stackrel{鈬赌}{\mathrm{v}}}_2\right){\stackrel{鈬赌}{\mathrm{u}}}_1||$to obtain the value of ${\stackrel{鈬赌}{u}}_2$.

Consider the equations below.

$$\begin{gathered} {\stackrel{鈬赌}{u}}_1.{\stackrel{鈬赌}{v}}_2=4 \\ \left({\stackrel{鈬赌}{u}}_1.{\stackrel{鈬赌}{v}}_2\right){\stackrel{鈬赌}{u}}_1=\left[\begin{array}{c}2\\ 2\\ 2\\ 2\end{array}\right] \\ {\stackrel{鈬赌}{v}}_2-\left({\stackrel{鈬赌}{u}}_1.{\stackrel{鈬赌}{v}}_2\right){\stackrel{鈬赌}{u}}_1=\left[\begin{array}{c}1\\ 9\\ -5\\ 3\end{array}\right]-\left[\begin{array}{c}2\\ 2\\ 2\\ 2\end{array}\right]=\left[\begin{array}{c}-1\\ 7\\ -7\\ 1\end{array}\right] \\ ||{\stackrel{鈬赌}{v}}_2-\left({\stackrel{鈬赌}{u}}_1.{\stackrel{鈬赌}{v}}_2\right)\stackrel{鈬赌}{u}||=\sqrt{1+49+149} \\ =\sqrt{100} \\ =10 \\ {\stackrel{鈬赌}{u}}_2=1/10\left[\begin{array}{c}-1\\ 7\\ -7\\ 1\end{array}\right] \\ \mathrm{Thus},\mathrm{the}\mathrm{orthonormal}\mathrm{vectors}\mathrm{are}\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right],\left[\begin{array}{c}-1/10\\ 7/10\\ -7/10\\ 1/10\end{array}\right]. \end{gathered}$$

Consider the projection matrix below.

$$\begin{gathered} P=Q{Q}^T \\ =\left[\begin{array}{cc}{\stackrel{鈬赌}{u}}_1& {\stackrel{鈬赌}{u}}_2\end{array}\right]\left[\begin{array}{c}{\stackrel{鈬赌}{u}}_1\\ {\stackrel{鈬赌}{u}}_2\end{array}\right] \\ =\left[\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right]\left[\begin{array}{c}-1/10\\ 7/10\\ -7/10\\ 1/10\end{array}\right]\right]\left[\begin{array}{cccc}1/2& 1/2& 1/2& 1/2\\ -1/10& 7/10& -7/10& 1/10\end{array}\right] \\ =\frac{1}{100}\left[\begin{array}{cccc}26& 18& 32& 24\\ 18& 74& -24& 32\\ 32& -24& 74& 18\\ 24& 32& 18& 26\end{array}\right] \\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{matrix}\mathrm{is}\frac{1}{100}\left[\begin{array}{cccc}26& 18& 32& 24\\ 18& 74& -24& 32\\ 32& -24& 74& 18\\ 24& 32& 18& 26\end{array}\right]. \end{gathered}$$