Let u1 and u2 be vectors in R^3: u1 = (1, 2, 3) and u2 = (1, -1, 1). Solve the following:
1. Show that dim(U) = 2 for U = Span{u1, u2}.
2. Verify that u1 and u2 are not orthogonal.
3. Find an orthogonal basis for U containing u1.
4. Let b = (0, 0, 1). Compute proj_U(b).
Let's work through the questions:
Show that dim ( U ) = 2 \text{dim}(U) = 2 dim ( U ) = 2 :
Vectors u 1 = ( 1 2 3 ) u_1 = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} u 1 = 1 2 3 u 2 = ( 1 − 1 1 ) u_2 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} u 2 = 1 − 1 1
Therefore, { u 1 , u 2 } \{u_1, u_2\} { u 1 , u 2 } U = Span { u 1 , u 2 } U = \text{Span}\{u_1, u_2\} U = Span { u 1 , u 2 }
Thus, dim ( U ) = 2 \text{dim}(U) = 2 dim ( U ) = 2
Show that u 1 u_1 u 1 u 2 u_2 u 2 :
Compute the dot product:
u 1 ⋅ u 2 = 1 ⋅ 1 + 2 ⋅ ( − 1 ) + 3 ⋅ 1 = 1 − 2 + 3 = 2. u_1 \cdot u_2 = 1 \cdot 1 + 2 \cdot (-1) + 3 \cdot 1 = 1 - 2 + 3 = 2. u 1 ⋅ u 2 = 1 ⋅ 1 + 2 ⋅ ( − 1 ) + 3 ⋅ 1 = 1 − 2 + 3 = 2.
Since the dot product is not zero, u 1 u_1 u 1 u 2 u_2 u 2
Find an orthogonal basis of U U U u 1 u_1 u 1 :
Use the Gram-Schmidt process:
Start with v 1 = u 1 = ( 1 2 3 ) v_1 = u_1 = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} v 1 = u 1 = 1 2 3
Compute v 2 v_2 v 2 v 2 = u 2 − u 2 ⋅ v 1 v 1 ⋅ v 1 v 1 v_2 = u_2 - \frac{u_2 \cdot v_1}{v_1 \cdot v_1} v_1 v 2 = u 2 − v 1 ⋅ v 1 u 2 ⋅ v 1 v 1
Calculate:
u 2 ⋅ v 1 = 2 , v 1 ⋅ v 1 = 14 u_2 \cdot v_1 = 2, \quad v_1 \cdot v_1 = 14 u 2 ⋅ v 1 = 2 , v 1 ⋅ v 1 = 14 v 2 = ( 1 − 1 1 ) − 2 14 ( 1 2 3 ) = ( 1 − 1 1 ) − ( 1 7 2 7 3 7 ) = ( 6 7 − 9 7 4 7 ) v_2 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} - \frac{2}{14} \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} - \begin{pmatrix} \frac{1}{7} \\ \frac{2}{7} \\ \frac{3}{7} \end{pmatrix} = \begin{pmatrix} \frac{6}{7} \\ -\frac{9}{7} \\ \frac{4}{7} \end{pmatrix} v 2 = 1 − 1 1 − 14 2 1 2 3 = 1 − 1 1 − 7 1 7 2 7 3 = 7 6 − 7 9 7 4
Orthogonal basis: { ( 1 2 3 ) , ( 6 7 − 9 7 4 7 ) } \left\{ \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \begin{pmatrix} \frac{6}{7} \\ -\frac{9}{7} \\ \frac{4}{7} \end{pmatrix} \right\} ⎩ ⎨ ⎧ 1 2 3 , 7 6 − 7 9 7 4 ⎭ ⎬ ⎫
Find proj U ( b ) \text{proj}_U(b) proj U ( b ) b = ( 0 0 1 ) b = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} b = 0 0 1 :
proj v 1 ( b ) \text{proj}_{v_1}(b) proj v 1 ( b ) proj v 1 ( b ) = b ⋅ v 1 v 1 ⋅ v 1 v 1 = 3 14 ( 1 2 3 ) = ( 3 14 6 14 9 14 ) \text{proj}_{v_1}(b) = \frac{b \cdot v_1}{v_1 \cdot v_1} v_1 = \frac{3}{14} \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} \frac{3}{14} \\ \frac{6}{14} \\ \frac{9}{14} \end{pmatrix} proj v 1 ( b ) = v 1 ⋅ v 1 b ⋅ v 1 v 1 = 14 3 1 2 3 = 14 3 14 6 14 9 proj v 2 ( b ) \text{proj}_{v_2}(b) proj v 2 ( b ) proj v 2 ( b ) = b ⋅ v 2 v 2 ⋅ v 2 v 2 \text{proj}_{v_2}(b) = \frac{b \cdot v_2}{v_2 \cdot v_2} v_2 proj v 2 ( b ) = v 2 ⋅ v 2 b ⋅ v 2 v 2
Calculate b ⋅ v 2 = 4 7 b \cdot v_2 = \frac{4}{7} b ⋅ v 2 = 7 4 v 2 ⋅ v 2 = 133 49 v_2 \cdot v_2 = \frac{133}{49} v 2 ⋅ v 2 = 49 133 proj v 2 ( b ) = 4 7 133 49 ( 6 7 − 9 7 4 7 ) = ( 24 133 − 36 133 16 133 ) \text{proj}_{v_2}(b) = \frac{\frac{4}{7}}{\frac{133}{49}} \begin{pmatrix} \frac{6}{7} \\ -\frac{9}{7} \\ \frac{4}{7} \end{pmatrix} = \begin{pmatrix} \frac{24}{133} \\ -\frac{36}{133} \\ \frac{16}{133} \end{pmatrix} proj v 2 ( b ) = 49 133 7 4 7 6 − 7 9 7 4 = 133 24 − 133 36 133 16
proj U ( b ) = proj v 1 ( b ) + proj v 2 ( b ) = ( 81 133 30 133 151 133 ) \text{proj}_U(b) = \text{proj}_{v_1}(b) + \text{proj}_{v_2}(b) = \begin{pmatrix} \frac{81}{133} \\ \frac{30}{133} \\ \frac{151}{133} \end{pmatrix} proj U ( b ) = proj v 1 ( b ) + proj v 2 ( b ) = 133 81 133 30 133 151
These solutions provide a complete analysis of the problem regarding the span and orthogonality of vectors in R 3 \mathbb{R}^3 R 3