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Sia

Question
Math
Posted 9 months ago

Let
A=[1−7−12−26−2−1−8]. 
A=\left[\begin{array}{ccc}
1 & -7 & -1 \\
2 & -2 & 6 \\
-2 & -1 & -8
\end{array}\right] \text {. }


Find an orthonormal basis of the column space of AA. \{[]],
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Answer from Sia
Posted 9 months ago
Solution by Steps
step 1
To find an orthonormal basis of the column space of matrix AA, we first need to orthogonalize the columns of AA. The matrix AA is given by: A=[1−7−12−26−2−1−8] A = \begin{bmatrix} 1 & -7 & -1 \\ 2 & -2 & 6 \\ -2 & -1 & -8 \end{bmatrix} We will use the Gram-Schmidt process to orthogonalize the columns
step 2
Let v1=[12−2]\mathbf{v}_1 = \begin{bmatrix} 1 \\ 2 \\ -2 \end{bmatrix}, v2=[−7−2−1]\mathbf{v}_2 = \begin{bmatrix} -7 \\ -2 \\ -1 \end{bmatrix}, and v3=[−16−8]\mathbf{v}_3 = \begin{bmatrix} -1 \\ 6 \\ -8 \end{bmatrix}. We start with u1=v1\mathbf{u}_1 = \mathbf{v}_1
step 3
Next, we compute u2\mathbf{u}_2 by subtracting the projection of v2\mathbf{v}_2 onto u1\mathbf{u}_1 from v2\mathbf{v}_2: u2=v2−v2⋅u1u1⋅u1u1 \mathbf{u}_2 = \mathbf{v}_2 - \frac{\mathbf{v}_2 \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1} \mathbf{u}_1 u2=[−7−2−1]−[−7−2−1]⋅[12−2][12−2]⋅[12−2][12−2] \mathbf{u}_2 = \begin{bmatrix} -7 \\ -2 \\ -1 \end{bmatrix} - \frac{\begin{bmatrix} -7 \\ -2 \\ -1 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 2 \\ -2 \end{bmatrix}}{\begin{bmatrix} 1 \\ 2 \\ -2 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 2 \\ -2 \end{bmatrix}} \begin{bmatrix} 1 \\ 2 \\ -2 \end{bmatrix} u2=[−7−2−1]−−7+(−4)+21+4+4[12−2] \mathbf{u}_2 = \begin{bmatrix} -7 \\ -2 \\ -1 \end{bmatrix} - \frac{-7 + (-4) + 2}{1 + 4 + 4} \begin{bmatrix} 1 \\ 2 \\ -2 \end{bmatrix} u2=[−7−2−1]−−99[12−2] \mathbf{u}_2 = \begin{bmatrix} -7 \\ -2 \\ -1 \end{bmatrix} - \frac{-9}{9} \begin{bmatrix} 1 \\ 2 \\ -2 \end{bmatrix} u2=[−7−2−1]+[12−2] \mathbf{u}_2 = \begin{bmatrix} -7 \\ -2 \\ -1 \end{bmatrix} + \begin{bmatrix} 1 \\ 2 \\ -2 \end{bmatrix} u2=[−60−3] \mathbf{u}_2 = \begin{bmatrix} -6 \\ 0 \\ -3 \end{bmatrix}
step 4
Now, we compute u3\mathbf{u}_3 by subtracting the projections of v3\mathbf{v}_3 onto u1\mathbf{u}_1 and u2\mathbf{u}_2 from v3\mathbf{v}_3: u3=v3−v3⋅u1u1⋅u1u1−v3⋅u2u2⋅u2u2 \mathbf{u}_3 = \mathbf{v}_3 - \frac{\mathbf{v}_3 \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1} \mathbf{u}_1 - \frac{\mathbf{v}_3 \cdot \mathbf{u}_2}{\mathbf{u}_2 \cdot \mathbf{u}_2} \mathbf{u}_2 u3=[−16−8]−[−16−8]⋅[12−2][12−2]⋅[12−2][12−2]−[−16−8]⋅[−60−3][−60−3]⋅[−60−3][−60−3] \mathbf{u}_3 = \begin{bmatrix} -1 \\ 6 \\ -8 \end{bmatrix} - \frac{\begin{bmatrix} -1 \\ 6 \\ -8 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 2 \\ -2 \end{bmatrix}}{\begin{bmatrix} 1 \\ 2 \\ -2 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 2 \\ -2 \end{bmatrix}} \begin{bmatrix} 1 \\ 2 \\ -2 \end{bmatrix} - \frac{\begin{bmatrix} -1 \\ 6 \\ -8 \end{bmatrix} \cdot \begin{bmatrix} -6 \\ 0 \\ -3 \end{bmatrix}}{\begin{bmatrix} -6 \\ 0 \\ -3 \end{bmatrix} \cdot \begin{bmatrix} -6 \\ 0 \\ -3 \end{bmatrix}} \begin{bmatrix} -6 \\ 0 \\ -3 \end{bmatrix} u3=[−16−8]−−1+12+169[12−2]−6+2445[−60−3] \mathbf{u}_3 = \begin{bmatrix} -1 \\ 6 \\ -8 \end{bmatrix} - \frac{-1 + 12 + 16}{9} \begin{bmatrix} 1 \\ 2 \\ -2 \end{bmatrix} - \frac{6 + 24}{45} \begin{bmatrix} -6 \\ 0 \\ -3 \end{bmatrix} u3=[−16−8]−279[12−2]−3045[−60−3] \mathbf{u}_3 = \begin{bmatrix} -1 \\ 6 \\ -8 \end{bmatrix} - \frac{27}{9} \begin{bmatrix} 1 \\ 2 \\ -2 \end{bmatrix} - \frac{30}{45} \begin{bmatrix} -6 \\ 0 \\ -3 \end{bmatrix} u3=[−16−8]−3[12−2]−23[−60−3] \mathbf{u}_3 = \begin{bmatrix} -1 \\ 6 \\ -8 \end{bmatrix} - 3 \begin{bmatrix} 1 \\ 2 \\ -2 \end{bmatrix} - \frac{2}{3} \begin{bmatrix} -6 \\ 0 \\ -3 \end{bmatrix} u3=[−16−8]−[36−6]+[402] \mathbf{u}_3 = \begin{bmatrix} -1 \\ 6 \\ -8 \end{bmatrix} - \begin{bmatrix} 3 \\ 6 \\ -6 \end{bmatrix} + \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix} u3=[−1−3+46−6+0−8+6+2] \mathbf{u}_3 = \begin{bmatrix} -1 - 3 + 4 \\ 6 - 6 + 0 \\ -8 + 6 + 2 \end{bmatrix} u3=[000] \mathbf{u}_3 = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} Since u3\mathbf{u}_3 is the zero vector, it does not contribute to the basis
step 5
Now, we normalize u1\mathbf{u}_1 and u2\mathbf{u}_2 to get the orthonormal basis vectors: e1=u1∥u1∥=13[12−2]=[1323−23] \mathbf{e}_1 = \frac{\mathbf{u}_1}{\|\mathbf{u}_1\|} = \frac{1}{3} \begin{bmatrix} 1 \\ 2 \\ -2 \end{bmatrix} = \begin{bmatrix} \frac{1}{3} \\ \frac{2}{3} \\ -\frac{2}{3} \end{bmatrix} e2=u2∥u2∥=145[−60−3]=[−250−15] \mathbf{e}_2 = \frac{\mathbf{u}_2}{\|\mathbf{u}_2\|} = \frac{1}{\sqrt{45}} \begin{bmatrix} -6 \\ 0 \\ -3 \end{bmatrix} = \begin{bmatrix} -\frac{2}{\sqrt{5}} \\ 0 \\ -\frac{1}{\sqrt{5}} \end{bmatrix}
Answer
The orthonormal basis of the column space of AA is $\left\{ \begin{bmatrix} \frac{1}{3} \\ \frac{2}{3} \\ -\frac{2}{3} \end{bmatrix}, \begin{bmatrix} -\frac{2}{\sqrt{5}} \\ 0 \\ -\frac{1}{\sqrt{5}} \end{bmatrix} \right\}
Key Concept
Orthonormal Basis
Explanation
An orthonormal basis of a vector space is a set of vectors that are both orthogonal and normalized. The Gram-Schmidt process is used to orthogonalize a set of vectors, and then each vector is normalized to form the orthonormal basis.

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