Back to Finding orthogonal bases

Activity 6.4.4

Avg difficulty 2.5

Suppose that AA is the 4×34\times3 matrix whose columns are v1=[1111],v2=[1322],v3=[1−3−3−3].\vvec_1 = \fourvec1111,\hspace{24pt} \vvec_2 = \fourvec1322,\hspace{24pt} \vvec_3 = \fourvec1{-3}{-3}{-3}\text{.} These vectors form a basis for WW, the subspace of R4\real^4 that we encountered in this activity. Since these vectors are the columns of AA, we have Col(A)=W\col(A) = W.

  1. When we implemented Gram-Schmidt, we first found an orthogonal basis w1\wvec_1, w2\wvec_2, and w3\wvec_3 using w1=v1w2=v2−v2⋅w1w1⋅w1w1w3=v3−v3⋅w1w1⋅w1w1−v3⋅w2w2⋅w2w2.\begin{aligned} \wvec_1 & = \vvec_1 \\ \wvec_2 & = \vvec_2 - \frac{\vvec_2\cdot\wvec_1}{\wvec_1\cdot\wvec_1}\wvec_1 \\ \wvec_3 & = \vvec_3 - \frac{\vvec_3\cdot\wvec_1}{\wvec_1\cdot\wvec_1}\wvec_1 - \frac{\vvec_3\cdot\wvec_2}{\wvec_2\cdot\wvec_2}\wvec_2\text{.} \\ \end{aligned} Use these expressions to write v1\vvec_1, v2\vvec_2, and v3\vvec_3 as linear combinations of w1\wvec_1, w2\wvec_2, and w3\wvec_3.

  2. We next normalized the orthogonal basis w1\wvec_1, w2\wvec_2, and w3\wvec_3 to obtain an orthonormal basis u1\uvec_1, u2\uvec_2, and u3\uvec_3.

    Write the vectors wi\wvec_i as scalar multiples of ui\uvec_i. Then use these expressions to write v1\vvec_1, v2\vvec_2, and v3\vvec_3 as linear combinations of u1\uvec_1, u2\uvec_2, and u3\uvec_3.

  3. Suppose that Q=[u1u2u3]Q = \left[ \begin{array}{ccc} \uvec_1 & \uvec_2 & \uvec_3 \end{array} \right]. Use the result of the previous part to find a vector r1\rvec_1 so that Qr1=v1Q\rvec_1 = \vvec_1.

  4. Then find vectors r2\rvec_2 and r3\rvec_3 such that Qr2=v2Q\rvec_2 = \vvec_2 and Qr3=v3Q\rvec_3 = \vvec_3.

  5. Construct the matrix R=[r1r2r3]R = \left[ \begin{array}{ccc} \rvec_1 & \rvec_2 & \rvec_3 \end{array} \right]. Remembering that A=[v1v2v3]A = \left[ \begin{array}{ccc} \vvec_1 & \vvec_2 & \vvec_3 \end{array} \right], explain why A=QRA = QR.

  6. What is special about the shape of RR?

  7. Suppose that AA is a 10×610\times 6 matrix whose columns are linearly independent. This means that the columns of AA form a basis for W=Col(A)W=\col(A), a 6-dimensional subspace of R10\real^{10}. Suppose that we apply Gram-Schmidt orthogonalization to create an orthonormal basis whose vectors form the columns of QQ and that we write A=QRA=QR. What are the shape of QQ and what the shape of RR?

Austin, Understanding Linear Algebra, §6.4 Finding orthogonal bases, Activity 6.4.4

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Adapted from Understanding Linear Algebra © 2023 David Austin. Licensed under CC BY 4.0. Source: Understanding Linear Algebra. Imported at revision 3e6dced.