Consider the vectors 7₁ = €₂ = T3 = 1 in R4. Use the Gram-Schmidt process to find an orthogonal basis for the subspace span{1, 72, 73}.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Note first that the vectors 71, 72, and 73 are linearly independent (a necessary precondition for performing
the Gram-Schmidt process); you can check this by row-reducing the coefficient matrix [₁
confirming that all three columns have pivots. Gram-Schmidt then produces the orthogonal basis
2 3] and
1/2
-2/3
{V1, U2, U3} =
2/3
181
2/3
0
If you like, you can scale the second and third vectors to clear fractions (the result will still be an orthogonal
basis).
Transcribed Image Text:Answer Note first that the vectors 71, 72, and 73 are linearly independent (a necessary precondition for performing the Gram-Schmidt process); you can check this by row-reducing the coefficient matrix [₁ confirming that all three columns have pivots. Gram-Schmidt then produces the orthogonal basis 2 3] and 1/2 -2/3 {V1, U2, U3} = 2/3 181 2/3 0 If you like, you can scale the second and third vectors to clear fractions (the result will still be an orthogonal basis).
Consider the vectors
€₁ =
₂ =
3: = 1
in R4. Use the Gram-Schmidt process to find an orthogonal basis for the subspace span{1, 72, 73}.
Transcribed Image Text:Consider the vectors €₁ = ₂ = 3: = 1 in R4. Use the Gram-Schmidt process to find an orthogonal basis for the subspace span{1, 72, 73}.
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