Find the orthonormal basis of the subspace Y=lin([1,-2,-3,4,1], [4,3,-2,1,2],[1,-3,2,4,-3],[5,2,3,4,-5]) in the space X=R^5. Using basis Y to find the orthonormal basis of X.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Find the orthonormal basis of the subspace Y=lin([1,-2,-3,4,1],
[4,3,-2,1,2],[1,-3,2,4,-3],[5,2,3,4,-5]) in the space X=R^5. Using basis Y to find the orthonormal basis of X.

 

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Orthogonalization
€₁,..., en € (X, (, )) linearly independent vectors
fj+1(x₁,...,xj) = ||ej+1 = x₁e₁
-
= 0
afj+1(x1,...,xj)
8x1
fj+1(x1, j
მე
=
0
ofj+1(x1,...,xj) = 0
?хj
• ₁ = ||1||
ej+1
of linearly independent vectors
B =
=
= {e₁,...,en}, Â = {ê1,…..‚ên}.
— x¡¤¡||², j = 1, ..., N – 1
....
⇒ X1 = Xj+1,1,..., Xj = Xj+¹,j, j = 1,. N-1
e2 = €2x2,1€1;
e2 = €3x3,1€1 - x3,2€2
ēj+1
||ēj+1||'
ēj+1 = €j+1 −Xj+1,1€1 — Xj+1,2€2-... -
-
· — Xj+1,j¤j, j = 1, ..., N − 1
Transcribed Image Text:Orthogonalization €₁,..., en € (X, (, )) linearly independent vectors fj+1(x₁,...,xj) = ||ej+1 = x₁e₁ - = 0 afj+1(x1,...,xj) 8x1 fj+1(x1, j მე = 0 ofj+1(x1,...,xj) = 0 ?хj • ₁ = ||1|| ej+1 of linearly independent vectors B = = = {e₁,...,en},  = {ê1,…..‚ên}. — x¡¤¡||², j = 1, ..., N – 1 .... ⇒ X1 = Xj+1,1,..., Xj = Xj+¹,j, j = 1,. N-1 e2 = €2x2,1€1; e2 = €3x3,1€1 - x3,2€2 ēj+1 ||ēj+1||' ēj+1 = €j+1 −Xj+1,1€1 — Xj+1,2€2-... - - · — Xj+1,j¤j, j = 1, ..., N − 1
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