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Q7E

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Linear Algebra With Applications
Found in: Page 224
Linear Algebra With Applications

Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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Short Answer

Using paper and pencil, perform the Gram-Schmidt process on the sequences of vectors given in Exercises 1 through 14.

7. [201],[-212],[1800]

The orthonormal vectors of the sequence [201],[-212],[1800] is [2/32/31/3],[-2/31/32/3],13[1-22].

See the step by step solution

Step by Step Solution

Step 1: Determine the Gram-Schmidt process.

Consider a basis of a subspace V of Rn for j=2,....,m we resolve the vector vj into its components parallel and perpendicular to the span of the preceding vectors v1,....,vj-1,

Then

u1=1||v1||v1,u2=1||v2||v2,.....,uj=1||vj||vj,.....,um=1||vm||vm

Step 2: Apply the Gram-Schmidt process

Obtain the value of u1 ,u2 and u2 according to Gram-Schmidt process.

u1=v1||v||.........(1)u2=v2-(u1.v2)u1||v2-(u1.v2)u1||..........(2)u3=v3-(u1.v3)u1-(u2.v3)u2||v3-(u1.v3)u1-(u2.v3)u2||....(3)

It can be observed that, v1v2=0v1v2.

Now, find u1.

role="math" localid="1659438380748" u1=122+22+1221 =13221 =2/32/31/3

Similarly,

role="math" localid="1659439075349" u1=122+12+22 =13-212 =2/32/31/3

Step 3: Find u⃗3      

Consider the equation below to obtain the value of u3 in equation (3).

Here is need to find out the values of v3-(u1v3)u1-(u2v3)u2and v3-(u1v3)u1-(u2v3)u2to obtain the value of u3.

  • role="math" localid="1659438669755" u1v3=12
  • u2v3=-12

u1v3u1=884u2v3u2=8-4-8v3-u1v3u1-u2v3u2=1800-884-8-4-8 =2-44 =21-22

Then,

role="math" localid="1659439148633" u3=131-22

Thus, the values of ,u1, u2 and u3 are role="math" localid="1659439162227" 2/32/31/3,-2/31/32/3,131-22.

Hence, the orthonormal vectors are 2/32/31/3,-2/31/32/3,131-22.

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