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Linear Algebra With Applications
Found in: Page 176
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

Find the set of all polynomial f(t) in P3 such that f(1)=0 and-11f(t)dt=0, and determine its dimension.

The dimension of P3 such that f(1)=0 and -11f(t)dt=0 is 3 which is spanned by span1,t2-23t,t3-t.

See the step by step solution

Step by Step Solution

Step 1: Determine the span set.

Consider the set of all polynomial such that and .

The set {1,t,t2,,t}is linear independent set of V if there exist constant aiR such that a0+a1t1+a2t2+...+antn=0 where a0=a1=a2=...=an=0.

Any polynomial f(t)P3 is defined as follows.


Substitute the value 1 for t in the equation f(t)=b0+b1t+b2t2+b3t3 as follows.


As f(1)=0, substitute the value 0 for f1 in the equation f(1)=b0+b1+b2+b3 as follows.

f1=b0+b1+b2+b3 b1=-b0+b2+b3

Take integration both side in the equation f(t)=b0+b1t+b2t2+b3t3as follows.

role="math" localid="1659407877530" ft=b0+bit+b2t2+b3t3-11ftdt=-11b0+b1t+b2t2+b2t3dt =b0t+b1t22+b2t33+b3t44-11-11ftdt=b0+b112+b213+b314+b0-b112+b213-b314

Further, simplify the equation as follows.


As -11ftdt=0, substitute the value 0 for-11ftt in the equation 11ftt=2bb+b223 as follows.

-11ft=2b0+b223 0=2b0+b223 b0=-b213

Substitute the value -b213 for b0 in the equation b1=-(b0+b2+b3)as follows.

b-1=-b0+b2+b3 =--b223+b2+b3 =-b223+b3b1=-b223-b3

Finally, substitute the value -b223-b3 for b1 in the equation ft=b0+b1t+b2t2+b3t3 as follows.


From the equation, the set 1t2-23t,t3-tspan P3 if f1=0 and 11ftdt=0.

Step 2: Determine the property of independency.

Compare the equations b0+b2t2-23t+b3t3-t=0 both side as follows.


By the definition of linear independence, the subset 1,t2-23t,t3-t is L.I.

Therefore, the set of all polynomial P3 such that f1=0 and 11ftdt=0 has dimension 3 and spanned by span1,t2-23t,t3-t.

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