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Linear Algebra With Applications
Found in: Page 414
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

51. If A is a symmetric 2×2 matrix with eigenvalues 1 and 2, then the angle between x and Ax must be less than π/6, for all nonzero vectors x in R2.

The given statement is TRUE.

See the step by step solution

Step by Step Solution

Step 1: Check whether the given statement is TRUE or FALSE

Let v˙1,v˙2be unit orthogonal eigenvectors with eigenvalues 1 and 2 respectively.

So, write localid="1659679849030" x˙=c1v˙1+c2v˙2 for some c1,c2 . Also x˙ =c12+c22

Ax˙=c1Av˙1+c2Av˙2Ax˙=c1v˙1+2c1v˙2 x˙ =c12+4c22x˙.Av˙=c12+2c22

Let is the angle between x˙and Ax˙. Then,

cosθ=x˙.Aa˙x˙Ax˙ =c12+2c22c12+c22c12+422 =c12+2c22c14+5c12c22+4c24 =c14+4c12c22+4c24c14+4c12c22+4c24+c12c221/2

=11+c12c22c14+4c12c22+4c24 =11+c12c22c14+4c12c22+4c24 = 11+c12c22(c12+2c22)21/2 = 11+c1c2(c12+2c22)21/2

Here, cos θ takes minimum value when localid="1659678852211" c1c2c12+c22takes maximum value.

To find the maximum value of c1c2c12+c22, differentiate it with respect to c1and c2.

c1c1c2c12+2c22=c12+2c22c2-c1c22c2c12+2c222 =-c12c2+2c23c12+2c222 =c2(-c12+2c22)c12+2c222c1c1c2c12+2c22=c12+2c22c2-c1c24c2c12+2c222 =c13-2c1c22c12+2c222 =c2(c12-2c22)c12+2c222

In order to find the extreme value, equate both the partial derivatives with 0. Ignore the solution c1,c2=0since it corresponds to minimum. The other condition we obtain is c12=2c22. Thus, the maximum is achieved when this condition is satisfied.

When c12=2c22, we have

c12c22c12+2c22=2c244c222 =2c2416c24 =18

Hence, cosθ=11+(1/8)21/2=8/9

Step 2: Final Answer

This means that the minimum possible value of is cos θ is 8/9 >3/2. Thus, in any case cos θ>3/2which implies that θ<π/6.

Thus, the given statement is TRUE.

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