CSIR NET MATHEMATICS | Linear Algebra | Inner Product Q.77(A) Q.68(B) Q.74(C)

Welcome friends, in this video we are going to discuss the problem of linear algebra from CSIR NET June 2019. Example number 74, booklet C, marks 4.75. Let Rn cross Rn to R denote the standard inner product on Rn. Function Tw defined from Rn to Rn by Tw v equal to v minus 2 times inner product v with w divided by w with w into w for v belongs to Rn. Which of the following are true? First option, determine Tw equal to 1. Second option, inner product Tw, Tw, V2 is equal to inner product V1 with V2 for all V1, V2 belongs to Rn. Third option, Tw is equal to Tw inverse and the fourth option, T2w is equal to 2 times Tw. First, we recall standard inner product on Rn, two vectors on Rn, U equal to u1 u2 un and v equal to v1 v2 vn so standard inner product on Rn is nothing but the dot product so u1 v1 plus u2 v2 plus un vn now for n equal to 2 we are going to define function Tw R2 to R2 which is Tw of v equal to v minus 2 times inner product v with w divided by w with w into w here we don't have given any basis on R2 or Rn so we consider the standard basis on Rn so for R2 the standard bases are e1 equal to 1 0 e2 equal to 0 1 now we consider any non-zero vector 1 1 on R2 we are going to find the image of bases with respect to Tw where w is equal to 1 1 so by the definition Tw e1 equal to so here v equal to e1 so e1 minus 2 times e1 inner product w divided by w inner product w into w subtracting the value of w 1 1 and e 1 equal to 1 0 we have 1 0 minus 2 times inner product 1 0 with 1 1 divided by 1 1 with 1 1 into 1 1 so inner product 1 0 with 1 1 1 and inner product 1 1 with 1 1 is 2 so 2 2 cancel into 1 1 so we have 1 0 minus 1 1 equal to 0 minus 1 so image of first basis vector e 1 with respect to tw is 0 minus 1 now similarly image of e 2 with respect to tw where e 2 is equal to 0 1 and w is equal to 1 1 so if we substitute the value then after the simple inner product we have image minus 1 0 for e 2 with respect to tw now corresponding matrix for this transformation first column is a image of e1 with respect to tw and the second column is a second basis vector e2 its image with respect to tw so e1 image is 0 minus 1 and image of e2 is minus 1 0 and its determinant value is minus 1 So, first option says that determinant value is always 1, but we are able to find the determinant value minus 1, so first option is wrong. Second option, which is the inner product between TWV1 and TWV2, so TWV1 equal to V1 minus 2 times V1W divided by inner product W with W into W and image of V2 with respect to TW is v2 minus 2 times v2 w divided by w inner product w into w now using simple properties of inner product so v1 inner product with the second component plus minus 2 times inner product v1 w divided by w w inner product into w with second component now inner product v1 with v2 minus inner product v1 with the second part where the this inner product ratio is a constant so take out from the inner product we have v1 inner product w now in the second term we are taking the inner product of first component with the v2 plus first component with the second term of the second component which is minus two times inner product v2 with a w divided by inner product w with w into w now in the last term we have two inner product v1 with w divided by w with w is a constant and the same similarly two times inner product v2 with w divided by w is a constant so if we take the multiplication and take out from the inner product we have four times inner product v1 with w v2 with w divided by square of inner product w with w into inner product w with now for the second and third term we have inner product two times v2 with w v1 with w divided by w with w so both the terms are the same always remember inner product w with v2 is equal to v2 with w so we have two equal terms coefficient are minus two minus two so it will convert into the four times inner product v2 with w v1 with w divided by w with w in the last term fourth term denominator has square of inner product w with w and the numerator has inner product w with w so one power cancelled and we have four times inner product v1 with w v2 with w divided by w with w so we can see the second and third term will give minus four times v1 inner product w v2 inner product w divided by w inner product w and the last term is a four times the same term So, these terms are cancelled and we have inner product V1 with V2. Second option is that inner product between TWV1, TWV2 equal to inner product between V1 and V2 is correct. Now, we are going to discuss inverse part. Suppose, image of any vector V with respect to TW is U. So, for the inverse, inverse of TW of U is equal to V. now twv equal to u so by the definition is v minus two times inner product v with w divided by w with w w equal to u now we take inner product both the side with w so inner product v with w and we have second term now we simplify then we have inner product v with w in the second term we have two times inner product v with w divided by w with w and w with w so inner product w with w cancel with the denominator inner product w with w and we have simplified form inner product v with w minus two times inner product v with w is equal to u with w and we have relation inner product minus times v with w is equal to u with w and v is equal to from this relation plus two times inner product v with w divided by w with w into w now for the inverse tw inverse u is equal to v but v equal to u plus two times inner product v with w divided by w with w into w but inner product v with w is equal to minus times inner product u with w substituting then we have t inverse u equal to u minus 2 times u with w divided by w with w into w which is the same definition of tw therefore tw is equal to tw inverse so third option is correct now we are going to check the fourth option t2w is equal to 2 times tw so t2w v equal to we replace w by 2w then we have v minus 2 times v in a product with 2w divided by 2w in a product 2w into 2w then we have 2 into 2 into 2 so we have 8 times in a product v with w divided by 2w with 2w is equal to 4 times in a product w with w into w so we have v minus 2 times v in a product with w w inner product w into w now 2 times tw v equal to 2v minus 4 times inner product v with w divided by w with w which is not equal to t2w v fourth option is wrong thanks for watching for more videos like share subscribe press the bell icon of next gen mess