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

Hi, hello friends, this Inside the video, we are linear The problem of LGP We will discuss CS Ein Ad June 2019 Example Number Seventy Four for Trauma Bookled C. M. for Punt Seventy Four A. R. and Tros A. R. to denote by R the standard inner product the Rn-valued function TW defined from Rn to Rn by TWV equals V minus two times the inner product of V and W divided by the inner product of W with W, all multiplied by W or V belongs to R, which of the The following are true: we have four options. The first option is to determine TW equal to 1 second. option is inner product TW V1 with TW V2 equal to V1 with V2 or all V1 V2 belong to Rn third The second option is TW equal to TW inverse, and the fourth option is T2W. is equal to 2 TW. So the standard inner product on ℝᴺ We have two vectors, U equal to U1U2UN and V equal to V1V2VN belongs to RN. In the X example, we are given Use the standard inner product So first, let's recall the standard inner product. We have two vectors. U equals U₁U₂Vᴺ, V equals V₁V₂Vᴺ Inside RN. The standard inner product will be the dot product between U and V. Then we will get u1, u1 plus u2, u2 plus u1 for n equal to 2 We will define a function tw, r2 to r2 tw equals v minus 2 times in a prop v with w, w with w into w We have not been given a basis. Therefore, we will use the standard basis vectors. R2 on so e1 equals 1, 0 and e2 equal to 0, 1 standard basis vector in R² Take any non-zero vector w equals 1, 1 With respect to this transformation We will find the images of E1 and E2. So, the image of E1 Inside the definition, we will replace V with even So E1 minus 2 times E1 inner product with W divided by W with W into W The value of E1 is 10. The value of W is 11. so 10 minus 2 times 10 inner product with 11 His answer will come today 1 inner product with 11 His answer will come today 2 2 and 2 will cancel out 10 minus 11 Its value will be 0 minus 1. The image of E1 will arrive. 0-1, in the same way To find the image of E2 We will replace V with E2 so E2-2 times E2 inner product W divided by the inner product of W W into W, the value of E2 is 0-1, then we'll put it here 0-1 -2 × (0 - 1) 1-1 The value of the inner product will be 1. and 1-1 inner product 1-1 The value will be 2-2-2 canceled 0-1-1-1 Then the answer will be minus 1 minus 0 Then the image of E2 will be minus 1. For the matrix of this transformation first column Image of E1 and image of E2 in the second column The value of the image of E1 is 0 minus 1. And the value of the image of E2 is minus 1, 0 Its determinant value will be minus 1. And the first option says that The value of the determinant should be 1. So our first option is wrong. in the second option, inner product TW V1 with TW V2 For TW V1 In the definition, we will replace v with v1. so v1 minus 2 times v1 inner product of w divided by the inner product of w into w, for v2 We will replace v with v2 so v2 minus 2 times v2 inner product w divided by w inner product w into, here we Simple properties of the inner product we will use, so v1's The inner product will be with the second component. And this is the second term. The inner product will be the second component. With it, in the same way inner product v1 of Will be with v2 The inner product of minus v1 will be with the second term And here in the second term, it is constant with respect to the inner product. We'll take him out then. Then the inner product of v1 with w will be plus the second term in The inner product of the first component will be with v2. And the first component is the inner product. Will be with the second term We will take this constant inner product outside within the third term. Then the inner product of w with v2 will be And in the last term, we have two coefficients appearing. The constant We will take it outside the inner product. And its multiplication will take place. -2 - 2 so plus 4 inner product V with W V2 with W divided by the square of the inner product of W with W into the inner product of W with inner product W with V2 is always the inner product of V2 with W Property of inner product Then these two terms will become the same. And the coefficient of both is minus 2 minus 2. Then it will come to us minus 4 times the inner product of v2 with w, v1 with w divided by w with w within the last term square of the inner product of w with w One power will be canceled out. inner product of w with w Then we will get 4 times the inner product v with w, v2 with w divided by w with w This is 4 times, and this is minus 4 times. Then these terms will be cut. So it will remain with us inner product of v1 with v2 so inner product T W V1 The answer for T W V2 will appear. V1 V2, which is our second option Then the second option will be correct here. The third option has an inverse. So we will find the inverse So T W V equals U. so its inverse image will be U to V T W V equals U From the definition of function V minus 2 times the inner product of V with W W with W into W Both sites will take the inner product with W We will then use the property of the inner product. Then the inner product of v with w will be And the inner product of the second term It will be with w. We will factor out the constant term So the inner product of w with w These 'w' and 'w' will cancel each other out. So it will come within the second term -2 times v with w v with w minus 2 times v with w So it will come today. minus times v with w equal to u with w And from the verification of this relation v equals will become u plus two times the inner product of v with w divided by w with w into w t w inverse for inverse function u equals v, the value of v we will substitute u plus two times inner product of v with w w with w into w but the inner product of v with w is equal to minus u with w, then we will substitute here So we'll get t w inverse u equals u minus two times the inner product of u with w divided by w with w into w But this is the definition It belongs to TW. Meaning of TW inverse equals TW That is our third option So it's been corrected now. Now we will check the fourth option T2W V equal to in the definition Jagir of W We will replace 2W So V minus 2 times V 2W 2W Inner Prabhi 2W 2W 2 by 2 by 8 Then 8 will come in the numerator. 2 times 2 in the denominator 4 will come Then it will come in simplified form. V minus 2 times the inner product V with W, W with W into W on the right side is 2 times TW, then definition of it we We will multiply by 2, then 2 volts minus 2, times 2, 4 times V with W, divided by W to W, but this T2, W is not So the fourth option will be incorrect. In this example, we have only 2 correct ones. The answer is second and third. Thanks for watching, and more videos Like, share, subscribe, press the bell icon Nauk Next Gen Mesh