#CSIRNET MATHEMATICS NOV 2020 | 30th Nov | Section C | Linear Algebra | Orthonormal Basis of R^4

Hello friends, this is Mahesh Yulekar and you are watching Next Gen Maths. If you have not subscribed my channel, then please subscribe the channel and press the bell icon for notification of new videos. In this video, we are going to discuss the problem of CSI NET, November 2020, problem of linear algebra. Question, let set Y1, Y2, Y3, Y4 be an orthonormal basis of R4, then which of the following sets are the orthonormal basis? We again have four sets in the four options. The first option is Y1 plus Y2, Y1 minus Y2, Y3, Y4. Second set is Y3, Y4 minus Y1, Y2. Third set is Y1 plus Y2 divided by 2, Y1 minus Y2 divided by 2, Y3, Y4. And the fourth set is 3Y1 plus 4Y2 divided by 5, 4Y1 minus 3Y2 divided by 5, Y3, Y4. If you want notes of this solution, then please visit our website www.nextgenmaths.com. In the description, link is available. It is a problem of orthonormal basis, so we begin with the definition of orthonormal basis. A set Y1, Y2, Y3, Yn is an orthonormal basis of Rn if it satisfy two conditions. First condition, inner product Y with Yj should be 0 when i not equal to j and norm of Y square that means inner product Y with Y should be 1. Here we have given that set Y1, Y2, Y3, Y4 is an orthonormal basis of R4. Therefore, by the definition of orthonormal basis, we can say that inner product Y with Yj equal to 0 when i not equal to j and norm of Y square that means inner product Y with Y equal to 1 for i equal to 1, 2, 3, 4, j equal to 1, 2, 3, 4. Now, we will check all the four options one by one. First option, we have given the set Y1 plus Y2, Y1 minus Y2, Y3, Y4. It will be the orthonormal basis of R4 if it satisfy both the conditions. First, we check norm of Y1 plus Y2 whole square which is inner product Y1 plus Y2 with Y1 plus Y2. If you simplify inner product Y1 with Y1 plus inner product Y1 with Y2 plus inner product Y2 with Y1 plus inner product Y2 with Y2. Inner product Y1 with Y1 is nothing but the norm of Y1 square which has value 1. Inner product Y1 with Y2 is 0 because i not equal to j. Similarly, inner product Y2 with Y1 is also 0 and inner product Y2 with Y2 is nothing but the norm of Y2 square which is 1. Therefore, 1 plus 0 plus 0 plus 1 equal to 2. That means norm of Y1 plus Y2 square equal to 2 which is not equal to 1 and it violate the second condition of orthonormal basis. Hence, the option number 1 is the wrong statement. Option number 2. Set Y3, Y4 minus Y1, Y2. In the data, we are given that set Y1, Y2, Y3, Y4 is an orthonormal basis of R4. Therefore, we have norm of Y2 square equal to norm of Y3 square equal to norm of Y4 square equal to 1 and inner product Y with Yj equal to 0 when i not equal to j for ij equal to 2, 3 and 4. Here, we have given fourth vector minus Y1. So, norm of minus Y1 square equal to norm of Y1 square equal to 1. That we can easily prove as norm of minus Y1 square is nothing but the inner product minus Y1 with minus Y1 which is nothing but the inner product Y1 with Y1 and it will give the norm of Y1 square which has value 1 and inner product minus Y1 with Y is equal to minus times inner product Y1 with Y which is 0 as i not equal to 1. It is given in the data. Therefore, this set Y3, Y4 minus Y1, Y2 satisfy both the condition of orthonormal basis and hence it is an orthonormal basis and option number 2 is the correct answer. Option number 3. We have set Y1 plus Y2 divided by 2, Y1 minus Y2 divided by 2, Y3, Y4. We check the norm of Y1 plus Y2. So, norm of Y1 plus Y2 whole square equal to inner product Y1 plus Y2 divided by 2, Y1 plus Y2 divided by 2. 1 by 2 and 1 by 2 is the constant. So, we can take out from the inner product. So, we have 1 by 4 inner product Y1 plus Y2 with Y1 plus Y2. As we discussed in the option number 1, repeat that inner product Y1 with Y1 plus inner product Y1 with Y2 plus inner product Y2 with Y1 plus inner product Y2 with Y2 and we have 1 by 4, 1 plus 0 plus 0 plus 1 which is 2 by 4 equal to 1 by 2 and norm of Y1 plus Y2 whole square has value 1 by 2 which is not equal to 1. Therefore, this set cannot be orthonormal basis of R4 and hence option number 3 is the wrong state. Option number 4. We have given the set 3Y1 plus 4Y2 divided by 5, 4Y1 minus 3Y2 divided by 5, Y3 and Y4. So, we begin with the square of norm of 3Y1 plus 4Y2 divided by 5 which is nothing but the inner product 3Y1 plus 4Y2 divided by 5 with 3Y1 plus 4Y2 divided by 5. Now, 1 by 5 is the constant. So, take out from the inner product part. Then we have 1 upon 25 inner product 3Y1 plus 4Y2 with 3Y1 plus 4Y2. Now, if we simplify then we have 9 times inner product Y1 with Y1 plus 12 times inner product Y1 with Y2 plus 12 times inner product Y2 with Y1 and 16 times inner product Y2 with Y2. As we know inner product Y1 with Y1 is 1, inner product Y1 with Y2 is 0, inner product Y2 with Y2 is 1. So, we have 1 upon 25 9 plus 0 plus 0 plus 16. 9 plus 16 is equal to 25. So, 25 by 25 equal to 1. Therefore, square of norm of 3Y1 plus 4Y2 divided by 5 equal to 1 and similarly we can prove the norm of 4Y1 minus 3Y2 divided by 5 square is also 1. In the data we have given that norm of Y3 and Y4 equal to 1. That means in this set each vector has norm 1. So, it satisfies the second condition. Now, we are going to check the first condition of the orthogonal basis which is inner product Y with Yj should be 0 when i not equal to j. So, we take the inner product of first with the second vector that means 3Y1 plus 4Y2 divided by 5 inner product with 4Y1 minus 3Y2 divided by 5. So, 1 upon 25 is constant. So, take out from the inner product part and if you simplify then we have 12 times inner product Y1 with Y1 minus 9 times inner product Y1 with Y2 plus 16 times inner product Y2 with Y1 and minus 12 times inner product Y2 with Y. So, inner product Y1 has value 1, Y1 with Y2 is 0, Y2 with Y1 is 0 and Y2 with Y2 is 1. And if we substitute that value then we have 12 minus 0 plus 0 minus 12. So, now inner product of first vector with Y3. So, we can rewrite 1 by 5 inner product 3 Y1 with Y3 plus inner product 4 times Y2 with Y3. But Y1 with Y3 and Y2 with Y3 are 0 as given in the data and hence it has also value 0. Now, inner product of first vector with the Y4 and if you simplify then we have 1 upon 5 3 times inner product Y1 with Y4 plus inner product Y2 with Y4. So, Y1 with Y4 and Y2 with Y4 is 0 that is given in the data and hence it has also value 0. So, inner product of first vector with all the remaining vector has value 0. Similarly, inner product of second vector with Y3 and Y4 is also 0 and inner product Y3 with Y4 equal to 0 that is given in the data. Therefore, the SAT in the option number 4 satisfy both the orthonormal basis and hence it is an orthonormal basis of R4 and option number 4 is the correct statement. Thanks for watching. For more videos please like, share, subscribe and press the bell icon of NextGenMaths.