ALA29: Gram Schmidt Process

laughter openview students in this video I would explain a very important topic related to applied linear algebra that is gram-schmidt process this process is actually used to convert a set of ordinary vectors into orthonormal vectors and to do that first I would explain what is an orthonormal vectors actually orthonormal is a term which is a combination of two terms one is orthogonal and the other one is normal we all know that orthogonal vectors are those vectors which are perpendicular to each other and their dot product is 0 suppose we have a set of vectors which contains V 1 V 2 V 3 up to VN then these V eyes would be called orthogonal if the dot product of any two different vectors is 0 like V 1 dot v2 is 0 V 1 dot V 3 is 0 and so on so mathematically we can write VI the dot the inner product are dot product of VI with V dot VJ is 0 if I is not equal to J that is if the these vectors are not same and for any different vectors the dot product should be 0 then we would call these vectors orthogonal vectors and the next term in ortho normalized normal vectors normal vectors are actually unit vectors and we know that if we want to convert a vector into its unit vector then we just divide whether its magnitude then we would get the normal vectors so in orthonormal vectors these two conditions should be satisfied 1 there should be orthogonal vectors and the second one is they should be unit vectors and mathematically we can write these two conditions in the form of these 2 equations so now let us come to the grabb math process and before starting this process let me recall another concept there is called projection of U onto another factor that is a that is the concept that I have already explained in one of my lectures I would just use it and if somebody is confused about these concepts then he or she can watch my lecture related to orthogonal projections suppose we have a vector U and we want to decompose this vector U with respect to another vector that is a then we would have to decompositions one is w1 and another one instead of Z 2 and this W 1 is component of U which is along two along a and W two is component of U which is perpendicular or orthogonal to a and to find these W 1 and W 2 the formulas are given over here W 1 is actually u dot a over a norm squared a and this u dot a can be replaced with the inner product that is u comma a within these brackets and in this problem these two notations would be equivalent because we are dealing with the vectors in RN so this w 1 is actually the component of the U along a and to get the vector W 2 we can use head-to-tail rule and W 2 would be u minus the value of W 1 that is U dot a over a norm square a and if you are not getting it then obviously you have to watch the lectures on orthogonal projections then you will know why we subtract this from you and we call this W 2 is actually projection of U which is orthogonal to it in a similar fashion when we convert uise into their counterparts that those are V is where u i's are ordinary vectors i mean non orthogonal vectors and v eyes would be orthogonal vectors we just fix one vector that is u 1 and we take u 1 equals to V 1 okay so when we take V 2 V 2 is actually the projection of you two on V 1 so like here we had the projection of U which is orthogonal to a here we will take the projection of U du which is orthogonal to V 1 and in this way these two vectors would become orthogonal to each other so u 2 V 2 would be u 2 minus the dot product of u 2 that V 1 over V 1 norm square V 1 so by this formula that we would get the value of V 2 similarly we would calculate V 3 and V 3 would be the projection of the U 3 which would be orthogonal to V 1 and V 1 and we do in a similar fashion we would calculate u 3 minus projection of U 3 on V 1 and projection of U 3 on V 2 in this way we would get V 3 and to calculate the value of V 4 we would calculate the component of U for which is orthogonal to V 1 we do n V 3 and in this way V 4 would be perpendicular to all these vectors all these vectors V 1 V 2 n P 3 and this way we would get such vectors which would be orthogonal to each other and the value of V 4 can be calculated by you for - projection of U for on V 1 - projection of U for on V 2 - projection of U for on V 3 in similar fashion we calculate we can calculate V 5 3 6 up to so on so by this process we convert U is into V is where you eyes were non orthogonal and V eyes are not now orthogonal so by this process we have converted non orthogonal vectors into orthogonal vectors and now to get orthonormal vectors all we have to do is just normalize these vectors that is divide them with their magnitude let's name the orthonormal vectors to be Q 1 Q 2 and Q 3 then Q 1 would be V 1 divided by its magnitude or norm Q 2 would be V 2 by V 2 norm and Q 3 would be V 3 by V 3 norm and this in this way Q 1 and Q 3 Q 1 Q 2 and Q through 3 would be called orthonormal vectors now let me explain this process with the help of an example suppose we have u1 u2 and u3 and we want to convert these vectors into they're orthonormal counterparts that is V 1 V 2 and V 3 and then we want to convert well then we want to find there are two normal counterparts as well which are Q 1 Q 2 and Q 3 so now let's start the process first we would write u1 u2 and u3 which are written over here u 1 is 1 1 1 u 2 is 0 1 1 and u 3 is 0 0 1 and you can check by multiplying these two that there would be non orthogonal vectors so now to get first orthogonal vectors that is V 1 we would fix it equal to u 1 that is V 1 equals to 1 1 1 now to find our vectors which is orthogonal to V 2 we would just calculate the projection of U du which is perpendicular to V 1 so we would convert we would use the value of U 2 from over here then the dot product of U 2 with V 1 can be calculated easily and if you are not familiar with the dista that has taught part at all in a product my previous videos are available which explains which explained everything related to dot products or inner products and the term which is in the denominator is called norm and there is also a lecture related to norm in my previous lectures so by using that cover those concepts we can plug in the values of u to V one in this formula and we would get the values something like this to get the dot product we would just multiply the corresponding entries and we would add them that is 0 times 1 then 1 times 1 then 1 times 1 that is this value and again this is the norm so 2 taken to take the norm of a vector we would just take the square of this component and square root take the square root because we have a norm square so we would ignore the square root so it would be just X square plus y square plus Z square that is 1 square plus 1 square plus 1 square so after the simplification the value of v2 would be you do - this would be actually the value of u dot V over that is u 2 dot v1 over u1 norm square and when we multiply this value inside and after subtracting these two vectors we would carry 2 which is minus 2 by 3 1 by 3 and 1 by 3 so we have calculated V 1 and V 2 now we would calculate the last vector last orthogonal vectors that is V 3 so V 3 is U 3 minus projection of U 3 on V 1 and projection of U 3 on V 2 so we would just plug in the values of U 3 V 1 and V 2 in this expression and we would get this value V 3 is actually this these values and after the simplification the V 3 the value of V 3 would be 0 negative 1 by 2 and 1 by 2 ok so now we have calculated V 1 V 2 V 3 and all of these vectors are and you can check the community by multiplying any of these two vectors that is V dot V 1 dot V 2 or V dot V 3 or v2 dot V 3 now they would turn out to be 0 and by this you can check that these vectors are orthogonal now to find that they're orthonormal counterparts you would just use these expressions or these formula that q1 is V 1 over V 1 magnitude or V 1 now we have already calculated the norm of V 1 so we would just use the values and we would get Q 1 Q 2 and Q 3 in this fashion Q 1 Q 1 is V 1 over V 1 which is equal to 1 over hundred 3 1 over unload 3 & 1 over under root 3 similarly we can calculate Q 3 Q 2 Q 2 is actually V 2 over V 2 norm and we would get this value when we when we would divide this we would have I would just solve one value that is minus 2 by 3 then if we divide it with under root 6 by 2 that would be minus 2 by 3 times 2 by under root 6 so we would have like minus 4 over 3 under root 6 so by using these values we would get Q 2 equals 2 minus 2 106 1 over under root 6 and 1 over under root 6 I'm sorry this value doesn't seem right let me calculate V 2 magnitude myself v2 was V 2 was actually minus 2 by 3 1 by 3 and 1 by 3 and if we take V 2 norm then it would be minus 2 by 3 squared plus 1 by 3 squared and plus 4 by 3 squared root so v2 norm would be that is 4 by 9 4 by 9 plus 1 by 9 plus 1 by 9 so that is under root 60 whatever 9 and this is equal to under root 6 divided by 3 so that will make the correction that is not under root 6 by 2 that is under root 6 by 3 and it's now if let me correct this this is root 3 then it would be 3 and this 3 would be cancelled with this and we won't get to or under root 6 and this is the same value now it seems correct so Q 2 is actually this factor and in the similar fashion we can calculate Q 3 which is v3 or v3 norm and by calculating we would get Q 3 equal to 0 and negative 1 over under root 2 and 1 over root and now these Q eyes would be orthonormal vectors and that that can be checked by taking the dot product of any of two different Qi with each other and there would be 0 and the norm of any Qi would be 1 and that you can check yourself and if you want to do some practice questions you can solve exercise 6.2 and the question 1 to 12 and the book is the linear algebra by Howard our Aunt

Channel: Dr. Naveed Ahmed

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Id: VE8iI46FUIQ

Length: 14min 58sec (898 seconds)

Published: Mon May 04 2020