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3-a) Show that Eigen values of Hermitian operators are real Ac.=a,9 b) Eigen Function of Hermitian operators are orthogonal < 4, /0>=0...

Question

3-a) Show that Eigen values of Hermitian operators are real Ac.=a,9 b) Eigen Function of Hermitian operators are orthogonal < 4, /0>=0

3-a) Show that Eigen values of Hermitian operators are real Ac.=a,9 b) Eigen Function of Hermitian operators are orthogonal < 4, /0>=0



Answers

Let V be the vector space of n-square matrices over a field K. Show that W is a subspace of V if W consists
of all matrices A = [aij] that are

Hello. So the question is taken from mattresses. And the question is if abc are free square my tricks of same order and then A B is equal to a C. That implies B is equal to see. So the condition on these metrics is so four B conditions will be is equal to see when a B is equal to this implies stated. We can write it as a In two B- which is equal to zero. In case of metrics we can do this because my treatment matrix multiplication is distributed. Okay, so we can write uh this A B is equal to a say in this way. Okay if this is the condition uh either is equal to zero or b minus, C is equal to zero. Okay? So if the minus is equal to zero, be strictly equal to see. Okay, so on first condition it says that is not equal to you. So for the condition of B is equal to see it can be zero. So okay, now check. The second condition is that asian, vertebral. It means A must be known singular matrix. That is true because if is singular metrics it this implies that in its inverse doesn't exist. So even if it isn't. Uh So uh it is not necessary that A involves AIDS and what we will if A is not convertible in that case. Also, even if a singular in that case, also B is equal to say, okay, if we take a note equal to zero. Okay, if A is not equal to you in that case B minus is equal to zero. This also implies that B is equal to see. So that second condition is also not necessary. Is all tonal is also no, I don't think that it is necessary because this implies a prime is equal to so if B minus C is not equal to zero then that can be B minus C must be equal to you because how then be the producing, so that the victims of discussion is I think a symmetric if there is no symmetry, so let us if this is the condition A B is equal to a C. Take that aid to the items. So we can be eh in worlds A C. So involves must be so B is equal to see. So from this it implies that A must be convertible. Okay, sorry for that, it must be in whatever. So the directors are called discussion is the option A must be able to hold this

And this problem we're calculating the Eigen vectors and the Eigen values for a given matrix. And we have our matrix A equal to the following. Uh huh, wow minus 71 minus 13 and minus 11, 2. And for okay so our strategy here is going to be well set up are characteristic equation where we take our matrix A. Subtract the identity matrix, uh find the determined determinant of that and saw that setting it equal to zero. And so that will give us a cubic equation. Uh So we just need to solve that cubic equation to find our three Eigen values. And then for each again value we check with our matrix A two and solve a set of three questions to find the I in vectors. Okay, so let's set up our characteristic equation. So are a minus lambda identity matrix. Going to look like the following. We're gonna have a minus lambda minus 21 minus seven minus one minus lambda three minus 11 two and minus four. Mine is lambda. Okay. And we want to set up determinant of our a minus, I am the I equal to and south or equal to zero and so are determinant. So first we'll do cross a first row first column. So we'll get ah minus lambda chimes to have a minus one minus and uh whoa for mine, islam and minus six, wow. Then next so we go move on to second column, first row. So crossing out the corresponding ah second column and first row. Oh and so that's gonna give us a plus two. We're gonna get a minus seven. Mhm. Four minus land. Ah Plus thirties three and then going to the third column and crossing out first row third column and we'll get plus a minus 14 minus one minus land. Uh It's a minus one minus lambda and a minus 11, horrible. Mhm. Okay And we've got adding all this up. All right, we got a minus lambda uh and a minus four minus three lambda. Yeah, plus lambda squared minus six. Uh Plus two. The minus 28 plus seven seven. Land plus 33 wow. Plus A When I was 14 minus 11 plus 11 lambda, wow, mm. All right. And if we sum all that up, we get a ah minus lambda, cute. Plus three. Land is squared plus 13 and ah minus 15 and we want to set that equal to zero. Ah So we have this cubic equation to solve and we could solve it ah numerically or graphically or whatever. If we need to solve it by hand, it's actually not too bad to solve. Ah We could see that ah If we're given that the Eigen values are all integers, so we're gonna have three roots and it's gonna be one, it has to be a one, a three and a five to get our landed to the zero term of 15. So we have to have something like ah one, A three and a five. Uh So where are possible routes are plus minus one plus minus three and plus minus five. And so we can play around with a little bit and we'll find that our ah lambda solutions for lambda are one minus three and five. So those are our Eigen values. Okay, so now we'll need to uh for each Eigen value will need to go through and find the corresponding Eigen vectors. And so that is according to are a so each a ah times are vector equals two uh lambda vector. Right? So ah well so we'll go through here for our first one, lambda one equals to one. And so we want to solve that. So taking are a uh so a times V. One equals two lambda one times V. One. So that's gonna be our our Eigen vector V. One. Uh And the corresponding with its corresponding Eigen value lambda one and so are a. Right? So we've got our a vector ah zero minus 21 minus seven minus 13 minus 11 2. For. And it's criminal applying that times are vector. And so I'll give those an X. And a. Y. And Z. And that equals to. Ah So our value here is just a. One, so X. Y. Z. Okay, so that's going to set us up with a system of three equations to solve. And so we get a minus two. Uh Of course the equals X. We get a minus seven, X minus Y. Plus three Z equals two Y. And a minus 11 X plus two. Y plus four, Z equals two Z. Okay. Ah We'll start here uh with the we've got a substitution for X from the first equation X minus two Y. Posey. So plug that into the second equation and we'll get a minus seven times a minus two. Ah Plus Z minus Y plus three, Z equals two. Why? And I'm gonna skip a few intermediate rearranging of equations here and go to that, giving us Z equals 23 Y. And then we can we can just plug that back into our first equation here and now we'll go X equals two minus two. I plus Z. When Z equals +23 Y. It's two Y plus three wise. So we get X equal to Y. Okay, so now we've got our solution so we've got uh X equals X. Yeah X equals X. Why equals X and z equals 23? Y. Okay, so that gives us our the one. So Eigen vector for our first agon value of one? Going to be 11 and three? Okay, so that will be our first solution. So we got lambda. Uh oh uh so we've got our lambda one equals to one and are corresponding Eigen vector of 11 and three. All right, next step, let's check. I mean vector for I can value of lambda too equal to minus three. Okay, so all right on our a at r zero minus 21 minus seven minus one. Three minus 11 minus to four, wow. And are again times are Eigen vector equals or I can value minus three times are again vector? Yeah. Okay. And so we'll set up those three equations, we get minus two plus C equals minus three X minus seven X minus y plus. Crazy. It's minus three Y minus 11 X. Uh Plus sweeps. Um This should be a positive too. Are they negative too? So we've got minus 11 X plus two. Uh uh Plus for Z equals two minus three. Z. Okay for this one I'll just take that first equation. Uh and use our Z as a substitution. Uh So take C equals to two y minus three X. And we'll plug that into our second equation here, which will give us a minus seven X minus Y plus three times to I minus three X. Uh huh. And skipping a few intermediate steps that leaves us with Y equals to two X. And then we can just put that back into our Z expression. So we've got Z equals to two times two X minus three X. Uh So Z equals two X. So we've got for this one, ah We had our wow, X. X equals X, Y equals two X. And Z equals two X. So are RV two. I can vector going to be a one 21 All right. So that so for our again our second Eigen value of minus three, the corresponding Eigen vector V two will be 1 to 1. Oh all right. Now we just got our third Eigen value to do. Okay, so lambda three equals 25 And our matrix a zero minus 21 minus seven minus 13 minus 11 to 4. And I in vector X Y. Z equals I can value five times Eigen vector X, Y. The animal. Mhm. And that gives us the equations a minus two plus Z equals five, X a minus seven, X minus Y plus three, Z equals five Y and a minus 11 X plus two. Y plus for z equals 25 Z. And I'll take that first equation and go Z equals five X plus two. Why? And plug that into our second equation, which will give us a minus seven, X minus Y plus 35 X plus two, Y equals 25 Y. And that gives us an X equal to zero. And then we can plug that into our first equation here, we get Z equals to two Y. And so we've got A. X equals zero, Why equals Y. And the equals to two Y. And so our V. Three, I am our third Eigen vector is going to be a zero 12 and that's going with our the corresponding Eigen vector for our, I can value lambda three equals 25 mm. So we've got our three Eigen uh three Eigen values, so we've got our lambda one equals to one corresponding Eigen vector of 11 and three. We've got our second agon value lambda two equals minus three, corresponding Eigen vector 1 to 1 and our third agon value. Lambda three equals 25 and corresponding Eigen vector 012

And this problem we're calculating the Eigen vectors and the Eigen values for a given matrix. And we have our matrix A equal to the following. Uh huh, wow minus 71 minus 13 and minus 11, 2. And for okay so our strategy here is going to be well set up are characteristic equation where we take our matrix A. Subtract the identity matrix, uh find the determined determinant of that and saw that setting it equal to zero. And so that will give us a cubic equation. Uh So we just need to solve that cubic equation to find our three Eigen values. And then for each again value we check with our matrix A two and solve a set of three questions to find the I in vectors. Okay, so let's set up our characteristic equation. So are a minus lambda identity matrix. Going to look like the following. We're gonna have a minus lambda minus 21 minus seven minus one minus lambda three minus 11 two and minus four. Mine is lambda. Okay. And we want to set up determinant of our a minus, I am the I equal to and south or equal to zero and so are determinant. So first we'll do cross a first row first column. So we'll get ah minus lambda chimes to have a minus one minus and uh whoa for mine, islam and minus six, wow. Then next so we go move on to second column, first row. So crossing out the corresponding ah second column and first row. Oh and so that's gonna give us a plus two. We're gonna get a minus seven. Mhm. Four minus land. Ah Plus thirties three and then going to the third column and crossing out first row third column and we'll get plus a minus 14 minus one minus land. Uh It's a minus one minus lambda and a minus 11, horrible. Mhm. Okay And we've got adding all this up. All right, we got a minus lambda uh and a minus four minus three lambda. Yeah, plus lambda squared minus six. Uh Plus two. The minus 28 plus seven seven. Land plus 33 wow. Plus A When I was 14 minus 11 plus 11 lambda, wow, mm. All right. And if we sum all that up, we get a ah minus lambda, cute. Plus three. Land is squared plus 13 and ah minus 15 and we want to set that equal to zero. Ah So we have this cubic equation to solve and we could solve it ah numerically or graphically or whatever. If we need to solve it by hand, it's actually not too bad to solve. Ah We could see that ah If we're given that the Eigen values are all integers, so we're gonna have three roots and it's gonna be one, it has to be a one, a three and a five to get our landed to the zero term of 15. So we have to have something like ah one, A three and a five. Uh So where are possible routes are plus minus one plus minus three and plus minus five. And so we can play around with a little bit and we'll find that our ah lambda solutions for lambda are one minus three and five. So those are our Eigen values. Okay, so now we'll need to uh for each Eigen value will need to go through and find the corresponding Eigen vectors. And so that is according to are a so each a ah times are vector equals two uh lambda vector. Right? So ah well so we'll go through here for our first one, lambda one equals to one. And so we want to solve that. So taking are a uh so a times V. One equals two lambda one times V. One. So that's gonna be our our Eigen vector V. One. Uh And the corresponding with its corresponding Eigen value lambda one and so are a. Right? So we've got our a vector ah zero minus 21 minus seven minus 13 minus 11 2. For. And it's criminal applying that times are vector. And so I'll give those an X. And a. Y. And Z. And that equals to. Ah So our value here is just a. One, so X. Y. Z. Okay, so that's going to set us up with a system of three equations to solve. And so we get a minus two. Uh Of course the equals X. We get a minus seven, X minus Y. Plus three Z equals two Y. And a minus 11 X plus two. Y plus four, Z equals two Z. Okay. Ah We'll start here uh with the we've got a substitution for X from the first equation X minus two Y. Posey. So plug that into the second equation and we'll get a minus seven times a minus two. Ah Plus Z minus Y plus three, Z equals two. Why? And I'm gonna skip a few intermediate rearranging of equations here and go to that, giving us Z equals 23 Y. And then we can we can just plug that back into our first equation here and now we'll go X equals two minus two. I plus Z. When Z equals +23 Y. It's two Y plus three wise. So we get X equal to Y. Okay, so now we've got our solution so we've got uh X equals X. Yeah X equals X. Why equals X and z equals 23? Y. Okay, so that gives us our the one. So Eigen vector for our first agon value of one? Going to be 11 and three? Okay, so that will be our first solution. So we got lambda. Uh oh uh so we've got our lambda one equals to one and are corresponding Eigen vector of 11 and three. All right, next step, let's check. I mean vector for I can value of lambda too equal to minus three. Okay, so all right on our a at r zero minus 21 minus seven minus one. Three minus 11 minus to four, wow. And are again times are Eigen vector equals or I can value minus three times are again vector? Yeah. Okay. And so we'll set up those three equations, we get minus two plus C equals minus three X minus seven X minus y plus. Crazy. It's minus three Y minus 11 X. Uh Plus sweeps. Um This should be a positive too. Are they negative too? So we've got minus 11 X plus two. Uh uh Plus for Z equals two minus three. Z. Okay for this one I'll just take that first equation. Uh and use our Z as a substitution. Uh So take C equals to two y minus three X. And we'll plug that into our second equation here, which will give us a minus seven X minus Y plus three times to I minus three X. Uh huh. And skipping a few intermediate steps that leaves us with Y equals to two X. And then we can just put that back into our Z expression. So we've got Z equals to two times two X minus three X. Uh So Z equals two X. So we've got for this one, ah We had our wow, X. X equals X, Y equals two X. And Z equals two X. So are RV two. I can vector going to be a one 21 All right. So that so for our again our second Eigen value of minus three, the corresponding Eigen vector V two will be 1 to 1. Oh all right. Now we just got our third Eigen value to do. Okay, so lambda three equals 25 And our matrix a zero minus 21 minus seven minus 13 minus 11 to 4. And I in vector X Y. Z equals I can value five times Eigen vector X, Y. The animal. Mhm. And that gives us the equations a minus two plus Z equals five, X a minus seven, X minus Y plus three, Z equals five Y and a minus 11 X plus two. Y plus for z equals 25 Z. And I'll take that first equation and go Z equals five X plus two. Why? And plug that into our second equation, which will give us a minus seven, X minus Y plus 35 X plus two, Y equals 25 Y. And that gives us an X equal to zero. And then we can plug that into our first equation here, we get Z equals to two Y. And so we've got A. X equals zero, Why equals Y. And the equals to two Y. And so our V. Three, I am our third Eigen vector is going to be a zero 12 and that's going with our the corresponding Eigen vector for our, I can value lambda three equals 25 mm. So we've got our three Eigen uh three Eigen values, so we've got our lambda one equals to one corresponding Eigen vector of 11 and three. We've got our second agon value lambda two equals minus three, corresponding Eigen vector 1 to 1 and our third agon value. Lambda three equals 25 and corresponding Eigen vector 012

To check this subspace. What we need to do is we need to verify three exams. The first axiom is zero, vector should be belonging to a subspace. W. I mean the set W If you won't, you too belongs to W. Then we should verify that you want to see. You too also belongs to W. What is the closure property? Third, if you belongs to W then land up. You should also belongs to W Where lambda. Is it number from the field? Okay, if all these three exams satisfied, then we call W is a subspace of the factor space week. No, these set off all environ mattresses over the field kit. All right now W is the set of all symmetric matrices. So what do you mean by a symmetric matrix? The matrix was transposed the same as it's A So that's called a symmetric matrix. All right. Now if is symmetric then that's null. Matrix belongs to W. Yes, of course. Because the non matrix transpose his narrative so yes, it's true. Second axiom some of two symmetric matters is A and B. Supposed to be a symmetric B is also symmetric. Then a place be whole transpose is equal to the property of transport. It is a transport transport. But he transposes A. Because asymmetric transpose is B. Because B symmetric diet implies a place to be. He's also symmetric so that implies equals B belonged to W. So second exam is very fine. Now coming to Torrance if is symmetric lambda E. We'll transport the property of transport is lambda and a transport is equal to the lambda. A. Because the symmetric it is lambda lambda. He will transpose islam day. So that means lambda is also symmetric so it belongs to doug. So all the three exams are verified. So yes. Set of all symmetric matrices is a subspace. The second one upper triangular mattresses. So what do you mean by an upper triangular matics? All the elements below the main diagnosed should be zeroes, for example A B C 000 123 So this is an upper triangular matrix below this. All the elements are zeros of is a pet Wrangler. He's a pet Wrangler and he's also a strangler. Then it let's be is it a veranda? Of course, yes. Because these zeros get started with the other metrics in the same place. For example 123 4045006 If you add with 178 089 00 10. So what do you guys? 000000000 So some of the upper triangular mattresses. Also an upper triangular markets. So this is very frank. Null matrix is not trying biometrics because null metrics by default, all the elements below the main diagnosis anyway, zeros. In fact all the elements are real zeros. So it belongs to going mathematics, The Time magazine. If you want to play any number with the matrix scale are multiple, zero into any number of zero. So these zeros will still be retained. So lambda you Orlando A is also an upper triangular matrix. So yes, the set of a particular mattress is a subspace. The third diagonal matrix, basically, we can just individually say that the diagonal matrices from the subspace because lower triangular mattresses also from subspace. Because what do you mean by lower strangler? All the elements about domain that loves you with the same reasoning as a crime. And what is the diagonal matrix? It is both the lower triangle ring up strangler. Since both our Wrangler and Wrangler mattresses from subspace set of diagonal mattresses. Also from subsidies. Simple reason. Right. And skill harm metrics is a special case of diagonal matrix, scalar matrix means all the diagnosed elements should be zero, sorry, All the diagnostic elements should be same and other diagnosed elements other elements should be basically zeros. So this is a special case of diarrhea magics. And since diagnosed mattresses from subspace, the set of the subset of it also forms the suspects. Alright, so this is a prison. So


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