61 - Properties of similar matrices

we defined similar matrices to be matrices that are representing matrices of the same linear transformation so two matrices are similar if they represent the same linear map with respect to different bases that was the definition and then we had a theory which we didn't prove that gave an equivalent definition of similar matrices in terms in the realm of matrices alone namely there was the existence of an invertible matrix matrix P such that B equals P inverse a peep okay everybody remember okay so I mentioned that similar matrices share many properties and this is what I want to discuss now and I want to start with defining a property that we haven't discussed yet which is the trace of a matrix so here's the definition the sum of the main diagonal diagonal elements of a square matrix a n by n is called the trace the trace of a and denoted by TR Ave so right now we're just gonna encounter this definition and not say much about why it's important and where it's used okay so for now it's just a reasonable thing to define motivation is yet lacking so the trace is just the sum of the diagonal elements so here's an example suppose you want the matrix PI 3 root 2 I'm just throwing in some interesting numbers that I know negative 1 e 1/2 and in fact I have no clue why I'm copying this off my page I can just invent it and put anything so what is the trace of this matrix right all you have to do very easy thing to calculate the trace of this matrix is just the sum of the diagonal elements so it's just PI + good ok so that's the notion of a trace and here's a little theorem regarding traces if you take two matrices square matrices and the trace of the product equals the trace of the product the other way okay so these are square matrices so they can be multiplied both ways this is not a claim of course that a B equals B a if it were then this would be an obvious statement it's not because matrix multiplication is not commutative in general ok so let's prove this proof so what I want to do is decipher what are the diagonal elements of each of these products and what is their sum okay so the diagonal elements in general let's start with the general statement what's the IJ element of the product a B the IJ element of the product a B I'm going to use Sigma notation whooping that you're already not too terrified when you see it is the sum over let's see K goes from 1 to N let's say these are matrices of size n by n ah a I K B K G remember okay so this is the formalism of what we know look up here a second going like this times like this right the eighth row of a time's the J's : of B so that's the formal way to write it so that's the IJ element of a B and therefore what is the diagonal element a diagonal element would be an element where you have I I right that's a diagonal element same row and column so it's the sum K goes from 1 to n e I K be K I good okay so that's the III element so now let's calculate what is the trace of a be the trace of a B is the sum of these guys so we have to sum over I I goes from 1 to n of all these guys so there's another sum of Sigma K goes from 1 to n a I K B K hi good ok what is the trace of B a let's think let's think let's do it slowly and carefully I agree that it is once here once you can see through the notation it's it's rather trivial but the notation is sometimes a barrier so let's do it carefully so the trace of B a is again the sum let's write an intermediate step the sum over I I equals 1 to N of these elements but for BA of b a.i I do you agree okay and what is Bai I so I'm carrying this sum over some I over I goes from 1 to N and be a III be AI I would be the same thing here but interchanging the B's and the ACE right so it would be again a sum K goes from 1 to N but now the first matrix is B so it's B i k a k i do you agree so in both cases we have a double sum of products and here i can afford of course reverse the order of the bees and da's these are these are products of numbers of entries and that is commutative so I'm summing over to summation indices I and K both range from 1 to N here I have the AI KS meaning BK eyes here I have a K eyes meeting bi case but that's the same thing ok do you agree that this is precisely the same number this sum and this sum are precisely the same ok now if you have a hard time viewing it what you have to do is open the Sigma's and convince yourself that what you're seeing is precisely the same elements ok so these two let's write it like this these our cool good okay so and by the way if if you're if you're scared because of this completely here's the best way to at least feel comfortable with it take a couple of two-by-two matrices concrete ones with numbers multiply them when we multiply them the other way you won't get the same thing in general but you can see that the traces are the same take a couple of 3x3 matrices and do the same thing okay everybody good okay so here's the theorem so this was the notion of the trace and an important property of the trace and here's the theorem that takes us back to similar matrices and it says the following suppose we have two similar matrices a is similar to be that was our notation for similar matrices okay then sure yep when years so we're back here a second when we do finite Sigma's okay in calculus you do infinite Sigma's but we're not there these are finite sums for finite sums you have associativity of numbers associativity commutativity distributivity all these properties we know from the field R or C okay so you can switch the order of the Sigma's you can rename I you can call I J and then change all the eyes to J's okay you can call I K that would be a bit tricky because you already have a key here so do a temporary switch call IJ it would be the same then call K I would still be the same and then call J that you have here K that's something you can do you're just changing names and you'll see that you get the same thing that's another way of seeing it okay okay back to our theorem then three things hold one the determinant of a equals the determinant of B to the trace of a equals the trace of B and three the rank of a equals the rank of B so these are the three properties of matrices that we already know that are shared by similar matrices okay what I want to do is prove one and two I'm not gonna prove three it requires some additional let's say lemmas some properties that we haven't discussed in our that would take us a bit off track and and i don't want to do it right now okay but but it is still a valid here so I want to prove one in two so let's start with proving the one about the trace because that's the one we just discussed so proof of two so I want to show that the trace of a in the trace of B are the same and I'm gonna resort to the definition of similarity via an invertible matrix p okay so the trace of a or let's do it the way we wrote it we wrote that be equal doesn't it of course doesn't matter but let's stick the previous notation so the trace of B is the same as B itself since it's similar to a there exists some invertible matrix p such that B is P inverse a P okay here I use the fact that B and a are similar therefore this equals therefore the traces are the same okay here I have a product I have P inverse a times P okay I'm this is a product of P inverse a times P I can rearrange the order that's the previous theorem that we had so I can write this as the trace of P times P inverse a do you agree by the previous theorem P inverse P sorry P inverse a times P and P times P inverse a these are not the same matrices but their traces are good and what's this P times P inverse is I I times a so this is just the trace of a good so very straightforward similar matrices have the same trace using of course the fact that this is an equivalent definition of being similar ok ok let's move on to showing that determinants are the same and it's going to be very similar very similar maybe maybe let's rewrite this and have it all fit on the same board proof so this is the proof of two and here's the proof of one so what's the determinant of B the determinant of B again using the fact that they're similar is the determinant of P inverse a P right that's using the fact that BN a are similar now a determinant of a product is the product of the determinants that's something we we know so this is the same as the determinant of P inverse times the determinant of a time's the determinant of P do you agree these are properties of the determinant another property of the determinant is we know what the determinant of the inverse matrix is the determinant of P inverse is 1 over the determinant of P remember that so this is the same as one over the determinant of P times the determinant of a times the determinant of P good now these are numbers the determinant of a matrix is a number I can calculate them in any order they're just numbers and of course that it's P and 1 over P cancel out and by the way how do I know that this determinant is not 0 I'm taking 1 over it right it's an invertible matrix and we know that for invertible matrices its if and only if the determinant is not 0 remember ok so this just equals I can cancel P and 1 over P and I get the determinant of it good so sure once you have them separately you can switch between them yeah ok so you're asking suppose I moved a over here and then said the determinant of P inverse times the determinant of P is the determinant of P inverse P which is the determinant of I which is one very good one make it shorter because this was short as is but would be an alternative proof which is just as correct very good very good good everybody followed the idea here ok very good ok as so as I said I'm not going to prove three the last thing guy do want to do is to show this on on an example show these properties really hold on an example and since we already had an example I'm just gonna continue that one so I wrote a reminder here we looked at a linear map from R 3 to R 3 which takes a BC 2 minus a b0 remember this we we spent a while on this right and in particular we found it's representing matrices with respect to the standard basis 1 0 0 0 1 0 and 0 0 1 and we found separately its representation matrix with respect to a we call death which was 1 1 0 1 0 1 & 0 1 1 everybody remember this right ok and these are the two matrices we got let's see that they indeed share these 3 properties of the determinant the rank in the trace okay so indeed what we have to do is calculate all 3 for each of the 2 ok so for this let's call this one a and let's call this one B they're similar by definition they're matrices that represent the same linear map and we even saw the similarity matrix the invertible p that does the job right we even calculated what P and P inverse are and showed that B is P inverse ap in this example ok remember everybody with me ok so once I have a different shirt it means that it's a different day we're recording gone I pay attention not to wear the shame the same shirt again and again so so it's a different day so people sitting here have seen this but a while ago okay so everybody good ok so let's calculate some stuff so what is for example the trace of a a is this matrix that no zero sum of the diagonal right minus one plus one plus zero so the trace of a is zero calculating the trace is really the easiest thing right what is the trace of B and don't let all the details get you off our focus right it's just summing the diagonal elements it's zero minus 1/2 plus 1/2 so it's again zero do you agree so that was prop oh that was property number two anyway good what is the rank of a what is the rank of a right a is already in echelon form right it has a row of zeros so the rank is 2 so the rank of a is 2 and by the way from this we know automatically what is the determinant of e0y okay that's one way to say it I wanted to say it from this the rank of a is not full it's not n and is 3 in this case therefore a is not invertible okay and therefore its determinant is zero okay so let's calculate the we have now two options either we calculate the rank of B and we'll find out that it's two and it will follow automatically that the determinant of B is 0 because b is not invertible right or we can calculate the rank of sorry the determinant of B it's going to be zero and it's going to follow that the right o that it's not full but we don't know that right so so it's better to calculate the rank of B once we know it's not three the determinant is going to be is going to be zero okay so let's do that so let's take right so let okay I'm def stinging do I want to really write this but yeah let's do it quickly zero negative 1/2 1/2 negative 1 please allow me please allow me to do this B goes to this is the first row operation okay good everybody understand what I'm doing this is B and the first operation row operation that I'm gonna do is just add Row two and Row three okay good so B by adding Row two and Row three this becomes a 0 this becomes a 0 and this becomes a 0 so it's pretty trivial so I have negative 1 negative 1/2 negative 1/2 and here's my row of zeros right so already I know already I know that the determinant of B is also 0 because it's not invertible because it's rank is not full good but this doesn't still doesn't mean that the rank is 2 but it does because these two rows are linearly independent so in fact I even know without going all the way to the Echelon form right without I already know that this is equal to the rank of B okay so I'm assuming that you in fact already feel comfortable with these ideas and and so I'm not doing all the details but I think you can follow right good so nothing difficult in this example but you can see that indeed two matrices that represent the same map therefore are similar share these three properties good okay so we're gonna stop this one here we're gonna see we're gonna see later that there are more properties shared by similar maps but we need to define them first so we'll do that later

Channel: Technion

Views: 19,398

Rating: 4.7412934 out of 5

Keywords: Technion, Algebra 1M, Dr. Aviv Censor, International school of engineering

Id: l2BEIOnG54k

Length: 22min 28sec (1348 seconds)

Published: Mon Nov 30 2015