 set of vectors and so on. So, I started by defining the set of vectors and I had a special notation for that. So, we have today we have to look at specific things like, so basis and dimension of a vector space or a subspace and rather than just giving definitions I want to give some examples today. So, that why these ideas are being perceived should become a little bit clear. Why we are looking at these funny spaces? Why would we look at the set of all continuous functions and then call it a space or why we would look at set of all long views and call it a subspace. So, yesterday we had developed this notation. This notation was a set S let us say was defined as x i where i is equal to 1 to this should be a finite set in some situations it could be a finite set. I am currently defining a finite set of n vectors. This means there are n vectors S belongs to some vector space. So, x is my vector space is a subset of x. Here one way of so one example that I gave was from r 5 I said that you have two vectors say x 1 this was 1 2 3 4 5 x 2 1 and putting some arbitrary vectors from r 5. So, I can define the set S which is x 1 and x 2 elements in this set these are two elements in the set S is subset of r 5 S is a subset of r 5. In some cases in some cases I will define a set S to be set of polynomials I define the set S which is set of polynomials. So, my first vector here my first vector x 1 z is 1 x 2 z. So, in this particular set in this particular set I have n plus n plus 1 vectors I have n plus 1 vectors which are denoted like this. This is the first function this is the second function x 2 z second vector this is the third vector this is the n plus 1 vector and so on. So, the notation here is generally it could be used in the context of n dimensional spaces it could be used in the context of function spaces it could be in the context of polynomials depends upon what context we are on. So, that is that is first thing that we yesterday we talked about linear combinations. So, F is my field and if I take some scalars say alpha 1 alpha 2 alpha 3 which the scalar set this belongs to F if the scalar set belongs to F then we define a set of vector which is linear combination which is linear combination of the original vector. So, I can construct a new vector x which is alpha 1 x 1 plus alpha 2 x 2 and so on this alpha n that is n. So, this linear combination in the later example where we took vectors as polynomials would be a nth order polynomial in n minus 1th order polynomial in the case of n dimension it will be a vector in r 5 I can talk about a new vector x which is alpha 1 1 2 3 4 5 5 4 3 2 1 when alpha 1 and alpha 2 are any two arbitrary scalars drawn from r. This is linear combination and then we said that we have this notion of span of a set span of a set is set of all possible linear combination span of s which is very times denoted as square bracket s is. So, I take all the linear combinations of elements of s and then what I get is span of if I take all possible linear combinations in this set and what I get is a span of this particular set. It is also example in three dimensions I do two vectors I set span of these two vectors will be a plane cutting through the original to be a plane cutting through the original. If I take three vectors in the same plane span will be again plane cutting through the original. So, it is not going to be different. Well this subject is very abstract. Let us let us get to some complete examples which probably you have seen in your undergraduate and then let us connect why do I need span why do I need to talk about subspaces where I do where we leave them and then we will move on to this concept of linear independence linear dependence and then talk about basis. Before I do that let us let us get some concepts clear about the span business. So, where do we leave span why implies it why is it important. So, something that we actually keep using all the time these concepts are used when you solve linear algebraic equations differential equations to keep doing it unknowingly without understanding what is actually happening because some of these concepts are not introduced in undergraduate. Let me start with an example which is simple solving linear algebraic equations. So, my example one is I want to solve a x equal to b where a is an invertible matrix. So, I am taking one simple matrix here 1 1 0 0 1 1 1 0 1 into x that is x 1 x 2 x 3 gives me b 1 gives me b 1 b 2 b 3 this equation I can rewrite slightly differently you will start getting insights why I am talking about the span. I will write this as 1 1 0 into x 1 plus 0 1 1 into x 2 plus 1 0 1 into x 3 b 1 b 2 I will just rewrite it this equation b is a vector b is a vector which is linear combination of first vector, second column vector and third column vector. What is span of these three vectors I want to show is same as three dimensional vector space. What is span all possible linear combinations? So, all possible linear combinations are these vectors linearly independent well by inspection for this particular case because you know about linear algebra you can see that these three vectors are linearly independent. What will be what will be span what will be all possible linear combinations of three linearly independent vectors in three dimensions? Entire space ok will be entire space. So, I can write a solution which is x this is a inverse b inverse I suppose you know this and this a inverse b turns out to be b 1 times half minus half b 2 times half half minus half plus b 3 times the computer solution of this system of equations ok. You can do it very easily because in the definition it was much order method in compute inverse. I am just writing the solution in terms of a inverse b except that I have chosen to write like this I have chosen to write this ok. Give me any b b vector I can write the solution like this ok. So, where does the solution belong to? Solution belongs to the span of the solution x belongs to span of this vector this vector and this vector. Again from undergraduate experience you know that these three vectors are linearly independent ok. So, the solution actually belongs to the span of the span of these vectors. Now, this is a well-behaved example matrix A is invertible I am going to move on to another example where matrix A is not invertible and I still want to find the solution matrix A is not invertible I still want to find the solution and then we will get this concept of the solution belonging to a space which is not R 3. In this particular case what is the solution space? So, x x belongs to minus half the solution actually belongs to span of this all possible linear combinations right. If you specify b I will get one particular x how many ways I can specify b? Infinite possible ways even b to b 3 can be chosen all vectors in R 3 can be put here I will get one particular solution. So, this is this is an example where you are look at a span now let us look at another example then the things will become more clear ok. I am just going to change this matrix to slightly different matrix minus 4 minus 1 minus 2 4 and 2 4 minus 8. We will look for a solution which is which is for 0 0 0 I am going to look for a solution which is 0 0 0 well. So, basically I want to solve for 1 minus 1 2 2 minus 2 4 and minus 4 4 minus 8 this becomes 0 0 0. Is this matrix invert people first of all? No sir. Why? Linearly. The columns are linearly dependent I just multiply this column by a factor I will get this I just multiply this column by another factor I will get this column. Is this problem solved? Is it going to 0? Number of unknowns are lesser than equations. Number of unknowns are suffering. We have known three equations. Linearly dependent. There is only one independent equation sir. Yeah linear dependence. So, these are linearly dependent and I should be able to get infinite solutions however get them. So, this matrix is non-invertible. So, as the columns are linearly dependent this matrix is not invertible. Can I find a solution? Can I find a solution? Simple solutions to this problem. So, what about to well I have not decided in 0 0 0 trivial solution. What I want to know is are there non-trivial solutions? Are there non-trivial solutions? So, what about x 2 is equal to 2, x 1 is equal to minus 1 and x 3 is equal to 0. Will this form a solution? Multiply this by 2, multiply this by minus 1. I will get 0 the last column is multiplied by 0. Other candidate solution of course is multiply this by 4, multiply this by 0 and multiply 2 possible candidate solutions. I get 2 possible candidate solutions here. My first solution well remember these are linearly dependent columns which means there are same equations represented again multiplied by scalars. Okay. I am not taking the equation view point. I am taking the column view point. I am not taking the row view point. What is different between row view point and column view point will come to that at some point later. Okay. So, now I have 2 possible solutions. I have one solution which is x 1. So, I would write this as 2 minus 1 0 and I have x 2 which is which is 4 0 1 2 possible solutions. But equations that are infinite solutions. How do I get infinite solutions? Any linear combination. Any linear combination. So, if I construct a vector which is alpha times 2 minus 1 0 and beta times 4 0 1. Just check. This also will be a solution because A times this vector is 0 plus B times beta times this vector is 0. So, 2 solutions add to 0. So, any vector which is what do I mean by any vector which is So, what I want to say here is that a solution lies in the span of these 2 vectors. So, solution are these linearly independent? These are linearly independent. Solution lies in the span of these 2 vectors. What is the dimension of the space created by the span of these 2? Because there are 2 linearly independent vectors. All possible linear combinations of these 2 vectors will give rise to a 2 dimensional subspace of 3 dimensions up to 2 dimensional subspace of R 3. So, span is something which you have been using without knowing what is happening now let us graduate from these examples to differential equations. So, that you get the feel of how we can visualize this concepts to some other spaces than what we have been looking for. So, my third example I already covered 2 examples. So, third example is not A x equal to b well conceptually it is not different from A x equal to b, but you will take some time to realize that what I am talking about is one and the same thing. Okay let us look at this particular example. We have a transition to an example which is different from linear algebraic equations. I am doing a differential equations. Now, while this equation is not too different from A x equal to b, A here would be this operator differential operator operating on vector u and giving me 0 vector okay giving me 0 vector. So, this is this is exactly like A x equal to 0 except I have to write as d 3 by d z q plus 6 d 2 by d z square plus d by plus 11 d by d z plus 6. This operator operating on u z is equal to 0 actually 0 0 is the 0 vector in this space. What is this funny space I have not talked about it earlier. This is set of trice differentiable continuous functions set of trice differentiable continuous functions on interval 0 to 1. So, z belongs to this is a set of z belongs to 0 to 1 and this particular set is trice differentiable continuous function c 3 0 to 1 is trice differentiable continuous function. Well you know how to solve this problem. I have not given initial conditions. I want to get a general solution. I want to get a general solution to this problem. How do you do that? You write the characteristic equation and then you write the solution. So, let us that what you already know. So, the characteristic equation is will be p 3 plus 6 p square plus 11 p plus 6 is equal to 0 and then I can write p plus 1 is equal to 0. So, there are three roots minus 1 minus 2 minus 3 and then all of you know how to write the solution. Point out here is something different. I will talk about a span. So, the general solution to this particular problem is given by u z is equal to alpha e to the power minus z plus beta e to the power minus 2 z plus gamma e to the power minus 3 z. Look at these three vectors. Where are the vectors coming here? The vectors are 1 2 3. So, this is my x 1 z this is my x 2 z and this is my x 3 this is my x 3 z. So, a general solution a general solution to this problem it belongs to a span of all possible linear combinations. What will fix alpha, beta, gamma? Moment you specify initial conditions. Moment you specify initial conditions alpha, beta, gamma will get fixed. Well, if you are not at this earlier in the VTech we will be doing these solutions later. This is linear combination of these three vectors. So, I can say that u z actually belongs to span of e to the power minus z e to the power minus 2 z and e to the power minus 3 z. So, this so span is something which I need to know something that I need to know. If you remember there are two solutions to a problem of this type. If there is if there is something on the right hand side there is something on the right hand side if there is a forcing function. Then you have a particular solution and you have the general solution which is the general solution will mean this can be a particular solution depending upon the forcing function. The general solution is actually governed by the initial condition and span of these characteristic vectors. Any solution any solution which is where forcing function in 0 will always lie in the span of these three vectors. Always lie in the span of these three vectors. Cannot leave this solution of this cannot reduce subspace. What is this? This is a what is this is a subspace right is it a subspace? It is a subspace all possible linear combinations of these three is a three-dimensional subspace underlying space which we are talking about the underlying space which we are talking about is twice differentiable continuous functions. Each one of them is twice differentiable Do you agree? All of them are twice differentiable. First of all I wanted to understand that matrix A operating on a finite dimensional vector x giving me 0 and this operator operating on this vector belonging to this space are not fundamentally different ideas. They are same things geometrically same things nothing different. If you start doing I do not do that you cannot visualize these spaces you can visualize only in three dimensions but believe me that if you understand in three dimensions what is happening very well. Okay we talked about a two-dimensional subspace of R3 just now right in the same sense see what is a two-dimensional subspace in R3 is it a very thin set there are too many vectors in R3 right and R2 is a plane passing through the origin it is a very thin set there are fewer vectors relatively smaller infinity than this bigger infinity right. Now in the same sense well I can construct vectors in this space which are you know minus 2, minus 3, minus 4, minus 5, minus 7 all of them belong to cryo distance right actually we will come to this idea of basis and what we show is that this particular space is an infinite dimensional space there are infinite vectors in the spaces linearly dependent vectors that we can find in this particular space are infinite at the right hand side not be 0 this would not have been a subspace which one solution if you will get a specific solution moment I specify a initial condition if I specify initial condition u 0 is equal to something say alpha not alpha I will take a and then d u dash 0 is equal to b and u double prime 0 is equal to some c moment I give you these three initial conditions these are some specific numbers you know you can take minus 1, 2, minus 3 I have taken some arbitrary numbers okay moment I fix this okay I will get one specific value of alpha beta gamma I will get one specific value of alpha beta gamma which is one element okay now if I change the initial condition what will happen I will get another value of alpha beta gamma okay I have to belong to sir but if the right hand side of the main ODE had not been 0 then 0 vector would not have been in the span then it would not have been a subspace when you have something here f of z how do you solve there are two solutions one is the characteristic solution this part will anyway appear plus something will appear here so this part is invariant whether there is something on the right hand side nothing other right hand side attack side can be 0 vector can be non-zero vector this will appear all the time so some part of the solution will always belong to this subspace you cannot escape it this is characteristic of this particular equation okay so so these three vectors these three vectors are kind of manic to this equation okay this can change the right hand side can change this can be psi in cause what is important okay but this is something like inherent nature of this differential equation you cannot you cannot change that may change only if you change these coefficients okay you change only if you change the coefficients okay so this is this is fine let us move on to another example which is again differential equation and it will sort of create what I am talking about as dimension that I have not introduced formally what is the dimension of the vector vector space but this particular example is leading you to that idea of dimension well there is one question one question these three vectors are they linearly independent yes why you have to prove it so I should prove that alpha e to the power minus z plus beta e to the power minus 2z plus gamma e to the power minus 3z is equal to 0 actually which means 0 function 0 is 0 function can be found only when alpha beta gamma are 0 so these are linearly independent okay can you prove it there is one example which is here I don't know whether the time to do it but you should look at this example this one example in my notes how you prove that these three vectors are linearly independent it's not it's not that straightforward you have to do some some amount of thinking okay let's move on to this fourth example so my example four is the boundary value problem now boundary conditions this is the differential equation which is satisfied in the domain z going from 0 to 1 okay then I have this boundary condition at z is equal to 0 so this this is u 0 is equal to 0 and then I have second boundary condition p is equal to 1 so this is u 1 is equal to 0 because this kind of problems will appear when you start solving partial differential equations and so-called strongly penetrating which will be covered in the other course so you might have seen this in your undergraduate this kind of problems we visit in p-transfer in master's work what's in the way u z is equal to alpha 1 sin pi z plus alpha 2 sin 2 pi z a general solution here cannot be expressed in terms of finite number of vectors for this particular problem a general solution cannot be a right that taking only finite number of vectors what are vectors here sin pi z yeah so this is my this is my x1z this is my x2z this is my x3z and so on how many such vectors I have infinite I have infinite vectors okay where does this solution belong to span of what is this what is this this is linear combination of vectors linear combination of vectors right same idea which was we talked about earlier take a set as okay what is the underlying space here somebody said it c2 c2 0 1 okay the underlying space here is c2 when you write this notation it has to be square vectors because two n's are included there are two boundary conditions not just so twice difference but continuous functions on intervals 0 to 1 that is the underlying space okay so where does this where does this solution belong to solution belongs to the span of these are vectors these are vectors these vectors belong to twice differentiable continuous functions okay a linear combination of them gives me a solution to this problem okay infinite linear combination not just one value these well in a partial differential equation these coefficients alpha 1 alpha 2 alpha 3 will be fixed by some other particular solution which this comes as a coupled problem in the partial differential equation so there will be one part of the partial differential equation the solution to this problem and the other part will be particular solution so that that is you can look at the book where was the push forward on because that's it that will tell you how these solutions are here but the point that I want to make here is that you see actually belongs to the span of belongs to span of this set of vectors belongs to span of this set of vectors so span is a very very important concept that we keep requiring everywhere not being introduced just for the sake of it in this particular case in this particular case the space or the subspace of subspace of twice differentiable continuous function which is which is fine by these vectors okay these infinite dimensions these infinite dimensions these are linearly independent vectors well coming to a point where we have to define what is the independence and then we have to also worry about the basis business is this clear are these four examples clear somebody has any doubt please please feel free to stop me at any point if possible if you take initial condition it depends upon the initial condition so if you take initial condition to be zero zero zero only possible solution that you will get this depends upon what is the initial condition of u zero u prime zero u prime zero okay all the three initial conditions are zero zero zero as for beta government is zero so the elements of this functions space c2 sir are they the functions are the value of those functions no they are functions they are the functions they are not value of the functions so then why do we need to specify zero to one or any elements for that zero to one in particular problem this particular problem would be heat transfer problem it will be something like conduction in a rod zero to one comes because you non-dimensionalize the space variables and one is it could be zero to l if you want does it matter sir why the limits why can't we just succeed fine if you if you no no no these are different spaces if you have if you you may have a situation see there are some problems in physics or in chemical engineering where you have some infinite dimension of rod and so you may have a scenario where you know you have space which is c2 minus infinity to infinity or c2 zero to infinity these are different spaces like this this space is different than the elements are no no a function defined over zero to one is not same as function defined over zero to five or you know minus five to five there are different functions there is some part which is not included here okay they are not same and it will not even be continuous sir no it will not even be continuous in some other region exact that is another problem it will not be continuous it might be continuous in in some region when not continuous in some other region so all kinds of all kinds of on to this basis and dimension business somehow what is basis what is dimension of the space or a subspace okay the formal technician of independent dependence that we will find in the notes so we are going to define abstract terms we start with this set s which is we start with this set s okay zero vector well then I am writing zero on the right hand side it means zero vector in that particular space because now if you are if these vectors are n-dimensional what will be this zero well let me let me qualify it where to take a bar over it zero bar it is zero vector okay zero bar is zero vector so if this is equal to zero bar that is zero vector in that space if it is function space if it is function space set of continuous functions over zero to one what is the zero vector zero function zero function everywhere okay fz is equal to zero on zero to one that is a zero function okay so that is what it means it does not mean deliberately zero at one point if it is upon one space okay so if there exists a linear combination such that there exists a non-zero linear combination there exists a non-zero linear combination which gives a zero vector which means if I can find some non-zero elements alpha and alpha 2 which will give me zero vector then these elements are linearly dependent else there if you cannot find if you cannot find a non-zero combination which means only way to get a zero vector is to put alpha 1 equal to 0 alpha 2 equal to 0 alpha 3 equal to 0 all of them are zero linearly then they are linearly independent what is basis linearly independent vectors which generates x yeah basis of a vector space nothing but the set of linearly independent vectors which generate the entire space minimal set minimal set okay which generates the entire space so the minimal linearly independent set okay minimal linearly independent set which generates the entire space yeah sir why minimal because because I can infer in three dimensions I can take four vectors whose span can generate the entire space sir but they won't be linearly independent huh they won't be linearly independent one will be dependent on yeah the minimal set okay the minimal is not really required the linearly independent set which is which generates the entire space when you say entire space actually it's the minimal set again when do you say that a space has finite dimension finite elements the number of elements in the set is finite okay number of elements in the set is finite it's called a finite dimension space if the number of elements in the basis set the number of linearly independent vectors in the basis set is infinite that is called infinite dimension space we saw an example of infinite dimensions space just now linear combination of sine sine sine phi x sine 2 phi x sine 3 phi x sine 4 phi x infinite okay all these vectors actually form a basis all these vectors actually form a basis we can show that they generate the entire sub space okay the dimension of that particular subspace is infinite why because number of linearly independent you cannot find a combination which will which you cannot find a non-zero combination which will give the sum of all those vectors equal to 0 linear combination of those vectors to equal to 0 so that means the dimension of the space generated by sine phi x sine 2 phi x sine 3 phi x sine 4 phi x is infinite okay so as we started with two dimensions spaces all that we are doing is just trying to you know generalize ideas from two dimensions just remember that we're just trying to generalize ideas for two dimensions visualization geometric ideas you should be clear in two dimensions you understand the dimensional world very well no difficulty in understanding all this infinite dimension and the n dimensional business okay so just just be clear about the three-dimensional geometry and then that's enough all visualization can be done using three dimensions the concepts are just being generalized so this grand generalizations will occur in mathematics 1850 to 1950 which is already a period with these grand generalizations you can actually view you can have a very different view of engineering or applied mathematics so you will see that you are just working with the if you start having this viewpoint then working with x equal to b infinite dimensions or working with some boundary value problem or some differential equation is not different operators change what we call is a vector change what we call is b on the right hand side change actually actually the problem that we talked about you know we talked about this problem of d3 by dz q plus something something 6 into uz if i put this equal to some fz okay this is fundamentally not different from ax equal to solving for x equal to b in three-dimensional spaces is same as solving for this equation in price differentiable continuous functions set of time fundamentally not different the same equation different different space underline space is different we are not doing something fundamentally different when you are solving this and solving this okay the specific solution which i wrote was for f of z equal to 0 that's a specific solution and then if you write if you take some fz whether you can solve this problem or not that depends upon that depends upon the span or the characteristic equation and so on so is it the same idea which will apply here when you solve x equal to b the same idea will apply here so we will continue in the next lecture with this is not enough this is not enough we are just defined these spaces we are defined span media dependence we need to go further and say now when in three dimensions you know i had something more important i had length of vector now how do i generalize length of vector and why generalize length of vector that is another thing because we will need to define something called conversions limit and so on that's why we have to define structures like norm we will talk about in the next lecture