 So, last class we have considered a optimal control problem with constraint input and constraint in the state that problem we could not solve completely. So, rest of the portion rest of the problem we will solve it now. So, before that we will just recall the our problems that what is this, we have a optimization function objective function is minimize this what is called the control effort subject to the dynamic equation of the system and initial condition this. Our job is to transfer the initial condition the 0 1 initial condition by using a suitable control strategy. So, that it satisfy the our input constraint as well as state constraint to transfer the state from x 0 to final state x 1 t f and x 2 t f is 0 means that origin this is our problem. So, what we did is this unconstrained optimization problem, we have now that this is the constraint in input side and this is a constraint in state this state x 1 and x 2 which we have written into this form less than equal to form this. So, this 2 equation in turn we will get 3 equation in turn from here we will get 2 in equal to condition less than equal to 0 and we assume this x of t is differentiable with time. Then in order to solve this problem we introduce a state new state variable x dot x 3 dot t we have a initially 2 state variable. Now, we have introduce a third variable x 3 of t dot and that is form with the knowledge of the constraint which is less than equal to 0 from there we have constructing this state equation. This if you see this one equations 3 is 0 when all the constraints as specified here for x 1 x 2 if satisfied then this right hand side of this equation will be 0. If one of the equation at least one of the one of the constraint is not satisfied then we can write the equation 3 is less than equal to 0 for all t when the constraint 5 either it is satisfied or it is not satisfied either when the constraint 5 either is satisfied or not constraint are not satisfied. Then it is less than equal to 0 for all t at least one condition is not satisfied then it will be less than equal to less than 0 if all condition is then this is equal to 0. Then we specify the boundary condition for the new state variable x 3 x 3 0 is 0 x 3 t f is 0 then x 3 t is 0 for all t that from the equation we can write it this equal to 0 for all t x 3 expressions. Now, what is we have to do we will form our standard procedure to form a Hamiltonian function. So, h is form just as before we have discussed earlier once h is form then next step is the our case is del h del x is equal to del h del x we will come later del h del u is equal to 0 from there we got it u is equal to minus lambda of t this expression and our constraint on the input is the mod of u must be less than equal to 0. So, which in turn from this equation we can write it the mod of lambda t lambda of t is must be less than 1 then it will be a input will be less than 1 when there is no guarantee that lambda of t may be greater than 1 in that situation what should be the control input and that control input we have decided on the basis of that point region minimum principle that means we have to select you see this Hamiltonian function that one that Hamiltonian function is here you have to select that Hamilton that u is associated with lambda 2 again u is associated with lambda 2 lambda 2 into this factor will come. So, you have to use the choice of u in such a way so that that function value is becoming this product is becoming negative. So, on that basis of this if lambda 2 is greater than 1 then u must be switched on minus 1 because our control input cannot be more than 1 even if it is more than 1 lambda 2 then when it is less than lambda 1 minus 1 lambda 2 is less than 1 then you switch to plus 1 when it is in between then you switch to that whatever is the minus of lambda 2 suppose lambda 2 is less than 0.5 positive then you give the input the opposite sign of lambda 2. So, that means minus so our set function is generally the set function is 0 when x is greater than equal to 1 this is 1 when a mod of x is less than equal to 1 x of t is same x of t when mod of x of t is less than minus 1 it is minus 1 that is the set function is nothing but a this type of function saturation function this is 1 and this is is minus 1 and when this is also 1 slope is 1 when this if you can say that this is x of t when you have x of t greater than equal to 1 the output that this is you can say y of t the or that is actually output will be 1 when this is less than minus 1 that output will be minus 1 when it is x is in between x is in between minus 1 to that that will be the value of x we have to consider that this is. So, in fact this is nothing but a u star if you see this one u star is nothing but a minus of set lambda t lambda 2 of t this is we can write it that you see when it is 1 greater than 1 this is minus. So, minus sign when it is a mod of this less than equal to 1 that this is the u x of t, but it is minus. So, it is opposite minus of set function this indicates. So, then what you did it that our first condition we is delivered del h del h differentiate with x of t u is equal to 0 from there we got it u star is equal to minus lambda 2 of t. Then another state equation we will get del h del lambda t is equal to x that from that we got the three equations agree that three state variables are there including the new state variables then costate equation del h del x of t is equal to minus of lambda t. So, again we will get three equations then next is we have to use the boundary conditions agree to solve this problems and x 1 of t we know is 0 is given in the problem x 2 of 0 is 1 as is given in the problem, but our that initial state we have to drive to a origin with a suitable control x n keeping in mind that input is constrained and the state also constrained as given in the problem. So, our final state is x 1 t is 0 x 2 t f is 0 and x 3 from our boundary condition of x 3 is 0 x t f is 0 and another boundary condition from the free end point if you just consider the free end point another boundary condition we will get it that one the boundary condition that boundary condition condition for free end point system is h Hamilton function del s of del t this put the value t is equal to t f into del t f plus del s this del x t minus lambda of t that whole t f is equal to t f that transpose into delta x f is equal to this is the boundary condition free end points of the system in terms of Hamiltonian. Now, you see our terminal cost that s is 0 if you note delta t f is free delta t f is free, but delta x f is fixed that means delta f f is 0 and we do not have any terminal condition x as of 0. So, this term will be 0 whole term is 0 and this term also 0 and we have a this term also 0 we have left with this one is h of t h of this at t is equal to t f plus minus that one t f that has 0 and this term delta x of this is fixed. So, this term is 0 whole term will not come into the picture. So, this will be a 0 then we will put what is the Hamiltonian function and there we put t is equal to t f that is next step. So, if you write this equation next that our what is Hamiltonian is a of u star square t is equal to t f of the input you have to find out then lambda 1 t f into x 2 t f plus lambda 2 t f plus lambda 2 t f plus lambda 2 t f plus lambda 2 t f into lambda 2 t f into minus x 1 t f plus u star t f this bracket closed then lambda 3 just Hamiltonian function I have written and putting t is equal to t f nothing else. So, this is t f delta t t is equal to x 1 square t f u s the unit step function x 1 t f plus x 2 t f plus 1 whole square u s minus x 2 t f minus 1 this plus lambda 2 t f plus lambda x 2 t f minus 2 whole square u s x 2 t f minus 2 this is equal to 0. Now, look at this expression this here x t f value is what we want to drive the state form initial condition to the final condition this value is 0 then here is x 1 t f is also 0. So, only this term is there. So, from this one it is a half u star square t f and from there this is equal to 0 this part then from this part is plus lambda 2 t f into u star. So, this is equal to u star t f and this thing you see x t f is 0. So, this term is not there and when x because it is a step input of x t f 0 this is also 0. So, this is 0 and this part is not 0, but whereas x t f is 0. So, this is u minus 1 that this the step input function is for the value of negative value of this one it is also 0 this is 0 and this is not 0 x t f is 0 this is not 0, but this value this is also 0, but this there is a minus 2 term t. So, this is also this quantity is if you see this quantity less than 0 this quantity also less than 0 and this quantity is 0 this quantity is 0. So, this will be the whole part will be 0 whole part will be a 0. So, now our is that one this boundary condition. So, from there you will see that is a lambda what is our lambda u is equal to what note u t f is equal to minus lambda t f that is we have our del h del h dot del u is equal to 0 from there we got it this one if you put this. So, this is the equation of this one our last equation if you recollect this one was that equation number 8 or 9 just check it. So, our last equation anyway this is the this is the equation from there you put it that is a lambda 2 t f square plus minus lambda 2 t f square is equal to 0 this is the lambda 2 t f square is equal to 0 this is the lambda t f is the lambda 2 square this. So, that is lambda t f lambda 2 t f is equal to 0. Now, you we can solve this problems we can solve this problem by considering the our boundary initial conditions of this one what is this one you see the from the basic state equation that our state equation we got it x dot when you are doing that what is called that x dot of t is x 2 of t then x 2 dot of t is equal to minus x 1 of t plus u plus u star of t which is equal to minus x 1 of t minus x 2 of t that we got it if you recollect this one was that you got it from the our state equation that this one if you see the del h del lambda is equal to that this equation this 2 3 equation and you see we have to keep the our state into the that in inside the constraint. So, if you put inside the constraint this then this quantity is 0 this quantity also 0 this quantity of the lambda x 3 dot is equal to 0 and this value is remain 0 for all t as we have considered the boundary initial and boundary condition of on x. So, we need only this 2 equation. So, that 2 equation you have written. So, this is the state equation we got it from del h del lambda of t is equal to x dot of t from there we got it now an x 3 dot if you write it you can write it is equal to 0 keeping that x 3 of 0 is 0 and x 3 of t f is equal to 0. So, this x t is 0 for all the time x 3 of this. So, we need basically this 2 equation then costate equation costate equation if you see the costate equation we also got 3 equation in terms of lambda 1 dot lambda 2 dot all these things and our costate equation finally, we will get it lambda 1 dot is equal to lambda 2 of t lambda 2 dot is equal to minus lambda 1 of t and now the third equation of lambda is lambda 3 dot is 0. So, now you see this 2 equals 2 set of equation and lambda 3 dot is equal to of t is equal to 0. Now see this also now x 1 x 2 lambda 1 lambda 2 dot all these things are you want to solve x 1 x 2 then it is you know you must have a information of lambda 2 then only you can solve it x 2. So, how to solve this one. So, our initial condition if you use this this one the solution of this 2 equation with the knowledge of this and this our solution is like this way x 1 of t is equal to twice sin t plus half lambda 1 of 0 into sin of t minus t cos of t. So, this is and this is this one minus lambda 2 0 t sin t solving this and this we will get it x 1 dot is this and x 2 of t similarly, you will get 2 cos t then half lambda 1 0 t sin t sin t t sin t then minus half lambda 2 0 sin t plus t cos t. So, this is this is the solution of x 1 and x 2 and lambda 1 and a lambda 2 solution we will get it lambda 1 lambda 1 of t is equal to lambda 1 0 cos t plus lambda 2 0 sin t and lambda 2 of t is equal to minus lambda 2 0 sin t and plus lambda 2 0 cos t. So, this now you see we cannot solve x 1 until unless we know the lambda 1 0 value lambda 2 0 values similarly, x 2 also. So, lambda 1 0 lambda 1 0 value lambda 2 0 value we must know it again. So, how to solve this equation so now and we have a one boundary conditions also there if you see this boundary condition we got it lambda 2 t f is equal to 0 the boundary condition Hamiltonian boundary condition free and point boundary condition we got it that one. So, now we have to solve this one you must know that this are two unknowns are there x lambda 2 0 lambda 1 0 lambda 2 0 and you do not know that what is called here also lambda 1 0 lambda 2 0. So, find next is find lambda 1 0 and lambda 2 0 and t f because you have to drive this straight from initial condition 0 1 to final condition 0 0 in time t f. So, t f also you unknown so how to solve so using this n condition what is this n condition this is one n condition another n condition x of 0 or x of 0 is equal to 0 x of 2 0 is also 0. So, now use x of 1 x of 1 t is equal to 0 this equation. So, let us call this equation number is some equation number is a this is the equation number b. Using t is equal to 0 in equation using t is equal to 0 x 1 0 is equal to 0 x 2 0 is equal to 0 in equation in equation a and b a and b we get two equations. So, another equation we will get it at t is equal to t f we know. So, you put the value of a 1 that this condition is a and b this we know x of t x 1 0 is 0 and x 1 this or not 0 at t is equal to t f t f at t is equal to t f we know t a t is equal to t f this x x t f is equal to 0 x 2 t f is equal to also 0. Using t is equal to t f this so in equation 1 and 2 we will how many equations will unknowns are there in equation 1 and 2 we have a three unknowns are there. You see one unknown is there t f is unknown lambda 1 0 unknown lambda 2 0 unknown, but we have a two equation we need another equation what is called that lambda t f lambda 2 t f values is equal to 0 and use the lambda 2 f value equal you will get another equation agree. So, using t is equal to t f in and x 1 t f this is t f x 1 t f x 2 t f is 0 in equation 2 1 and a and b in this equation a and b we will get two equation, but three unknowns what is the three unknowns lambda 1 0 lambda 2 0 and t f is three unknowns, but we have a two equation then another equation I can get it lambda t f is 0 then that means we can write it this equal to minus lambda 2 0 sin t f sin t f plus lambda 2 0 cos t f is equal to 0. So, from this equation a and b two equation I am getting from this is another equation let us call this three equation solving these three equation we will get the values of lambda 1 0 we will get the value of lambda 1 0 lambda 2 0 and t f then easily we can find out the solution of x 1 x 2 and x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 all these things and we and also lambda t lambda 2 t. So, we monitor the lambda 2 when it is a the mod of lambda t is less than equal to 1 then we will switch that u of t when mod of lambda 2 of t is less than equal to 1 then we will take u star of t is equal to minus lambda 2 of t then when it is a greater than 1 this is when mod of lambda 2 is less than equal to 1 then this equal to lambda is equal to minus 1 when lambda 2 of t is greater than 1 then it is minus 1 and you will switch to 1 when lambda 2 is less than minus 1 see that equation according to this we have just switched this one if you recollect this one here this thing we have used it here once you know the lambda t expression. So, lambda t expression I can know once you know lambda 0 to lambda 0 expression then you know what is lambda 0 t of this I think this is the lambda 1 of 0 that that one lambda 1 of 0. So, this you can do it and then you can implement the control that will drive at t is equal to t f that will drive the initial state to a final state at origin. So, this is all about your what is called the solution of that optimal control problem with constant input as well as constant in the state how you have to solve it. So, next we will discuss about the norms the new topics is a norms the norms of a norms for vectors matrices then your signals and systems means linear systems linear systems this norm. So, the norms basically the norms gives an overall measure of the vector size of the vector. So, that norms is a norms it gives an overall measure of the size of a of a one vector matrix signal systems then is systems this norm is often useful for measuring for analysis the for analyzing the growth of numerical examples that norm we generally use for analyzing the growth of what is called numerical errors involved in the process of iterative process again. So, let us call the norm of a vector norm of vectors it is often useful when the norm analyzing growth of numerical errors. Let us call how to find out the norm of a vectors there are different norms different context it is used. So, let us call we have a vector let us call we have a vector whose dimension is n which is written as x 1 x 2 elements of these vectors is x n. So, this is a vector is dimension n cross n n rows one column and this form form a vector space vector space called r cross 1. That means we can short we can write x belongs to vector space r cross one dimension agree that dimension is n cross one. So, now this are the elements of this vector whose dimension in n cross one and that x 1 x 2 is a constant for the time being is a constant elements. So, what a definition of p norm of a vector is denoted by double line whose it is a vector x n 1 by n double line and this p is defined as summation of i is equal to how many elements are there and elements are there. So, x i mod absolute value of each element x 1 x 2 dot that x n you take it then to the power of p whole to the power of 1 by p. So, this is called p norm of a vector that means how many elements are there you take the absolute value sum of absolute value of the elements whole to the power of p then 1 by p p th square root of this one you do it. Now, let us call most frequently we use the norm 1, norm 2 then it is called infinity norm. So, most frequently we use l 1 norm which is also called that one norm this is also called one norm of a vector of a vector of a vector and this symbolically it is represented by x of this suffix 1. This is nothing but a when you put p is equal to 1 is nothing but a norm 1 or l 1 norm l 1 norm. So, it is nothing but a summation of i is equal to 1 to n mod of x i this p is 1 or here you can write for p greater than 1 greater than equal to 1 less than equal to 1 or if or you can write it p is equal to 1 to dot dot dot this p is equal to 1 to this. So, here for p is equal to 1 that it is called norm of a vector. Now, this is you can think of a norm of is suppose you have a set of equation is there you want to solve it by iterative method. Then you the variables are let us call x 1, x 2 dot dot x n variables are there you are solving the a set of simultaneous equation. Now, you absolute value of x 1, x 2 whether it is a convergent or not here. So, present value of x 1 let us call it is at this e 1 is the present value of x 1, present value of this is the present value of x 1 minus the old value of x 1 this is the old value of and that is I denoted by e 1. Similarly, for x 2 now I am putting i, i is where is 1 to n. So, if I take the norm 1 and l 1 norm that means 1 norm of a vector e, e how many components of e are there u 1, e 2 dot dot e n because we have n variables. So, some of the all errors you make it and if you less than 10 to the power of minus let us call 4. Then you stop your iterative process that means it indicates that all the values at least it is to the value 10 to the power minus 4 all the values of e 1, e 2 because I am putting some of all e 1, e 2 there. So, some situation if it is a iterative method and norm 1 also they use it. Next is your norm 2, when is norm 2 it is nothing but a p is equal to 2 this l 2 norm or it is called or it is called 2 norm of a vector of a vector or you can say or l 1 norm or it is called 1 norm of a vector. So, our 2 norm of a vector is what this denoted by x 2 this 2 and which is nothing but a your summation of i is equal to 1 to n mod of i this square whole square root 1 by 2. Here for p is equal to 2 this has a some meaning and this is also called this norm also called is a eclinion norm or it is called distance is basically if it is a 2 dimensional case x 1 square plus x 2 square square root of this 1 and it is nothing but a the distance or length of this vector. Suppose in a 2 dimensional case if your x 1 is here x 2 is this x is if it is this point was x 1 and x 2 and this is nothing but a denoted by x norm of 2 that means length of this vector is nothing but a eclinion norm or the 2 norm of a vector. Again this is also one can use for what is called to find out the what is called in iterative method also you can use this norm to check it whether the algorithm has converged or not or this next is your l infinity l infinity norm or it is simply called infinity norm. What is this infinity symbolically it is represent by x is this one and this when p is infinity that means in that expression p norm expression you what you write x 1 to the power of infinity x 2 to the power of infinity. If x 1 let us say that x 1 x 2 I told you the numerical value if x 1 is less than 1 then x 1 to the power of infinity is 0. Suppose x 1 is greater than 1 this is infinity out of this which is x 1 x 2 all this thing out of this which magnitude is the greatest and it is greater than 1 that information is enough to find out the infinity norm. So, the infinity norm I am repeating is max i mod of x i and i varies from 1 1 2 1 2 dot dot n out of this which one is maximum you take that is the infinity norm of that one logic is this one if you just see the this situation of that one again suppose we have a 3 variables are there x 1 x 2 we have a 3 variables x 1 x 2 x 1 is less than 1 x 2 is let us call is 2 x 3 is 5. Now, you see x 1 is less than 1 to the power of infinity it is 0 x 1 is greater than 1 let us call 2 to the power of infinity and then it is a x 2 which x 3 which is a let us call 10 greater than 1, but then to the power infinity also infinity. So, you are what we are considering the maximum value of x y is considered to find out the infinity norm of a vector this and you see this power multiplied by the 1 by p. So, this this cancelled of this one if you can think of it some of all these things because basically k u is you make it some of the all these things which one is highest you keep it then you pit it highest magnitude of this one p into p p p cancelled. So, it is a ultimately it is coming max i 1 to p dole absolute below of x i you have followed my point this one this if I write it let us call x 1 x 2 and x 3 x 1 value is 0.5 x 2 value is 1.5 x 3 value is 3 then what we will write it that p values of this is summation of x 1 means 0.5 whole to the power of p 1.5 whole to the power of p 3 3 0 whole to the power of p whole 1 by p this is 1 by p. Now, this is equal to 0 when p is infinity this is compared to this this is is because this is the greater than this one. So, you take it the whatever the absolute value of maximum that one. So, this and this cancel only 3 maximum value of from all the components what is that is the infinity norm that is why you are writing it that one. So, this is the you call it the infinity norm of a vector. So, next is called the norm of a matrix similar way we can define the norm of a matrix. So, first is is Frobenius norm it is defined for matrices by treating the matrix elements the matrix elements. As a vector suppose we have a matrix m cross 1 is a matrix which elements are a 1 1 a 1 2 dot dot a 1 n a 2 1 a 2 2 a 2 n we have a m rows are there a m 1 a m 2 dot dot a m n this is the matrix whose dimension is m cross 1. So, the Frobenius norm is defined you that matrices all elements you put it into vector form. That means this vector below this one you keep it this column vector below this one you keep it the next column vector in this. So, whole thing will be vector of dimension n cross this whole thing will be the vector of dimension m into n cross 1. That means this vector below this you keep it this one below that one keep it third column below this one keep fourth column in this way and you will get a matrix m into cross 1. Then Frobenius norm is nothing but a sum of absolute value of all elements square sum of all elements square square root of sum of all elements square. So, Frobenius norm is nothing but a we denoted by this a m cross n of f stands for Frobenius norm is equal to summation of i is equal to 1 to m summation of j is equal to 1 to n m is the number of rows this one and a i j square. So, absolute value is not necessary because it is a square of that one this summation of all elements you take it then take the square root of this one. So, it is like if you see it is like is what L 2 norm it is it is L 2 norm of a matrix you can say L 2 norm of a matrix just like L 2 norm of a vector it is L 2 norm of a matrix is called the feminist norm. Next is your that L 1 norm of a matrix the same as the L 1 norm of a vector. So, that is denoted by a m cross n of 1 is equal to maximum j then summation of mod a i j this i is equal to this is j i is equal to 1 to m and j is equal to what 1 to n because our dimension m cross n. So, you take j is equal to 1 i varies 1 to m you calculate all summation then j is equal to 2 means second column all the elements absolute sum you do it. Third column j is equal to 3 third column absolute sum of all these things you do out of this which one which column absolute sum is maximum that is called the L 1 norm of a matrix. So, this is nothing but a maximum column norm column sum maximum column sum. So, if your matrix if you see what will that L 1 norm this column you absolute sum of this you do it that column absolute sum of this all elements you do it and in this way you proceed up to nth column out of this which one is the maximum value that will give you the L 1 norm of a matrix that is why it is called maximum column sum. Then L 2 norm L infinity norm next is L infinity norm of a matrix L infinity norm of a vector is L infinity norm of a vector is what we have seen L infinity norm of a vector the maximum value of this element of a vector. Similarly, here also L infinity norm of a matrix m cross n is equal to this L infinity is equal to maximum i is equal to 1 comma 1 2 comma dot dot m summation of j is equal to 1 to n absolute value i j this. Now, i is equal to 1 where is j is equal to 1 to n that row first row i is equal to second row out of this all row wise your sum out of the absolute sum then what are out of this which row is the maximum absolute value of the sum that is called the h infinity norm of a matrix or it is that is called L infinity norm of a matrix. So, that is called your also called maximum row sum this has a similarity with the L infinity norm of a vector L infinity norm of a vector L infinity norm of a vector what is telling you take the absolute value of all the elements then you picked up which value which element have the maximum absolute value that is the L infinity norm of a vector. Similarly, here also your row wise you take the absolute value of this one out of this which one is maximum that is the L 2 norm of a matrix. So, next is L infinity norm sorry L 2 norm of a matrix is defined as a m cross n 2 is equal to lambda max a transpose a then square root of that one that means a is a rectangular matrix you see a is a rectangular matrix and if you do a transpose that dimension is to be n cross m and this dimension I know it is a m cross n. So, it will be a square matrix n by n. So, n by n you will get the square matrix of that one you find out the absolute maximum value of this matrix and take the square root of this is a symmetric matrix all Eigen values are will be positive non negative number. In other words all Eigen values are non out of this which Eigen value is maximum you take the square root of that one. So, this is nothing but a this lambda max means maximum Eigen value of this value of a transpose a this will show later this is nothing but a maximum singular value of the matrix a transpose a this is nothing but a maximum singular value of the matrix a transpose a. So, this is all about the vector and matrix norm what are the norms you have to P norm then a feminist norm is nothing but a L 2 norm agree and this or you can do not write it this L 2 norm we write it feminist norm and this is nothing but a L 1 norm h infinity norm L 2 norm this. So, you take an example and verify all the norms how to compute example and this is you can write it is equivalent to a L 2 norm of a vector the way you are doing agree the way I told you that whole elements of the matrix is whole elements of the matrix you convert into vector form then find out that vector L 2 norm. So, this is equal to L 2 norm of a L 2 norm of a vector instead of you write it L 2 norm of a vector when the matrix elements are arranged into a vector form. So, do the example of that one x we have a vector whose dimension is 4 cross 1 is given 0 minus 10 to 1 and a is given 1 0 0 0 5 minus 10. So, this is the value of the matrix agree. So, find out if you find out the one norm of a vector one norm of a vector means some of all elements absolute value that will come 13 that means 10 absolute value then 2 absolute value 1 absolute value is 1 that will be 13 then x 2 norm is nothing but a square of this one then 0 square plus 100 square 10 square then 2 square plus 1 square whole then square root of this one that will come root over 105 agree and next is infinity norm of a that infinity norm that infinity one out of this which magnitude is maximum that means that transpose you can write it that will be the 10. And next is that your matrix A matrix norm A fabulous norm agree this one is you have arranged in the all vector one and some of the all things then it will be coming if you see this one and this will be summation of i is equal to 1 2 3 summation of j is equal to 1 2 3 then mod i 1 j square and whole to the power of this is the fabulous norm to the power of this is the fabulous norm and that value is coming near about 126 to the power of half then a 1 norm a 1 that means 1 norm of a vector l 1 norm or 1 norm of a vector is equal to what maximum column sum maximum column sum is what this is 1 this is 15. So, this will be a 15 or you can say maximum maximum column sum then this is a 2 a 1 and this is a 2 a 1 and this 2 infinity a 1 2 infinity is your maximum row sum is 10 maximum row sum then l 2 norm of a matrix what is this square root of lambda max a transpose a that means that quantity and this quantity a 2 norm this quantity if you do the multiplication this will come your 125 square root of this one that this lambda maximum lambda maximum again value of a transpose a is 25. So, that square root of this one will come 11.18. So, this is the example form how you will find different norms l 1 norms l 2 norms and a l infinity norm always and in case of matrix is a Frobenius norm is what is same as l 2 norm of a vector. So, this is I will stop it here today and next class I will discuss the what is signal and what is system how signal norms physical interpretation of signal norms and system norms that we will discuss next class.