Basis, Dimension, Kernel and Image

okay so in this video we want to consider the concept of a basis of a space of vectors and let me explain why I'm using or I'm saying that the space a space of vectors and not explicitly RN as we will see in this video we can consider subsets of RN such as say a line passing through the origin or a plane passing through the origin in r3 of course RN itself but we will assume the next few videos extend our concept of vectors to more general objects will consider any rectangular M by n matrix as a vector and will even consider functions as vectors so just keep this in mind that what we'll talk here about our N and its subsets will apply to more general sets of objects which we'll call vector spaces so here's a definition of what the basis is so let V be a space of vectors so again V could be say a line through the origin or plane through the origin and r3 or RN for any n but later on we'll see that we can even be more general than that it can be a space of rectangular matrices it can be a space of functions and so on but for now think of V as being our N or a subset of RN so let V be a space of vectors and B be a subset of V so B is a set that contains say M vectors V 1 V 2 up to V M and all these vectors are contained in the space of vectors V right this notation means is a sub set up so be the set of these M vectors is a subset of V the space of vectors which means every element in B is an element of V therefore every vector in our set is a vector in the space of vectors V with this now we can define what a basis of the specific vectors V is so we say that B is a basis of V if B satisfies two conditions so if condition a the elements of B are generators for the entire space and therefore generate means taking all possible linear combinations therefore the span of the vectors in B so the span of these M vectors has to be the entire space right the span of a set of vectors is the set of all linear combinations of these vectors and B will be a basis if the span of the vectors in B gives you the entire space of vectors V so if a that's our first condition and B the vectors in B are linearly independent you and these are the two conditions to say that B is a basis of V and now again what is a basis well condition a says a basis is a set of generators for the space of vectors right if you compute if you find all linear combinations of these M vectors you get the entire space so the vector is in B generate the entire space but the idea is again we want a set of generating vectors for the space to be as small as possible right and think of this if you can generate the space say with four vectors why would you use say five six or more it doesn't make sense you want to generate the space with as few vectors as possible and that's what condition B guarantees if the vectors are linearly independent then we know that none of the vectors would be linear combinations of the others and so you do need all of these M vectors so if you lose one of those you will actually lose part of your space so condition again guarantees that the set of vectors in B generates the entire space and the fact that the vectors are linearly independent guarantees you that the set of generators you have is as small as possible let's consider let's say three examples we'll consider an example for RN for any n then for a plane in r3 and then we'll look at two new definitions regarding rectangular matrices which will give us two new subspaces spaces of vectors let's do one example so consider our end for any end and you'll see that we'll have a very simple and yet fundamental basis of our n for any n so our n is the space of vectors with n components the space of vectors of length n so X 1 X 2 up to xn and again the x i's x1 through xn range over all possible real numbers that's what our n is well again let's split those and components and we'll get a linear combination if you factor in X 1 you will have X 1 times the vector 1 and everything else 0 plus if you factor X 2 X 2 times the vector now 0 1 and everything else 0 up to X n oops plus X n factoring X n out of this vector times the vector all zeros and the one at the very last entry where each X I can range over all real numbers and if you look here what we have is the span of these n vectors we have X 1 times this vector plus X 2 times this vector up to plus X n times this vector and because x1 through xn can range over all real numbers we have here the set of all linear combinations of these and vectors and this is by definition the span of these n vectors so the vector 1 0 0 all the way down the vector 0 1 is 0 all the way down up to the enth and final vector 0 is 0 0 1 at the very end and we give these vectors very special names we call this vector e1 call this vector e 2 up to this is the F 1 the vector e n and these are very simple vector to describe rate if in general e aí the is vector will consist of all zeros except for the I entry which will be a 1 right II won is all zeros but the first entry is 1 e 2 all the entries are 0 but the second entry is 1 up to en all the entries are 0 except the nth entry is equal to 1 so now we have a set of generators right the set e 1 e 2 3 n spans all of our n so this is a candidate for a basis right the first condition for a basis must be that the span of the vectors is the entire space of vectors and there's here is our n and indeed the span of these end vectors is all of our n but there is one last condition we have to check the vectors must span the space which they do here but the vectors must be linearly independent well why is this obvious here well remember that vectors are linearly independent if the only way to obtain is your vector through a linear combination of these vectors is the trivial solution therefore if you construct the linear system the reduced row echelon form of the matrix will be so that every column will contain a leading one therefore every variable is leading which implies a unique solution but think of building the matrix where the columns are a 1 e 2 up to e n this will build I and therefore every coefficient of the linear combination will have a leading one which implies that all variables are leading therefore a unique solution being the trivial solution so automatically here we also have linear independence so indeed this is a basis of our N and it's also called the standard basis of our N and that's a first example of a basis these n vectors form a basis of our n and they're very special they're called the standard basis of our n now one last little definition before we consider our next few examples and that is the concept of dimension if you recall in the previous video we said that if we use our intuition and if you build back from our one just a real line the dimension we think of it as one if you think of our to the XY plane we think of it as having a dimension equal to two if you think of our three the XYZ coordinate system we think of it as being a three dimensional space so in general because our n consists of all vectors with n components and every entry could be anything you want we think of our M being in n dimensional space and if you see here our n is spanned by n linearly independent vectors and so will now define this but the dimension of a space of vectors is simply the number of elements in a basis so you can see here if we say well what is the dimension of RN this will be the number of elements in a basis this happens to be n so the dimension of our n is exactly n let's define this more generally and then we'll just prove very easily that a basis of our n has to always contain and linearly independent vectors so this definition makes sense no matter which basis of RN you choose and no matter which one I would choose we will always have exactly n linearly independent vectors so let's define the dimension properly and then write down what we've just said about our n so if be a subset of V is a basis of the space of vectors V then we denote them to find the dimension of V by the following rewrite simply as we did before the I M for dimension of the space of vectors V and all this is is the number of elements in any basis that you find let's now prove that this makes sense and is well defined in RN but we will prove later on when we consider more general spaces of vectors that this is always true if you have a space of vectors even where the vectors may be rectangular matrices or even functions if you find one basis and I find another they will always have the same number of elements therefore this definition makes sense let's now make a remark about our n sorry about that so any basis of our n must contain exactly n linearly independent vectors so if I give you a set and I ask you is this a basis of our n if you have fewer than n vectors or more than n vectors you don't have to do anything guaranteed it cannot be a basis because the basis for our n has to contain exactly n linearly independent vectors and think of why that is right there are two conditions for B to be a basis of our n the vectors in B must be condition be linearly independent and now we're assuming that it is the case but the vectors in B must also span our n and think of it vectors will spend our n if when you put them in the matrix and you will reduce the matrix you get exactly n leading ones but think of it if you have fewer than n vectors then you will have fewer than n leading ones therefore the vectors will not spend our n so you have to have at least n vectors but if you have more than n vectors now think of it you will lose linear independence if you have more than n vectors when you were reduced since you only have n rows that means that one column will not have a leading one therefore you will have at least one free variable therefore an infinite number of solutions therefore the vectors will be linearly dependent so you see if you have less than n vectors you cannot span our n but if you have more than n vectors the vectors are dependent so the only way to have a basis of our n is to have exactly n linearly independent vectors which proves that the dimension is well-defined because any basis of our n will contain exactly n vectors and so the dimension of RN is equal to n let's consider now a simple example of a plane passing through the origin in r3 and we'll ask to find its basis and the dimension of the space of vectors so this is our second example so find a basis and again there are many bases and bases is the plural of bases but given any space of vectors there's always or almost always an infinite number of different bases so find a basis and the dimension of the plane will call this plane PI and just to make it simpler we'll take 1 times X as the first term let's go with plus 6y minus go with minus that it really matters 9z is equal to 0 so we have this plane in r3 and again it is a plane passing through the origin because the origin the point 0 is 0 is 0 is a solution to the equation replace all XYZ by 0 and it solves the equation so this is a plane passing through the origin we want the basis for it and a dimension of the plane well let's see every time you look for a basis of any space that means that you have to describe the space explicitly well here what we have is implicit right we're at all at the points on this plane are the solutions to this equation what are the solutions well we have to find them explicitly therefore we have to solve the linear system as you will see though this will be quite easy to solve because it already will be in reduced row echelon form so we can write the solution set why is that our three so you can let @y be say s set bt and X will be negative six Y plus 9z therefore negative 6s plus nine T where s and T range over all real numbers now think of it we write the points on this plane or vectors right because points are vectors and vectors are points as a column matrix X Y Z so the points on the plane must be of this form for some choice if as some choice of T so negative 6s plus 90 s and T split the free variables Smt and you will get s times negative 6 1 0 plus T times 9 0 1 and if you look now every point on our plane is a linear combination of these two vectors some multiple of this vector plus some other multiple of this vector and because S&T convention for all real numbers we see that the span of these two vectors will give us every possible point on the plane therefore the plane is equal to the span of these two vectors negative six one zero nine zero one we can call this vector v1 and this vector v2 sorry and now we have the first condition write a basis of its space and the space fee that we consider is the plane in r3 so we have a subspace of r3 right this plane is living inside of r3 but clearly it's not the whole space so it's a subspace of r3 and we now have two generators right the span of these two vectors gives us the entire plane so a question is now do we have a basis well a candidate would be thinking the vectors v1 v2 and B will be basis of by the plane if first condition the span of these vectors give you the whole space which we have just proved so these vectors are generators of the plane and the second condition is the vectors must be linearly independent well remember what we have proved in an earlier video two vectors are linearly dependent if they are parallel if one is a multiple of the other well is v1 a multiple of v2 well if it were zero will be equal to a multiple of one well what times 1 is 0 that's 0 but 0 times v2 clearly is not v1 these two vectors are not parallel so clearly they are linearly independent because neither is the zero vector so let's write that down so since we only have two vectors as generators and they are not parallel and keep in mind this argument only works for two vectors if we had more vectors we would have to look at the matrix we reduce it and make sure that every column contains a leading one but because we only have two vectors no need for it and because they are not parallel the vectors are linearly independent so we have a basis the vectors in B span the entire space and the vectors in B are also linearly independent so we have a conclusion B is a basis of Pi well we also ask and of course the vectors here are negative 6 1 0 9 0 1 but of course we also asked for not only a basis of 5 but the dimension of pi as well well we're color definition of dimension given a space of vectors V the dimension of in space is the number of elements in any basis what we have our bases here so the dimension of the plane which is our space of vectors will be the number of vectors in a basis and we have your basis with two elements so we get a dimension of two and this fits perfectly with our geometric intuition of a plane right a plane in space no matter where you put it is a two dimensional flat surface so this matches our intuition perfectly let's consider now two more definitions which will give us two examples of subspaces of our n and these will be subspaces that will come from fixing a rectangular matrix so here's our definitions and here's a setup we let a be a fixed M by n rectangular matrix and recall that we can view a matrix as a function right we can view the matrix a as a function from RN to RM if you take a vector and you can check it here if you take a vector in RN say X so again a column matrix of length n so if this is an N by 1 vector a column matrix of length n and you multiply the vector by a where a is a M by n matrix this is the action of the matrix we take points or vectors in this space and we multiply them by our matrix a first of all is ax defined well N equals n check the multiplication between a and vector X is defined and the result is a M by 1 vector so call a matrix of length M so this does belong to our M and now with this idea in mind we can form one subspace of RN and one subspace of RM let's first define the subspace of RN this is called the kernel of matrix a and the colonel is quite simple so we write usually k ER for Colonel of a it's the set of all vectors in RN all vectors X that satisfy one property and the property is the vectors in the kernel of a are the vectors are sent to the zero vector in RM by the matrix a so the property is if you multiply the vector X by a you will get the vector in RM but we only take those where a times X is the zero vector in RM and that is the kernel of the matrix we also have the image of matrix a and we notice by im4 image of a and this is simply every possible vector in RM that are of the form a times X so we just multiply every point of our n by a and that's the image it's just multiplying every point of our n by the matrix a well if you remember if you were to write the matrix a one column at a time multiplying by a vector of length n what you will get is all and this you should check as an exercise this will be simply the set of all linear combinations of the columns of a so this will be the span of the columns of it and that's what the image is and if you think of it well a is an M by n matrix the columns have length m so they are vectors in RM and if you take the set of all linear combinations of vectors in RM you will get the vectors in RM and there's exactly what the matrix a times our n is and this would be a subspace of our n so it's worth noting right this is a subspace the kernel of a is a subspace a subset of RN but the image of a will be a subspace subset of our M just for good measure let's just check this it will take us two lines and then we'll consider an example of a matrix and we'll find its kernel a basis and the dimension of it and we'll also find the image as well let's just prove this so we have our matrix a which we said was an M by n matrix so it will consists of n vectors right there are n columns so the matrix a will be say column 1 column 2 up to the nth column and now if we multiply a vector in RN by this matrix let's see what we get we'll have matrix a v1 v2 up to VN then we are multiplying vectors in RN therefore column matrices with n components say C 1 C 2 up to CN and we know from the sheet on linear combinations that if you multiply a column matrix a column vector by a matrix what you will get is simply the linear combination of the columns of a where the entries of your column matrix are the coefficient this will be c1 times v1 plus c2 times v2 up to CN times VN and because we're multiplying all of our N by a the coefficient C 1 through CN can take on any real value of your choice therefore a time's our n gives you all linear combinations of the vectors v1 v2 through VN and this is the span of the vectors v1 through vn therefore the span of the columns of matrix a as we have claimed so now let's do an example let's take see a two by five matrix will find its kernel its image dimension and a basis for both in both cases so let a be equal to say 2 3 4 negative 5 8 2 3 5 4 and say 10 and the size is important right this is a 2 by 5 matrix so the matrix a will be a function from our 5 to r2 let's find the kernel and then we'll find a basis for it and a dimension and then we'll move on to the image but you will see that once we still work for the kernel will actually actually have already have done all the work for the image as well ok so the kernel of a remember consists of all the vectors in R 5 let's write this explicitly so vectors with n components say X 1 through X n so X 1 X 2 X 3 X 4 X 5 but we only keep those vectors are killed by matrix a so a times the vector you equals the zero vector in r2 and again it's easy to check because this is 2 by 5 this is a 5 by 1 the result would be a 2 by 1 matrix so it is d 2 by 1 0 matrix and R 2 and that's your kernel right the set of vectors in RN that are killed by matrix a so once you multiply the vector by the matrix a you get the zero vector in our case in R 2 now think of this equation this is a times x equals B right so think of what we have here the kernel of a consists of the vectors of length 5 that are solutions to the equation ax equals B so we're just looking at finding all solutions to this linear system so we can write this of course as an Augmented matrix and the solutions of our system will be the elements in the kernel so in a sense there is really nothing new here and as always whenever you look for a basis you have to solve for the elements of your space explicitly and here what we have is implicit the vectors in the kernel are the solutions to this linear system well you will never find a basis unless you actually find the elements of your space explicitly so here we have to solve this linear system so X 1 X 2 X 3 X 4 X 5 are the variables the constant terms are 0 and the matrix of coefficients is the matrix 8 so 2 3 4 negative 5 8 two three five four ten well first operation before we multiply this by 1/2 let's do Row 2 minus Row 1 then we'll get 0 0 5 - 1 5 - 4 1 4 minus negative 5 4 + 5 9 10 - 8 positive - this is still 0 now we'll multiply Row one by 1/2 so we'll get 1 3 over 2 - negative 5 over 2 4 we want the reduced row echelon form of our matrix let's finish by doing Row 1 - 2 Row 2 we are only changing rule 1 the zeros will remain zeros so we can recopied Row 2 as is 1 minus 2 times 0 1 3 1/2 minus 2 times 0 3 1/2 2 minus 2 times 1 2 minus 2 is 0 negative 5 over 2 minus 2 times 9 negative 18 but negative 18 if you want this over 2 is negative 36 minus 5 is negative 41 and finally 4 minus 2 times 2 4 minus 4 is 0 now we have the reduced row echelon form of our linear system so we can write the general solution set we have 5 variables X 1 through X 5 as always we handle the free variables which are X 2 X 4 and X 5 let's call those RS and T and now solve for the leading variables X 3 will be negative 9 X 4 negative 9s negative 2 X 5 negative 2t X 1 will be negative 3 half of X 2 which is our it's negative 3 half R plus 4 T 1 over 2 times X 4 which is s and now we have the complete solution set where our s and T I'm running out of space here but they range over all real numbers and now from this we can write our basis if we wanted we could knock off these over twos by replacing R by 2 R so if you replace R by 2 R you've knocked off the over 2 here so get just negative 3 R if you replace and here be careful there's an S here too if you replace s by 2 s 2 s over 2 is just s and here you'll have negative 9 times 2 s this will become negative 18 times s and this will give us a basis without fractions and as always if you can try to avoid introducing fractions but now we're good to go we have found the elements of our space explicitly and you'll see that once again to get the basis we'll just add the split the three free parameters so the elements in the kernel where the solutions to this linear system we have found all of them so we can replace now x1 was negative 3r plus 41 s x2 was next to are sorry x3 was negative 18s negative 2t X 4 is 2 s and X 5 ways T let's not split up all 3 free variables so R times the vector negative 3/2 there's no R no I know R so 0 0 is 0 plus s times 41 0 negative 18 2 & 0 plus T times 0 0 as there is no T negative 2 0 positive 2 and look what we have now every vector and the colonel must be equal to this for some types of r SN t therefore is a linear combination of these three vectors so this is v1 say call this v2 and call this v3 and because our s and T are three variables they are allowed to range over all real numbers therefore the span of these three vectors recall the span is the set of all linear combinations of given vectors but because our essence you're free we do have here all linear combinations of these three vectors so the span of v1 v2 v3 gives you all the elements in the kernel so the kernel of a is the span of these three vectors v1 v2 v3 so we now have here three generators the vectors v1 v2 v3 through all possible linear combinations give you all the vectors all the elements in the kernel the only question is condition B are these three vectors linearly independent right so we have here a possible choice of a basis condition a is satisfied the span of B is the whole space we have to check now for linear independence and this I will leave up to you because remember to check for linear independence put these three vectors in the matrix row reduce and if every column contains a leading one therefore we have a unique solution and so we have linear independence and this is really easy to check using row reduction if you wanted to you could be sneaky and you could avoid the row reduction by looking at this because this linear combination can be collapsing to this vector and you see if this vector were equal to the 0 vector well look at this entry T would have to be 0 look at this entry this would also have to be 0 therefore if 2 s is 0 s would have to be 0 and this entry would have to be 0 but if 2 R is 0 or we'd have to be 0 and you see the only way that this vector equals the 0 vector is if R equals 0 x equals 0 T equals 0 but if our s and T are all equal to 0 and that is the only solution by definition these three vectors are linearly independent so either you notice this shortcut or even if you don't put the three vectors in the matrix will reduce and check that you have in each column a leading one either case this is a basis of the kernel so we had to dimension the dimension of the kernel or any space for that matter is the number of elements in any basis and here our base is contained three generators therefore the dimension of the kernel of our two by five matrix is equal to three and that completes our look at the kernel we found a basis and we have the dimension let's now look at the image of matrix a find a basis for it and the dimension and you'll see that we already have done all the work if we only look at the reduced row echelon form of the matrix eight so recall that the image let me just rewrite the matrix a it was the matrix two three four negative five eight two three five four positive 10 and recall that all the image is is the span of the columns of the matrix a so it is the span of the vectors two two three three four five negative five four eight ten so right away we already have a set of generators right right away by definition of the image it is the span of the columns of a so we already have a set of generators the only question is are the vectors linearly dependent or independent if they're independent we're done if they're dependent the question is which vectors do we throw away to reduce our set through a basis let's give these vectors name this is vector v1 v2 v3 and don't confuse those as the ones we've used in the kernel right we call these vectors v1 v2 v3 and I we're moving on forget this and now we call these vectors v1 through v4 whoops v4 and v5 so we have five generators the question is do we need all of them well we already know that they are dependent right the vectors are in r2 they only have two components and so if you're rubbing through this matrix we know that we can have at most two leading ones but we have five columns so at least one column will have no leading one therefore an infinite number of solutions therefore automatically these vectors are linearly dependent the only question is which one can we throw out and not lose anything and the answer lies in the reduced row echelon form of eight remember the intuition if you have a leading one in a column this means that this column generates one dimension if a vector does not have a leading one it generates nothing new and you don't need it so let's put this with what we have at this point we have reduced the matrix a to this form the first column does contain a leading one which has that vector one vector v1 will span one dimension so you have to keep this vector vector v2 that's not container leaving 1 therefore this second vector gives you no additional dimension and you can just throw it out and it's clear from just by looking at the two vectors because foot vector v2 is just three half of vector v1 so V through is just a multiple of V one so if you already have v1 you don't need v2 and this was already clear from the reduced row echelon form of the matrix no loading one this vectors gives you nothing new you can drop it and you lose nothing now v3 does contain a leading one so we will allow you to build to span an additional dimension so you have to keep v3 but v4 and v5 don't have leading ones therefore they don't spend any additional dimension so you don't need v4 and v5 you can drop them and now if you leave if you are left with these two vectors they will form a basis because if you reduce the matrix format these two vectors you will get in the N these two columns so you'll have two columns two leading ones which proves linear independence and they will be a spanning set for the image and now you have your basis and that's it so the dimension of the image actually is simply too and if we notice here there is something interesting right the dimension of the image sorry is the number of elements in its basis but the business contains two vectors therefore the dimension of the image is two this is a parenthesis but this happens to always be true if you recall a was a - well it's right there we need to recall but a is a two by five matrix and if you notice five is two plus three this will always be the case if you add the dimension of the kernel of a matrix with the dimension of the image of a matrix it will always be equal to the number of columns of the particular matrix and this is not an accident proving this is a little bit more subtle so we'll just make this observation and we won't go any further one last mention or something worth mentioning is the image of a is something very simple if you think of it the image of a is the span of these two vectors but these two vectors are in r2 and they are not parallel to each other right and recall that vectors will span r2 if and only if the reduced row echelon form of the matrix that they give you will contain two leading ones but as we have just done before if you start with just the vector v1 and the vector v3 and you will reduce you get a leading one in each column so because the reduced row echelon form of the matrix two two four five contains two leading ones these two vectors span r2 so the image of a is actually simply r2 the entire XY plane so if you think of it we could give now a even simpler basis for our image because the image is just R to the whole of the xy-plane any basis will do so we could give instead just the standard basis of our 2 1 0 0 1 and this would also be an appropriate basis for the image as the image is all of our two because the basis vectors span all of our two and now this is a simpler basis being the standard basis of r2 and that's it

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Id: 2lafIhB7ajk

Length: 56min 57sec (3417 seconds)

Published: Mon Oct 14 2013