we've just learned about null spaces and now we're going to take a look at column spaces what is a column space well it is the set of all linear combinations of the columns of matrix a so keep in mind matrix a is comprised of vectors and the column space of a is the span of those vectors so this all linear combinations of the columns of a shouldn't look real familiar too because we've talked about that before and that just means it's the span of those columns so again we keep working on these same concepts over and over and they might look a tiny bit different so it's difficult to keep everything straight but we know that if we can write a linear combination it's in the span and that's what we're looking for is the span for the vectors in matrix a another way we could write this is that the column space of a is equal to all be such that B equals ax for some X and RN so again it's every single value that provides us with a solution which would mean that it's in the span a theorem in your textbook says if a is an M by n matrix call the column space of a is a subspace of RM and we're actually not going to prove that one because we know based on a proof I think we proved it in a previous video either that or we just talked about it that the span of vectors is in fact a column space of RM so we're not going to prove that but just know that is a theorem in your text that if the column space of a is a subspace of RN here's a great practice for us not only in null space but also in column space so we've got one vector a and I'm sorry one matrix a and one vector U and I'm asking if u is in the null space and of U is in the column space so this would be a good practice to help us to know exactly how to go about solving these so as a reminder if we're looking at null space we're saying that a times u would have to be 0 for this to be true so that's pretty straightforward all I have to do is do some multiplication 2 negative 6 12 make sure you multiply in the proper order negative 3 and 9 9 0 0 1 x vector u which is 3 1 0 that's going to give me 6 plus negative 6 plus 0 and then negative 9 plus 9 plus 0 and then 0 plus 0 plus 0 and I end up with 0 0 0 and because this is the zero vector which is what I was looking for then I can say therefore u belongs to null a so now let's switch gears and talk about the column space for the column space we're going to take a and augment it with you and then we're just going to do row reduction and determine if it's a consistent system or inconsistent so if the system is consistent then you would be in the column space of a and if it's inconsistent then it's not so that means I've got 2 negative 6 12 negative 3 9 9 0 0 1 augmented with 3 1 0 and I'm just going to do my row operations so again there's really no right or wrong way to do your operations I'm just trying to get it to the point that's either in reduced row-echelon form or I find out that I don't have enough pivots and therefore it's inconsistent so the first thing I'm going to do is I'm just going to reduce my first row by 1/2 oops that is not for half of 12 still six and then I'm going to reduce my second row by negative one third so that gives me positive one negative three negative three negative one-third and zero zero one zero so it's pretty clear I'm going to have a pivot here and it looks pretty clear I'm going to have a pivot here probably but let's go ahead and keep going if I do another row operation I'm going to keep this one three six three halves and then I'm just going to subtract the two rows so one minus one is zero negative 3 minus negative three is zero 6 minus negative 3 is 9 three-halves minus negative 1/3 is 11 6 and this is zero zero one zero and then hopefully at this point you can see that there is no pivot in the second column and therefore this is inconsistent you could obviously keep going and if you kept going you would end up with a zero zero zero zero but it's really not necessary because once I got to this point it's pretty clear that there are not enough pivots so because this is inconsistent and then therefore you does not belong to the column space of a here's another practice for us to try this one asks us to find a matrix a such that W is equal to the column space of a and then it gives us this matrix W which is sort of defined in a weird way but this is actually a fairly straightforward question I know that W can be rewritten as a linear combination so I can say that this is a times 4 1 0 plus B 2 negative 1 negative 4 and then of course a and B are real numbers and so essentially what I'm saying is this is the span of 4 1 0 2 negative 1 negative 4 but I still haven't answered the question because the question says find a matrix a so what they want me to do is find a matrix a well no problem a is then 4 1 0 2 negative 1 negative 4 there's my matrix a that exists such that W is equal to the column space of a here's a great practice for you to try on your own so I would like you to try this one and when you're ready press play to see how you did if you were not quite there yet that's okay you can just wait and I'll go over with you all right let's take a look I'm asking you to find both a non-zero vector in the column space of a and then one in the null space of a and you'll find that the nonzero vector in the column space of a is any one of these columns written as a vector so I'm going to choose say 4 negative 1 6 that is a perfectly good solution I could have chosen any one of those columns and that would all be the correct solution for the second part a non-zero vector in the null space of a requires me to take a and row reduce it and augment it with the zero vector so I'm going to take 1 0 3 4 0 1 4 negative 1 0 2 0 6 augmented with the 0 vector and do some row reductions so I like my first row I've already got a 1 and I've already got a 0 and I actually like my second row because I have a 0 and a 1 so far I've got a pivot and a pivot and I need to turn this guy into a 0 so let's take 2 times Row 2 and add it to Row 3 so I get 0 I'm sorry subtract Row 3 so 2 times 1 is 2 minus 2 is 0 2 times 4 is 8 minus 0 is 8 2 times negative 1 is negative 2 minus 6 is negative 8 and then of course 0 continue with my row operations so my first column is great my second column is great I'm going to leave the first and second rows exactly as they are for now and in that last one I'm going to take it times one eighth so that gives me one negative one and then I'm almost where I want to be I've got a pivot I've got a pivot and I've got a pivot but what I don't have is I need a zero and a zero so let's keep going that's going to give me first row I'm going to take my third row times negative three and add it to my first row so I get 1 0 0 and then 3 plus 4 is 7 0 and then for my second row I'm going to take my third row times negative 4 and add it so I get 0 1 0 and then that's positive 4 plus negative 1 so that's positive 3 and then this guy I didn't do anything to at all so what am I going to do from here well I know that my solution is that X 1 is equal to negative 7 X 4 and because obviously X 4 is my free variable and so that's the only one I should have here and then X 2 is equal to negative 3 X 4 and X 3 is equal to 1 X 4 and X 4 is free so really my solution here if you'll recall is that I can write it negative 7 negative 3 1 1 and from here I'm looking for again a non-zero vector and so what that tells me to do is essentially come up with some x4 so whatever x4 I want and then multiply it so let's just say X 4 is 2 so a nonzero vector in the null space of a would be if I turn this into a 2 which would give me negative 14 negative 6 2 2 so that's one of the many many solutions I could have a solution for any value of x for that I chose one last practice for us again two parts so press pause try it both parts of this question and then press play to see how you did so the first thing to talk about here is a with respect to you so the first question says is you in null a and of course we already know how to do this this is 1 0 4 0 0 1 2 0 0 0 3 1 times 2 negative 1 3 4 and I'm essentially asking is this 0 so I would get 2 + 0 + 12 + 0 and do I even need to continue because that's 14 and so this is not equal to 0 and I could continue I would get 0 plus negative 1 plus 6 plus 0 which is 5 and I would get 0 plus 0 plus 9 plus 4 which is 13 again not the zero vector therefore you does not whoops wrong notation you does not belong Wow no a there we go the second part of this question says could you the vector u B belong to the column space of a and hopefully hopefully it's very clear that this answer is no because we talked before about the column space of a obviously that would be these vectors so could you which is in our for be in the column space of a no because column a column space of a is a subspace of r3 let me get rid of those little markings now so let's move on to the next one is V in the column space of a so here's V they're asking is this in the column space well are any of these that exact column no so no these columns none of those matched column that is the vector V so that's a big fat no so the second question says could V be in the null space of a so good question what do we know about the null space of a well the null space of a is a subspace of our four so again the column space of a is a subspace of r3 the number of rows the null space of a is a subspace of R for the number of columns so could V be in the null space of a and again this is a no because it has to be a subspace of R 4 and obviously V is in r3 I want to sort of rewind back to something that we've talked about before and that is linear transformations that's not going to be our focus in this section but it was in the textbook and I just wanted to make sure that I brought it up in the videos so we have talked before about a linear transformation from a vector from a vector space V to a vector space W now previously we did not talk about a vector space but essentially what we talked about was that we had some set and that linear transformation is just a transformation on to another set so this is a function so if you'll recall a linear transformation and again we're looking at it from a vector space to a vector space which is not what we did previously is a rule that assigns a unique vector TX to each vector X so again X TX again a function such that the transformation of the sum of vectors U and V is equal to the transformation of vector u plus the transformation of vector B again for all U and B in the vector space V and that the transformation of some scalar times vector u is equal to the scalar times the transformation of vector u so that was just our same definition that we talked about before this time applied to vector spaces a couple of new terms for you know space of T is called the kernel and again I'm just spraying that we're not going to focus on that in this lesson but you will see again throughout your mathematical education so I'm going to let you know now the null space of T is called the kernel and again here is the Mathew definition and essentially we're just saying that if just like before we talked about the null space we said a X was equal to zero so notice I'm saying all of the vector use in the vector space such that the transformation of vector U is equal to zero so that is called the kernel and then the range again just thinking back to this transformation this is our domain and this is our range the range of T is the set of all vectors in W of the form T X for some X in V so again that's the range what is going to map back to our original X