Nullspace Column Space and Rank

alright thanks for watching and I don't want to say this video is important but this video is super important especially if you're crowding from linear algebra exam and you have one in an hour or something and you need to know all the concepts well turns out this one problem has all the concepts in it or at least most of them so make sure to watch it I'm not even exaggerating so in particular let's take this matrix here a is 1 3 2 - 2 - 6 - 4 1 4 0 - 5 - 24 8 + 6 xx + 1 well we'd like to talk about a water called matrix spaces so now space of a column space of a and maybe others if I have time particularly let's consider this matrix and first of all let's find a basis for the null space of words are not an awesome column space first find a basis for the column space because it's a bit easier to find a basis for the column space and in order to do that let's row reduce this matrix so you know the answer to every question in linear algebra is row reduction so let's do that we would like to get rid of this 3 in the two so then subtract 3 times the first row from the second row and let's subtract 2 times the first row from the third row and then you get 1 - 2 1 - 5 6 and then let's see 0 0 1 - 9 + 2 + 0 0 - 2 80 - eleven all right almost done so we still need to get rid of this - - so let's add two times of second row to the third row and we get 1 - 2 1 - 5 6 zeroes there are 1 - 9 2 & 0 0 0 0 - 7 I can any question of linear algebra boils down to row reducing and also finding the pivots notice here there are three pivots 1 1 and minus 7 in particular your pivots in columns 1 3 & 5 to find the basis for the column space of a very important not do one mistake people doing those kinds of problems you're not allowed to choose those 3 things you have to go back to your original matrix and pick the first third and fifth columns of a and the reason for this is turns on row reduction they preserve linear independence so vaguely speaking if you reduce notice those three vectors are linearly independent therefore we also get that those three vectors are linearly independent so we have those three linearly independent vectors and those are automatically linearly dependent and I put quotation marks because technically a set is linearly independent or not vectors however row reduction destroys the span so the span of those three vectors is not the same as the span of those three vectors so in particular the basis for the column space of a is 1 3 2 1 4 0 & 6 20 and 1 and I totally forgot to say what is the column space it's just the span of the columns of your matrix so it's the span of this vector this vector this vector in this vector but a particular basis for the column space is that and question what is this a subspace of well here's a subspace of r3 so remember subspaces are nice spaces and here they're subspaces of R 3 and notice precisely a is a three by five matrix so in general it's a subspace of R and month where a is n by n and I'd like to think of it in terms of input and mouth put so column space Jews with outputs with Mouse books or another way to remember this we have called London space and I think column okay that's watching so the basis for the column space of a then the second thing is find what's called the rank obey all that rank of a is is just a fancy word for the dimension of the column space how do you find a dimension just count the number of vectors in the basis so here there are three vectors in this basis so the dimension is three that's three and it turns out it's exactly equal to the number of pivots here so we have three pivots so the rank is three and sort of the rank tells you how good matrixes to pivot around the better the matrix for example the zero matrix has rank zero it tells you nothing but the identity matrix has the highest rank possible all right so that's one thing and then then that's fine on the one hand we found the column space now let's find a basis for the null space for now away and again all of a that's just the set of solutions of AX equals 0 which means you have to find X very important the column space deals with the outputs the null space deals with inputs then throwing up and but that's good in order to do that while we reduce the matrix to the row echelon form but to find a null space you need to go one step further and do the reduced row echelon form so let's start with this so 1 0 0 minus 2 0 0 1 1 0 minus 5 minus 9 0 6 2 minus 7 and then bonus of 8 x equals 0 so it would bunch of 0s here and then you want the reduced row echelon form which first of all means the pivots has to be 1 so turn this minus 7 into a 1 and we get maybe over there we good so 1 minus 2 1 minus 5 6 zero zero zero one minus 9 to 0 and then 0 0 0 0 1 0 that next thing is we want to turn those numbers into 0 after the numbers above the pivots have to be 0 but that's not too bad let's subtract 2 times the third row from the second row and subtract 6 times the third row from the first row and we get 1/2 1 minus 5 0 0 and 0 0 1 minus 9 0 0 and then 0 0 0 0 1 0 next we want to turn this non pivot into 0 so let's just subtract the second row from the first row and we get 1 minus 2 0 4 0 0 and then 0 0 1 minus 9 0 0 & 0 0 0 0 1 0 so indeed it is it row reduced row echelon form because the pivots are 1 and the stuff above the below the pivots aren't 0 then let's just solve this so one little thing this will be very important in a thing I'll talk about next their pivots in the first 3 and fifth column what are the free variables there in the non-pivot columns so here Y and T they're the free variables the second in the fourth variable and this is all so they're useful when you want to solve an equation before solving this do that so then we get X minus 2y plus 4 T equals 0 Z minus 9 T equals 0 and the last pair believe zero let's call it s and then that's all in terms of our free variables X is two Y minus 40 z is 90 and s is zero okay why are you doing this because again we want to find X such an ax equals zero so X then just becomes again everything X Y Z T s and that becomes 2y minus 40 y 9t t and zero and you can separate out the Y's so 2y y 0 0 0 plus minus 40 0 90 t 0 and you get Y times 2 1 0 0 0 plus T times minus 4 0 9 1 0 so you know space is really the set of all linear combo to things and it turns out that those things will always be a basis for you now States so the miracle of row reduction is that row reduction all matically gives you a basis for knowledge so the basis for knowledge a is in this case 2 1 0 0 0 and minus 4 0 9 1 0 and the question is what is this a subspace of well here it's a subspace of r5 on our five well how does five relate to our matrix or a was a three by five matrix so it's in general it's a subsidies of art and then does it know of a this is how you remember this alright once we found the basis we can find a dimension so let's find dimension of knowledge a well you just count the number of vectors which is 2 okay which leads us to a very important theorem in linear algebra called the rank nullity theorem so note what happens if you add the dimension of the null space plus the rank well let's see rank of a plus the dimension of the null space of a rat was the number of pivots which was three there are three pivots here plus dimension of nor space that's 2 + 3 + 2 that's fine how does that relate to our matrix well our matrix had 5 columns so rank of a plus the dimension of the null space equals to n and you know where a is n by n so this is always true it's always true that the dimension of the null space of a plus the rank of a equals and this thing I remember it confuse the hell out of me when I took a linear algebra but hopefully it won't confuse the hell out of you because this actually makes sense remember null space of a measures how bad a matrix is it's sort of the fat of a snake matrix or something the bigger and all of a the worse is rank which is the dimension of the column space the column space measures how good of matrixes so the bigger the rank of a matrix the better so in other words it's a snake part of the of a state matrix like it's the meat part of it well if you add the null space in the column space things should balance out because the null space tells you how bad a matrix is the column space tells you how good a matrix is and this is precisely what this is saying here is to add the two dimensions then you get a fixed number which is n also how another way to figure this out remember the dimension of the null space of a gives us the number of free variables hey the rack obey you see for example here we had three pivots but the free variables were in columns 2 & 4 the pivots they give us the non free variable so the prisoner variables no so here were it was like you know columns to it 4 plus number of non free variables which recall is 1 3 & 5 well if you add the number of free variables and the number of non free variables you get the total number of variables which here is 5 so in fact you see if you add you know if your so add all the columns together so that not the free variables when it comes to it for the non free ones are in 1 3 & 5 well the total number you'll see now it's just a total number of columns that's all we do add them we don't do Emma because that's the number of rows okay that's all I wanted to talk about here okay so if you're taking my class at least math 3a at Irvine this is the stuff we care about just in case you taking more than that just a row space let me talk a little bit about the row space again let's take this matrix and find a basis for the row space of it turns out it's even easier so I think it was number E or something find a basis for Row fortunately I raised this matrix but what the row space is is simply you take all the rows as vectors and you take the span of it so here row of a would be this bad oh 3 minus R the 1 matrix R 1 minus 2 1 minus 5 6 & 3 minus 6 4 minus twenty four twenty P and 2 minus 4 0 8 one that's a row space and it turns out the nice thing about the row space is in order to find your basis you don't need to go back to the original matrix just see in which rows the pivots are well the pivots are in rows 1 2 & 3 so indeed a basis for the row space of a are simply you knew vectors so 1 minus 2 1 minus 5 6 0 0 1 minus 9 2 & 0 0 0 0 minus 7 and here notice the row space in this case it's a subspace you know what I eat each component has each vector has five components so it's a subspace of our five year which in this case is and so general it's a subspace of RN okay where is careful the column space was a subspace of RN and I don't know how to remember that but one thing is maybe roh-roh sounds like rose-like knows that's why it's a subspace of RN Oh No okay lastly just then what is the dimension of probe a well the nice thing is we found the number we know we found a basis risk on the number in the basis which is three and it turns out incidentally that the same as the rank of a because that's just the same as the number of pivots so even though the row space and the column space are two different things the dimensions are the same but a bit surprising but I think it's pretty cool and also why don't you have to go back to your original matrix because raw operations are nice with rows they preserve span and linear independence so it gives us an easier basis if you like alright so I hope you like this linear algebra craziness extravaganza if you want to see more math please click like and make sure to subscribe to my channel and yeah I think it's time to go

Channel: Dr Peyam

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Keywords: linear algebra, peyam, math, dr peyam, nullspace, nulspace, NulA, Nul(A), kernel, row-reducing, row reduction, ref, rref, basis, span, row-echelon, row-echelon form, reduced row-echelon form, zero, Rn, input, linearly independent, independent, free variable, pivot, freedom, subspace, vector space, Nul A, matrix, homogeneous, column space, Col(A), Col A, ColA, column, Row(A), Row A, RowA, Row space, row, rows, rank(A), rank A, rank, rank nullity, rank-nullity, nullity, rank theorem, output, Rm

Id: XYLtYmlbh3Y

Length: 20min 58sec (1258 seconds)

Published: Wed Mar 13 2019