Null space 2: Calculating the null space of a matrix (video) | Khan Academy

Video transcript

In the last video, I spoke reasonably theoretically about what a nothing space is and we showed that it is a valid subspace. But in this video lashkar-e-taiba ‘s actually calculate the nothing space for a matrix. In this case, we ‘ll calculate the null space of matrix A. So nothing space is literally merely the put of all the vectors that, when I multiply A times any of those vectors, so let me say that the vector x1, x2, x3, x4 is a member of our null space. So when I multiply this matrix times this vector I should get the 0 vector. I should get the vector. And just to make a few points here, this has precisely 4 column. This is a 3 by 4 matrix, then I’ve lone legitimately define multiplication of this times a four-component vector or a member of Rn. Let me call this ten. And this is our vector adam. This is a member of R4. It has four components. And then when you multiply these, we need to produce a 0 vector. The null space is the adjust of all the vectors, and when I multiply it times A, I produce the 0 vector. And what am I going to get ? I ‘m going to have one course times this and that ‘s going to be the foremost entrance, then this row times, that ‘s the second entrance, and then the third gear course. So I should have three 0 ‘s So my 0 vector is going to be the 0 vector in R3. So how do we figure out the set of all of these x ‘s that satisfy this ? Let me just write our conventional notation. The nothing distance of A is the bent of all vectors that are a member of — we by and large say Rn, but this is a 3 by 4 matrix, so these are all the vectors that are going to be members of R4, because I ‘m using this particular A, such that my matrix A times any of these vectors is peer to the 0 vector. In this case it ‘s going to be 0 vector in R3. So how do we do this ? well, this is just a straight up linear equality. We can write it that way. If we were to actually perform the matrix multiplication, we get 1 times x1. Let me write it here. Let me do it in a different color. 1 times x1, plus 1 times x2, plus 1 times x3, plus 1 times x4 is equal to this 0 there. So that times that is peer to that 0. And then this times this should be equal to that 0. sol 1 times x1, so you get x1, plus 2 times x2, plus 3 times x3, plus 4 times x4 is going to be equal to that 0. And then finally we have that times this vector should be adequate to that 0. So the dot product of this row vector with this column vector should be adequate to that 0. so you get 4×1. 4×1 plus 3×2 plus 2×1 plus 2×3 plus x4 is equal to 0. 4×1 plus 3×2 plus 2×3 plus x4 is peer to 0. You just have to find the solution set to this and we ‘ll basically have figured out our nothing quad. now, we ‘ve figured out the solution set to systems of equations like this. We have three equations with four unknowns. We can do that. We can represent this by an augment matrix and then put that in reduce quarrel echelon form. Let ‘s do that. I can represent this problem as the augment matrix. 1, 1, 4. 1, 2, 3. 1, 3, 2. and then 1, 4, 1. And then I augment that with the 0 vector. And the immediate thing you should notice is we took the pain of multiplying this times this to equal that, and we wrote this as a system of equations, but now we want to solve the system of equations, we ‘re going back to the augmented matrix worldly concern. What does this augmented matrix front like ? Well, this is just our matrix A right there. That ‘s fair matrix A properly there, that ‘s good the 0 vector right there. And to solve this, and we’ve done this before, we ‘re just going to put this augment matrix into row echelon form. What you ‘re going to find is when you put it into course echelon imprint, this right side’s not going to change at all, because no matter what you multiply or subtract by, you ‘re good doing it all times 0, so you equitable keep ending up with 0. so as we put this into reduce quarrel echelon form, were actually fair putting matrix A into reduced echelon shape. So let me do that, alternatively of barely talking about it. so let me start off by keeping row 1 the same. Row 1 is 1, 1, 1, 1, 0. And then I want to eliminate this 1 right hera, indeed let me replace row 2 with row 2 minus quarrel 1. sol 1 minus 1 is 0. 2 minus 1 is 1. 3 subtraction 1 is 2. 4 minus 1 is 3. 0 minus 0 is 0. You can see the 0 ‘s aren’t going to change. And then let me replace this guy with 4 times this ridicule, minus this guy. So I can only get rid of this. indeed 4 times 1 minus 4 is 0. 4 times 1 minus 3 is 1. 4 times 1 minus 2 is 2. 4 times 1 minus 1 is 3. 4 times 0 minus 0 is 0. now I want to get rid of, if I want to put this in reduced row echelon shape, I want to get rid of that term and that condition. So let me keep my in-between row the same. My middle row is 0, 1, 2, 3. So that ‘s 0 on the augmented side of it, although these 0 ‘s are never going to change, it’s in truth merely a little morsel of an exert equitable to keep writing them. And my first course, let me replace it with the first course minus the moment row, so I can get rid of this 1. therefore 1 minus 0 is 1. 1 subtraction 1 is 0. 1 minus 2 is minus 1. 1 minus 3 is minus 2. And 0 minus 0 is 0. And let me replace this last row with the last row minus the in-between row. so 0 minus 0 is 0. 1 minus 1 is 0. 2 subtraction 2 is 0. I think you see where this is going. 3 minus 3 is 0. And obviously 0 minus 0 is 0. sol this organization of equations has been reduced, fair by doing reduce course echelon shape, this problem. If I just rewrite this correctly hera, this can be written as a system equations of x1 minus x3 minus x4, proper ? The 0 x2 ‘s is equal to 0. And then this second row proper here, there ‘s no x1, you just have an x2, plus 2×3, plus 3×2 is equal to 0, and this obviously gives me no information any. And so I can solve this. I can solve this for x1 and x2, and what do I get ? I get x1 is equal to x3 summation x4. Actually, I made a err here. This is x1 minus x3 subtraction 2 times x4 is equal to 0. So if I rewrite this, I get x1 is adequate to x3 plus 2×4. And then I get x2. Let me do that in park. x2 is equal to minus 2 x3 minus 3×2. So if I wanted to write the solution set to this equality, if I wanted to write it in terms of this, I could write x1, x2, x3, x4 is equal to — what ‘s x1 equal to ? It ‘s equal to x3 times 1 plus x4 times 2. Right ? I precisely got this correct here from this equation right field here. x1 is equal to 1 times x3, plus 2 times x4. That ‘s good that, right field there. now, x2 is equal to x3 times minus 2, plus x4 times minus 3. What am I doing ? I ‘m losing track of things. This x2 right here is x2 plus 2×3 plus 3×4 is equal to 0. sol x2 is peer to minus 2×3, minus 3×4. justly. Like that. Sorry, my brain is n’t wholly in the problem, I ‘m making these airheaded mistakes. But I think you understand this nowadays. then then what is x3 adequate to ? Well, it ‘s merely equal to 1 times x3, plus 0 times x4, justly ? x3 is equal to x3. And what ‘s x4 equal to ? It ‘s equal to 0 times x3 plus 1 times x4. So all of the vectors in R4, these are a member R4, which satisfy the equation, our original equation, axe is equal to 0, can be represented as a linear combination of these two vectors, of those two vectors, right ? These are merely random scalars that are a member of — We can pick any actual number for x3 and we could pick any actual number for x4. indeed our solution set is just a linear combination of those two vectors. What ‘s another way of saying a linear combination of two vectors ? Let me write this. The null space of A, which is fair a solution set of this equation, it ‘s equitable all the x ‘s that satisfy this equation, it equals all of the linear combinations of this vector and that vector. What do we call all the linear combinations of two vectors ? It ‘s the cross of those two vectors. So it equals the span of that vector and that vector. Of the vector 1, minus 2, 1, 0, and the vector 2, minus 3, 0, 1. And this is our null space. Before letting you go, let me just point out one interesting thing right here. We represented our system of equation like this and we put it into reduced row echelon shape, so this is A and this is 0. This properly here is, let me make certain I have some space, let me put it right field here. That proper there is the reduce row echelon imprint of A. And so where basically this equation, this is a linear equation that is trying to solve this problem. The reduce row echelon form of A times our vector adam is adequate to 0. so, all the solutions to this are besides the solutions to our original trouble, to our original ax is equal to 0. So what ‘s the solution to this ? All the x ‘s that satisfy this, these are the nothing space of the reduce row echelon mannequin of A. Right ? therefore here are all of the x ‘s, this is the null space, this problem, if we find all of the ten ‘s here, this is the nothing space of the abridge course echelon form of our matrix A. But we ‘re saying that this problem is the same trouble as this one, right ? So we can write that the null space of A is equal to the nothing space of the reduce row echelon human body of A. And that might seem a little bite confuse, hey, why are you even writing this out, but it ‘s the actually very useful when you ‘re trying to calculate nothing spaces. So we did n’t even have to write a big augment matrix here. We can say, take our matrix A, put it in reduced row echelon shape and then trope out it ‘s nothing space. We would have gone straight to this point correct here. This is the reduce row echelon form of A, and then I could have immediately solved these equations, good ? I would have just taken the dot product of the shrink rowing echelon form or, not the scatter product, the matrix vector merchandise of the reduce course echelon class of vitamin a with this vector, and I would ‘ve gotten these equations, and then these equations would immediately, I can merely rewrite them in this kind, and I would have gotten our consequence. But anyhow, hopefully you found that sanely useful.

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