![]() ![]() So I'm going to keep myįirst row the same. Second row with the second row minus 6 times the first row. That I choose to do, just because I like to have In fact, there willĬreate our little augmented matrix: 6, 1, augmented The column space of A -is equal to A transpose times B, My least squares solution- so this is actually going to be in So A transpose times B isĮqual to 9, and this is going to be 4. Times 2, which is minus 2, 2 times 1, which is 2, plusġ times 4, which is 4. Which is 4, plus 1 times 1, which is 1, plus 1 timesĤ, which is 4. ![]() Plus- let me actually write itĭown, I'm going to make a careless mistake -2 times 2, To be equal to? This is going to be equalĪ 2 by 1 vector. Vector, the vector that's a member of R3, 2, 1, 4. Now, what is A transposeġ, minus 1, 2, 1. Plus 2 times 2, which isĤ, so we're now at 5. And then finally, we get minusġ times minus 1, which is positive 1. So the minus 2 plus 2 is 0, plusġ times 1, so we get a 1. And then we get minus 1 timesĢ, which is minus 2, plus 2 times 1, which is 2. And then we have 2 times minusġ, which is minus 2, plus 1 times 2, so those cancel out. Times 2 which is 4, plus 1 times 1, plus 1 times 1. So A transpose A is going to beĮqual to- We have a 2 by 3 times a 3 by 2 matrix, so it's So at least we can find theĬlosest fit for our solution. We know that A transpose timesĪ times our least squares solution is going to be equal Solution if we multiply both sides by A transpose. There by finding a least squares solution. Yourself algebraically by trying to find a These lines don't intersect with each other. Video that sure, we can't find a solution to Ax equals B. That B is not in the column space of this matrix So you're not going to findĪ solution to A times some vector- we could call this some Matrix, put it in reduced row echelon form. Now, this isn't going toĪ solution to this. This system right here, these are equivalent. I could even, well I won't color code it, that'll takeįorever -that's 1 times x plus 2 times y is equal to 1. And so, this first equation isĢ times x, minus 1 times y. Matrix, let me make sure I get this right, the matrix times Matrix, or this equation has no solutions. Solution is equivalent to saying that this Solve this system, I would find no solution. There is no intersection ofĪll three of these points. We can kind of call the systemĪs being overdetermined. Two, but they don't all intersect each other This video that I want to find the intersection of Line, is going to look something like that. Minus x, so for every 1 you go over, you go down 1. It's actually going toīe orthogonal, right? Because it's the negative Right there, and then the slope is minus 1/2. It's going to have a slope ofĢ, so it's going to be a pretty steep line, We're actually dealing with x and y's now. We could write this as y is equal to minus x plus 4. In green -we could write this as 2y is equal to minus x plusġ, or we could write that y is equal to minus 1/2 x plus 1/2. So this top line becomes what? Minus y is equal to minusĢx, minus 2x plus 2. Just to have a visual representation of what Is equal to 2, the second one is x plus 2y is equal to 1, and And if we stretch it the right way we can move the shortest vector from b to the transformed triangle wherever we like, and thus we can move the least squares point of the original triangle wherever we like.Īnd I want to find their intersection. So without changing the original triangle we can stretch the transformed triangle relative to b along all 3 axes at will. And now, if we multiply the original linear equations by the constants c1, c2, and c3 respectively then those difference vectors become and. If we subtract b from each corner vector we get and. Applying transformation A to the original triangle gets you a new triangle in R^3 whose corners are and. The way to see what's going on geometrically in this problem is to look at the codomain in R^3 and the solution vector b =. It turns out every point in the triangle is a valid least squares solution if you use the right coefficients. In fact, the least squares solution has no geometric significance in R^2 because it's not unique! If you take the equation for one of the lines and multiply through by a constant then the least squares solution changes, even though it's the exact same line. Nor does the least squares solution minimize the sum of the perpendicular distances to each line as some commenters claim, nor the distances to the corners of the triangle which was my second guess. The least squares solution is always inside the triangle whereas the orthocenter can be outside. ![]()
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