00:01
All right, hello.
00:01
So since we have that the rank of our matrix a is equal to m, and that is we have that all m singular values are non -zero, then using the svd where we have that a is equal to u, sigma, v transpose, the least square solution is going to be x star is equal to, a plus b here.
00:35
Now a to the little plus, this is going to be the pseudo -inverse.
00:39
For full column rank, we have that the pseudo -inverse is going to be equal to v sigma plus u transpose.
00:54
And then we have that x star is going to be equal to v sigma u transpose b.
01:04
And now we have that u transpose b gives the dot products with the left singular vectors, where we have that u transpose b is going to be equal to, well, b.
01:21
Dot u1, b.
01:24
Dot u2, and so on, down to b...