Hi MHF,
We have the following vectorial subspace pertaining to such that .
1)Find a basis of W.
2)Find a basis of W orthogonal.
My attempt :
1)I just wrote the 2 equations defining W as a matrix and chosen the 2 first column vectors of the matrix as being a basis of W.
Precisely I got that a basis of W is {(2,1),(0,-1)}. (0,-1) being the last column vector of the matrix I formed.
2)I'm not 100% sure about what I've done here. I chosen the basis {(2,1),(1,0)}.
How can I check if I'm right on this? I guess I could form a matrix with the vectors in the basis I gave and try to get a null row or column when reducing it. But I'm not sure, and not sure how to form the matrix too.
Can you help me?
,normal vector , normal vector
, so is intersection line of two hyperplanes.
We solve equation
, so there are two linear independent vectors in basis.
by the way, dimension is four. if it's 3, it's easy to calc by W = span(v1 x v2)
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is hyperplane orthogonal to intersection line . Its basis is two linear independent normal vectors
There is no union solution. This answer is the same with **NonCommAlg's Avatar** above.
Without using matrices: You have two equations, 2x-z+w=0 and x-y-z=0 in 4 variables. The equations are independent so we can solve for two of the variables in terms of the other 2 meaning that this is a two dimensional subspace of the four dimensional .
We get z= x- y from the second equation and, putting that into the first, 2x-(x-y)+ w= x+ y+ w= 0 so w= -x-y. If we take x= 1, y= 0, z= 1, w= -1 so (1, 0, 1, -1) is a vector in that subspace. Taking x= 0, y= 1, z= -1, w= -1 so (0, 1, -1, -1) is another vector in that subspace. Taking x=1, y= 0 and x= 0, y= 1 ensures that they are independent vectors so {(1, 0, 1, -1), (0, 1, -1, -1)} is a basis for the subspace W.
Vectors in the orthogonal subspace to W must be perpendicular to every vector in W. Since we already know that 2x- z+ w= 2(x)+ 0(y)+ (-1)z+ (1)w= 0 and x-y-z= 1(x)+ (-1)y+ (-1)z+ 0(w)= 0, it is clear that (2, 0, -1, 1) and (1, -1, -1, 0) are in that orthogonal orthogonal space and it is clear that they are independent so {(2, 0, -1, 1), (1, -1, -1, 0)} is a basis for the orthogonal subspace to W