How can I find an orthonormal basis for column space of the following matrix without using Orthogonalize?
$$ \begin{bmatrix} -10 & 13 & 7 & -11 \\ 2 & 1 & -5 & 3 \\ -6 & 3 & 13 & -3 \\ 16 & -16 & -2 & 5 \\ 2 & 1 & -5 & -7 \\ \end{bmatrix}$$
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Domen
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m = {{-10, 13, 7, -11},
{2, 1, -5, 3 },
{-6, 3, 13, -3 },
{16, -16, -2, 5 },
{2, 1, -5, -7}}
Orthogonalize[m\[Transpose]]*{10, 2, Sqrt[3], Sqrt[2]}
{{-5, 1, -3, 8, 1}, {1, 1, -1, 0, 1}, {1, 0, 1, 1, 0}, {0, 1, 0, 0, -1}}
In order to adapt the basis to the Nullspace and its orthogonal complement: there is a 1d nullspace
x0= NullSpace[m\[Transpose]][[1]]
x0 . m
{0, 0, 0, 0}
Next, take a unit vector and subtract it's projection to x0
x1 = (#2-#1 * #1 . #2 / Norm[#1]^2 &)[x0, UnitVector[5, 2]]
{1/5, 19/25, -(7/25), 2/25, -(6/25)}
x1.x0
0
for the following vectors taken from the Set of UnitVector list, you have to subtact the sum of the projections to the foregoing.
Write a Do loop or a FoldList procedure running over the UnitVector List. Divide each vector by its norm, if an orthonormal system is sought.
Roland F
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Transpose@GramSchmidt@Transpose@m- you can getGramSchmidtfrom halirutan's answer - the columns of the resulting matrix will be orthonormal - remove the first transpose to get the basis as mathematica 'vectors' – flinty Apr 26 '23 at 18:04Orthogonalizeforbidden here? – David G. Stork Apr 26 '23 at 21:30Normalize. That's really not difficult. – flinty Apr 27 '23 at 10:15