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Linear algebra question: Orthogonal subspaces

  • Thread starter Mdhiggenz
  • Start date
1. The problem statement, all variables and given/known data

For each of the following matrices, determine a basis for each of the subspaces N(A)

A=[3 4]
[ 6 8]

2. Relevant equations

3. The attempt at a solution

So reducing it I got [1 4/3]
[0 0]

I know x2 is a free variable

I set x2 = to β

and found my N(A)=(-4/3β,β)T

However the book has simply (-4,3)T

Am I incorrect?
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Science Advisor
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What would have happened if you had set ##x_2=3\beta##?
We get -4B


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To find a basis for the nullspace is to find a minimal set of vector(s) that span your nullspace. If the nullspace is [itex] \binom{-4}{3} [/itex] as the book says, then any member of the nullspace can be written as a multiple of [itex] \binom{-4}{3} [/itex]. This means that the general form of a vector in N(A) is [itex] \binom{-4β}{3β} [/itex], which is equivalent to what you have. So you have written out the general form of a vector in the nullspace (assuming that your β is a free variable), whereas your book just gave the basis vector.
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Awesome thanks!

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