Linear algebra question: Orthogonal subspaces

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Homework Help Overview

The discussion revolves around determining a basis for the null space of a given matrix in the context of linear algebra. The original poster presents their calculations and compares them with a reference solution from a textbook.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to find a basis for the null space by reducing the matrix and expressing the solution in terms of a free variable. Some participants question the choice of the free variable and suggest alternative representations.

Discussion Status

The discussion is ongoing, with participants providing insights into the nature of the null space and clarifying the relationship between different representations of the basis vector. There is an acknowledgment that both the original poster's and the textbook's forms are valid, leading to a productive exchange of ideas.

Contextual Notes

Participants are navigating the nuances of expressing vectors in the null space and the implications of choosing different free variable values. The original poster's calculations and the textbook's solution are being compared, highlighting potential differences in representation rather than correctness.

Mdhiggenz
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Homework Statement



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

A=[3 4]
[ 6 8]

Homework Equations


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?
 
Last edited by a moderator:
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What would have happened if you had set ##x_2=3\beta##?
 
We get -4B
 
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.
 
Last edited:
Awesome thanks!
 

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