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Exercise 23

Avg difficulty 3.0

One of the exercises in the Subspaces subsection shows that the set

{(xyz)∣x+y+z=1}\{\hskip 1.00006pt\begin{pmatrix}x\\ y\\ z\end{pmatrix}\mid x+y+z=1\hskip 1.00006pt\}

is a vector space under these operations.

(x1y1z1)+(x2y2z2)=(x1+x2−1y1+y2z1+z2)r(xyz)=(rx−r+1ryrz)\begin{pmatrix}x_{1}\\ y_{1}\\ z_{1}\end{pmatrix}+\begin{pmatrix}x_{2}\\ y_{2}\\ z_{2}\end{pmatrix}=\begin{pmatrix}x_{1}+x_{2}-1\\ y_{1}+y_{2}\\ z_{1}+z_{2}\end{pmatrix}\qquad r\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}rx-r+1\\ ry\\ rz\end{pmatrix}

Find a basis.

Hefferon, Linear Algebra, Vector Spaces, Basis, Exercise 23

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Adapted from Linear Algebra © 2016 Jim Hefferon. Licensed under CC BY-SA 3.0 US. Source: Linear Algebra. Imported at revision df2262e.