Kathleen Rausch

2022-01-07

Let u and v be distinct vectors of a vector space V. Show that if {u, v} is a basis for V and a and b are nonzero scalars, then both {u+v, au} and {au, bv} are also bases for V.

Orlando Paz

Given, u and v be distinct vectors of a vector space V and {u, v} is a basis for V and a and b are nonzero scalars.
We have to show: both {u + v, au} and {au, bv} are also bases for V.
$u\ne v,a,b\ne 0$
$\left\{u,v\right\}$ is a basis for V $⇒\left\{u+v,au\right\}$ and $\left\{au,bv\right\}$ are bases for V
From basis $\left\{u,v\right\}$ we get dimension of V
$\left\{u,v\right\}$ is a basis for V $⇒$ V is a 2-dimensional vector space and $\left(\cdot \right)\alpha u+\beta v=0⇒\alpha =\beta =0$
Show $\left\{u+v,au\right\}$ is a basis for V:
$\left\{u+v,au\right\}$:
$A\left(u+v\right)+B\left(au\right)=0⇔Au+Av+Bau=0⇔$
$\left(A+Ba\right)u+\left(B\right)v=0$
By the step (in second step)
$A+Ba=0,a\ne 0$
$B=0$
$⇒A=B=0⇒\left\{u+v,au\right\}$ is linearly independent set of two vectors
$⇒\left\{u+v,au\right\}$ is a basis for V
(Proved)

xandir307dc

here is the continuation of the solution:
Show $\left\{au,bv\right\}$ is a basis for V
$\left\{au,bv\right\}:$
$A\left(au\right)+B\left(bv\right)=0⇔\left(Aa\right)u+\left(bb\right)v=0$
By the step (in second step)
$Aa=0,a\ne 0$
$Bb=0,b\ne 0$
$⇒A=B=0⇒\left\{au,bv\right\}:$ is linearly independent set of two vectors
$⇒\left\{au,bv\right\}:$ is a basis for V
(Proved)

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