MHB Saqifriends's Question from Math Help Forum

[Sudharaka]

Title: Find a standard basis vector

Find a standard basis vector that can be added to the set {v₁, v₂} to produce a basis for R³ where;
v₁ = (-1, 2, 3), v₂ = (1, -2, -2)

Hi saqifriends, :)

We can use the determinants to see which standard basis vectors are linearly independent with the given two vectors.

\[\begin{vmatrix} 1 & 0 & 0\\-1 & 2 & 3\\1 & -2 & -2 \end{vmatrix}\neq 0\]

\[\begin{vmatrix} 0 & 1 & 0\\-1 & 2 & 3\\1 & -2 & -2 \end{vmatrix}\neq 0\]

\[\begin{vmatrix} 0 & 0 & 1\\-1 & 2 & 3\\1 & -2 & -2 \end{vmatrix}=0\]

Therefore, \(\{(1,0,0),\,(-1, 2, 3),\,(1, -2, -2)\}\mbox{ and }\{(0,1,0),\,(-1, 2, 3),\,(1, -2, -2)\}\) are linearly independent sets.

Now we shall show that, \(\{(1,0,0),\,(-1, 2, 3),\,(1, -2, -2)\}\) spans \(\Re^{3}\).

Take any, \((x,y,z)\in\Re^{3}\).

\[(x,y,z)=\alpha(1,0,0)+\beta(-1, 2, 3)+\gamma(1, -2, -2)\]

\[\Rightarrow \alpha=x+\frac{y}{2},\,\beta=z-y,\,\gamma=z-\frac{3\,y}{2}\]

\[\Rightarrow \alpha,\,\beta,\,\gamma\in\Re\]

Therefore, \(\{(1,0,0),\,(-1, 2, 3),\,(1, -2, -2)\}\) spans \(\Re^{3}\).

Similarly it could be shown that, \(\{(0,1,0),\,(-1, 2, 3),\,(1, -2, -2)\}\) spans \(\Re^{3}\).

Hence both \(\{(1,0,0),\,(-1, 2, 3),\,(1, -2, -2)\}\mbox{ and }\{(0,1,0),\,(-1, 2, 3),\,(1, -2, -2)\}\) are bases of \(\Re^{3}\).

Kind Regards,
Sudharaka.

 

[chwala]

Title: Find a standard basis vector
Hi saqifriends, :)

We can use the determinants to see which standard basis vectors are linearly independent with the given two vectors.

\[\begin{vmatrix} 1 & 0 & 0\\-1 & 2 & 3\\1 & -2 & -2 \end{vmatrix}\neq 0\]

\[\begin{vmatrix} 0 & 1 & 0\\-1 & 2 & 3\\1 & -2 & -2 \end{vmatrix}\neq 0\]

\[\begin{vmatrix} 0 & 0 & 1\\-1 & 2 & 3\\1 & -2 & -2 \end{vmatrix}=0\]

Therefore, \(\{(1,0,0),\,(-1, 2, 3),\,(1, -2, -2)\}\mbox{ and }\{(0,1,0),\,(-1, 2, 3),\,(1, -2, -2)\}\) are linearly independent sets.

Now we shall show that, \(\{(1,0,0),\,(-1, 2, 3),\,(1, -2, -2)\}\) spans \(\Re^{3}\).

Take any, \((x,y,z)\in\Re^{3}\).

\[(x,y,z)=\alpha(1,0,0)+\beta(-1, 2, 3)+\gamma(1, -2, -2)\]

\[\Rightarrow \alpha=x+\frac{y}{2},\,\beta=z-y,\,\gamma=z-\frac{3\,y}{2}\]

\[\Rightarrow \alpha,\,\beta,\,\gamma\in\Re\]

Therefore, \(\{(1,0,0),\,(-1, 2, 3),\,(1, -2, -2)\}\) spans \(\Re^{3}\).

Similarly it could be shown that, \(\{(0,1,0),\,(-1, 2, 3),\,(1, -2, -2)\}\) spans \(\Re^{3}\).

Hence both \(\{(1,0,0),\,(-1, 2, 3),\,(1, -2, -2)\}\mbox{ and }\{(0,1,0),\,(-1, 2, 3),\,(1, -2, -2)\}\) are bases of \(\Re^{3}\).

Kind Regards,
Sudharaka.

Correct! one has to first solve the simultaneous equation:
$x=α-β+ϒ$
$y=2β-2ϒ$
$z=3β-2ϒ$

$x=α-β+ϒ⇒α=x+β-ϒ$
and from
$y=2β-2ϒ$⇒
$\dfrac{y}{2}$=$β-ϒ$
on substituting into
$α=x+β-ϒ$
$α=x+\dfrac{y}{2}$ as shown...the other constants can be found in a similar approach (by use of elimination method).