My solution follows the exercise, is this the right procedure? thanks beforehand for your time.
We have p as a parameter, and the bais U which is a subspace of the three-dimensional plane R, spaned by the independent vectors $V1 = (-2,-p,3)$ and $V2= (2p^2, -p^2,p)$
- Find the values for p for which V1 and V2 form an orthogonal basis of U
- Choose one basis that you found in (1) and turn it into an orthonormal basis of U
So, we know that an orthonormal set is a set of vectors that are orthongonal (They are all perpendicular to each other) and vectors that are normalized (The length of every vector is equal 1). We also know that if a set of vectors is orthonormal, then the vectors must be linearly independent. But the opposite cannot be always true; a set of linearly independent vectors may or not be orthonormal.
If we have a set of linearly independent vectors and we wish to form an orthonormal basis with them, we must first make them orthogonal, and then normalize their lenght.
The vectors $V1 = (-2,-p,3)$ and $V2= (2p^2, -p^2,p)$ will be orthogonal if:
$V1 \cdot V2 = 0$
So then, we find the values of p which make this possible.
$-2(2p^2)+(-p)(-p^2)+3p = 0$
$-4p^2 + p^3 + 3p = 0$
$p^3 - 4p^2 + 3p$
We factorize, and find the roots of p
$p(p^2 - 4p + 3) = 0$
$\begin{array}{*{20}c} {p= \frac{{ - (-4)\pm \sqrt {-4^2 - 4(1)(3)} }}{{2(1)}}} \end{array}$
$p1 = 3$
$p2 = 1$
So we can form 2 orthogonal basis here, one using the value 3 and another using the value 1.
Basis with value 3: $V1 = (-2,-3,3)$ and $V2 = (18,-9,3)$
Basis with value 1: $V1 = (-2,-1,3)$ and $V2 = (2,-11)$
Now we choose one of these basis, and turn into an orthonormal vector by replacing each vector $Vi$ with:
$\frac{1}{\lVert Vi\rVert}Vi$
And I get for the basis with value 3, the following orthonormal vectors:
$V1 = (\frac{-2}{\sqrt22},\frac{-3}{\sqrt22},\frac{3}{\sqrt22})$
$V2 = (\frac{118}{\sqrt114},\frac{-9}{\sqrt114},\frac{3}{\sqrt114})$
I have another exercise requesting the Gram-Schmidt process, so I guess that my solution was the right procedure. Let me know otherwise, thank you.