# Orthonormal and/or Orthogonal Basis of a Pair of Vectors

I was hoping someone could verify if this is the correct way to answer this problem:

Let $\mathbb{R^{2}}$ have the standard dot product. Classify the following pair of vectors as (i) basis, (ii) orthogonal basis and/or, (iii) orthonormal basis:

$$\vec{v_{1}} = \begin{bmatrix} -1 \\ 2 \end{bmatrix} \quad \vec{v_{2}} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$$

The way I did it was:

$$\textbf{Basis:} \qquad \text{rref}\left( \begin{bmatrix} -1 & 2 \\ 2 & 1 \end{bmatrix} \right) \,\,=\,\, \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \quad \checkmark \\ \textbf{Orthogonal Basis:} \qquad \vec{v_{1}} \cdot \vec{v_{2}} \,\,=\,\, \begin{bmatrix} -1 \\ 2 \end{bmatrix} \cdot \begin{bmatrix} 2 \\ 1 \end{bmatrix} \,\,=\,\, 0 \quad \checkmark \\ \textbf{Orthonormal Basis:} \qquad \Vert\vec{v_{1}}\Vert \,\,=\,\,\sqrt{5} \qquad \Vert\vec{v_{2}}\Vert \,\,=\,\, \sqrt{5} \qquad \textbf{X}$$

so this pair of vectors is classified as an $\boxed{\text{orthogonal basis}}$

I know that the final answer is correct from the back of the book that I got this problem from, but if I didn't get the solution the right way I would like to know where I went wrong.

• Also, for technical purposes should I say $\vec{v_{1}}^{T} \cdot \vec{v_{2}}$ for the orthogonal basis? Commented Apr 1, 2015 at 5:52
• No, is there a reason for this? Note that $\vec v_2^T \vec v_1$ is the matrix with one entry, $\vec v_1 \cdot \vec v_2$. Commented Apr 1, 2015 at 6:08
• I was concerned about the "legality" of multiplying a $2 \times 1$ matrix with another $2 \times 1$ matrix. Maybe I am just confusing the dot product with matrix multiplication. Commented Apr 1, 2015 at 7:31