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.

Thank you in advanced


1 Answer 1


Your answer is correct. Just note that in order for a set to form a basis, it must be linearly independent, and span the given space. It is easy to see in this example, but in further problems, you may need to verify manually.

  • $\begingroup$ Okay, my textbook is pretty vague so I followed PatrickJMT's video on Finding a Basis for a Set of Vectors (youtube.com/watch?v=0utd-Noc_Fs ) and he simply augments the set of vectors and puts it into reduced row echelon form. What I took away from the video is that if the rref of the set of matrices is the identity matrix then the set forms a basis on the vector space, if not only the pivot columns of the original matrix form a basis. $\endgroup$
    – user227915
    Commented Apr 1, 2015 at 5:37
  • $\begingroup$ Yes, take the columns from the original matrix corresponding to the pivot columns to form a basis. $\endgroup$
    – MathMajor
    Commented Apr 1, 2015 at 5:41
  • $\begingroup$ Also, for technical purposes should I say $\vec{v_{1}}^{T} \cdot \vec{v_{2}}$ for the orthogonal basis? $\endgroup$
    – user227915
    Commented Apr 1, 2015 at 5:52
  • $\begingroup$ No, is there a reason for this? Note that $\vec v_2^T \vec v_1$ is the matrix with one entry, $\vec v_1 \cdot \vec v_2$. $\endgroup$
    – MathMajor
    Commented Apr 1, 2015 at 6:08
  • $\begingroup$ 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. $\endgroup$
    – user227915
    Commented Apr 1, 2015 at 7:31

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