Assuming we have a basis for a set $\mathbb{R}^n$, would any set of linearly independent vectors that form a basis for $\mathbb{R}^n$ also be orthogonal to each other?

Take the trivial case of $(1,0)$ and $(0,1)$. Now any set of linear independent vectors would be a scalar multiple of these two vectors that form a Basis for $\mathbb{R}^2$ hence they have to be orthogonal. Right?


3 Answers 3


No. The set $\beta=\{(1,0),(1,1)\}$ forms a basis for $\Bbb R^2$ but is not an orthogonal basis. This is why we have Gram-Schmidt!

More general, the set $\beta=\{e_1,e_2,\dotsc,e_{n-1},e_1+\dotsb+e_n\}$ forms a non-orthogonal basis for $\Bbb R^n$.

To acknowledge the conversation in the comments, it is true that orthogonality of a set of vectors implies linear independence. Indeed, suppose $\{v_1,\dotsc,v_k\}$ is an orthogonal set of nonzero vectors and $$ \lambda_1 v_1+\dotsb+\lambda_k v_k=\mathbf 0\tag{1} $$ Then applying $\langle-,v_j\rangle$ to (1) gives $\lambda_j\langle v_j,v_j\rangle=0$ so that $\lambda_j=0$ for $1\leq j\leq k$.

The examples provided in the first part of this answer show that the converse to this statement is not true.

  • $\begingroup$ but they are not linearly independence right? Can you also please explain Gram-Schmidt because I have trouble understanding from videos and my book? $\endgroup$
    – iordanis
    Commented Apr 29, 2014 at 19:32
  • $\begingroup$ Yes, $(1,0)$ and $(1,1)$ are linearly independent. Soliciting information on Gram-Schmidt might be better suited for a different question. $\endgroup$ Commented Apr 29, 2014 at 19:34
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    $\begingroup$ @μακακας to use very imprecise terminology, for a vector $\in\mathbb{R}^n$ to be linearly independent only means that it must be "partly orthogonal" to the other vectors in the set - in other words, that it must have some component that is orthogonal to the other vectors, but it can also have components parallel to one or more of those other vectors. To be orthogonal means it cannot have any component parallel to any of the other vectors. So orthogonality is a more restrictive criterion than linear independence. $\endgroup$
    – David Z
    Commented Apr 29, 2014 at 19:37
  • $\begingroup$ I am still confused about orthogonality but thanks anyways. I found a theorem that says An orthogonal set of non zero vectors in R^n is linearly independent so I guess it doesn't work the other way around? $\endgroup$
    – iordanis
    Commented Apr 29, 2014 at 19:37
  • 1
    $\begingroup$ @μακακας What would it mean for two different bases to be "interchangeable"? $\endgroup$ Commented Apr 29, 2014 at 20:35

In general: No! Any vectors ${\bf u} = (u_1,u_2,u_3)$, ${\bf v} = (v_1,v_2,v_3)$ and ${\bf w} = (w_1,w_2,w_3)$ will span $\mathbb{R}^3$ if, and only if, the following determinant is non-zero: $$\left|\begin{array}{ccc} u_1 & v_1 & w_1 \\ u_2 & v_2 & w_2 \\ u_3 & v_3 & w_3 \end{array}\right| \neq 0$$ This is equivalent to the single polynomial condition $$u_1(v_2w_3-w_2v_3) - v_1(u_2v_3 - w_2u_3) + w_1(u_2v_3-v_2u_3) \neq 0$$

For a set of non-zero vectors to be orthogonal, we need them all to be at right angles:

$$\langle {\bf u}, {\bf v}\rangle = \langle {\bf u}, {\bf w}\rangle = \langle {\bf v}, {\bf w}\rangle =0$$ For this to happen, we need three simultaneous polynomial conditions:

\begin{eqnarray*} u_1v_1 + u_2v_2 + u_3v_3 &=& 0 \\ \\ u_1w_1 + u_2w_2 + u_3w_3 &=& 0 \\ \\ v_1w_1 + v_2w_2 + v_3w_3 &=& 0 \end{eqnarray*}

There are many sets of linearly independent vectors $\{{\bf u},{\bf v},{\bf w}\}$ which are not orthogonal.


Nope. $(1,1)$ and $(1,0)$ form a basis of $\mathbb R^2$.


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