Is there any connection between the fact that a set of vectors are mutually orthogonal and the same set of vectors are linearly independent.


I have learned that if a self-adjoint mapping of a finite real vector space has $n$ eigenvector, then that set of eigenvectors form a orthogonal basis for the vector space, so in general, is there any connection between these concepts, i.e being orthogonal and being linearly independent, in the sense that one implies the other.

  • $\begingroup$ What is the reason for the down vote ? $\endgroup$ – onurcanbektas Feb 8 '18 at 7:21

A set of vectors being orthogonal implies that it is linearly independent. Suppose that $V$ is an inner product space with inner product $\langle \cdot,\cdot \rangle:V\times V\to \mathbf{R}$. Then suppose that $\{v_1,\ldots, v_k\}$ is an orthogonal set. Suppose that for coefficients $a_i$ in $\mathbf{R}$, $$ \sum_{i=1}^ka_i v_i=0.$$ Taking inner products, we have that $$ \bigg\langle v_j,\sum_{i=1}^k a_iv_i\bigg\rangle=\langle v_j,0\rangle=0.$$ Simplifying the expression on the left, we have $$0=\bigg\langle v_j,\sum_{i=1}^k a_iv_i\bigg\rangle=\sum_{i=1}^k\langle v_j,a_iv_i\rangle =\sum_{i=1}^ka_i\langle v_j,v_i\rangle=\sum_{i=1}^ka_i\delta_i^j=a_j$$ so that $a_j=0$ for all $1\le j\le k$. Thus, $\{v_1,\ldots, v_k\}$ are linearly independent. To see that the converse does not hold, take vectors $(1,0),(1,1)$ in $\mathbf{R}^2$ equipped with the standard dot product. Then we have that these vectors are linearly independent, but $$ (1,0)\cdot (1,1)=1$$ so that the vectors are not orthogonal.

  • $\begingroup$ Thanks for your answer Antonios-Alexandros Robotis :). Even though I have understood the proof, I was going to object the sentence "Fix chef. $a_i$ and suppose the sum is equal the zero", because what you mean by fixing $a_i$ is not clear, so I advise you to edit that part. I mean you don't have to say that $a_i$s are fix, you can just directly suppose that the sum is equal to zero. $\endgroup$ – onurcanbektas Feb 8 '18 at 7:12
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    $\begingroup$ Updated the solution. $\endgroup$ – Antonios-Alexandros Robotis Feb 8 '18 at 7:35

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