Consider a set of $n$ linearly independent $d$-dimensional vectors $\left\{\vec{a}_i\right\}_{i=1}^{i=n}$ that span the vector space $V$ and that are not in general orthogonal with respect to the standard dot product.

Question : Is there in general a linear transformation $L$ : $ V\rightarrow V$, $L(\vec{a}_i)=\vec{b}_i$ such that the mapping yields to an orthogonal basis $\left\{\vec{b}_i\right\}_{i=1}^{i=n}$ : $\vec{b}_i\cdot\vec{b}_j=c_{ij}\delta_{ij}~\forall~i,j$ where $c_{ij}\in\mathbb{C}$ and $\delta_{ij}$ is a Kronecker delta ?

  • $\begingroup$ Yes; in fact, given any two bases, you can find a linear transformation mapping one to the other. $\endgroup$ Feb 8, 2014 at 22:26
  • $\begingroup$ First one is said to span the set, second one is a basis. Is the terminology correct. (span can have redundancy) $\endgroup$
    – Maesumi
    Feb 8, 2014 at 22:28

1 Answer 1


Yes. It's easily seen that a linear map is determined by specifying the images of any basis. In particular in $\mathbb{C}^n$ an arbitrary basis $\{b_1,\ldots,b_n\}$ can be matched up with the "orthogonal" (orthonormal) standard basis vectors.

The mapping of the standard ordered basis $\{e_1,\ldots,e_n\}$ (considered as columns) back to $\{b_1,\ldots,b_n\}$ is simply multiplication by matrix $M = [b_1|\ldots|b_n]$ taking the $b_i$'s as columns. Therefore the inverse mapping $L:\mathbb{C}^n \to \mathbb{C}^n$ sought in the Question is explicitly multiplication by $M^{-1}$.

If the vectors $b_i$ are real-valued, so are $M$ and $M^{-1}$.


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