Linear Operators

Change of basis¶

Let $V$ be a finite-dimensional vector space of dimesnion $n$. Let $\beta=\{e_1,...,e_n\}$ and $\beta'=\{e_1',...,e'_n\}$ be both basis for $V$. We are interested in change reprentation of $x\in V$ from using $\beta$ to $\beta'$. To do this we need to know how $e_j\in \beta$ can be represented using $\beta'$:

Here coefficientcs $a_{\nu}^{\mu}$ can be orgainized into a matrix. The matrix $a_{\nu}^{\mu}$ looks like:

Contravariant Transformation¶

There are two interpretations of $a_{\nu}^{\mu}$:

1. ($a_{\nu}^{\mu}$ transform $\beta'$ to $\beta$) This is the definition above:

   $$e_\nu = a_{\nu}^{\mu} e_{\mu}'$$

   (3)

2. ($a_{\nu}^{\mu}$ transform representation in $\beta$ to $\beta'$) Now for arbitrary $x\in V$, we have:

   $$x=c'_{\nu}e'_{\nu}=c_{\nu}e_{\nu} =c_\nu (a_{\nu}^{\mu} e_{\mu}' )=(c_\nu a_{\nu}^{\mu}) e_{\mu}'$$

   (4)

   from which we recognize:

   $$c_{\nu}'=(c_{\nu}a_{\nu}^{\mu})$$

   (5)

We note that the matrix has opposite effect on basis and representation. This is actually intuitive: in order to represent the vector in new basis $\beta'$, we will have to write the old basis $e_\nu \in \beta$ using linear combinations of the new basis $e_{\mu}'$. The later which effectively transforms basis in $\beta'$ to $\beta$. For this reason, the representation is said to be transformed contravariantly.

Triangularization¶