# Mathematics Prelims

## June 18, 2009

### Linear Transformations and Matrix Representations

Filed under: Algebra,Linear Algebra — cjohnson @ 2:22 pm

A common theme in mathematics is (or seems to be) looking at sets with a particular structure, and then looking at functions between those sets which preserve that structure.  In groups we have homomorphisms; in topological spaces we have continuous maps; in general categories we have morphisms.  In the particular case of vector spaces, though, there are two particular “structures” we want to preserve: vector addition and scalar multiplication.  The maps which preserve these are what we refer to as linear transformations.

Specifically, suppose $V$ and $W$ are vector spaces over the field $\mathcal{F}$.  A function $\tau : V \to W$ is called a linear transformation if for all scalars $\alpha, \beta \in \mathcal{F}$ and for all vectors $u, v \in V$ we have

$\displaystyle \tau(\alpha u + \beta v) = \alpha \tau(u) + \beta \tau(v)$

Note that because of this linearity, a linear transformation is completely determined by how it maps the basis vectors of the domain.  Suppose that $\beta = \{ \beta_1, \beta_2, ..., \beta_n \}$ is a basis for V.  Let $u$ be any vector in $V$ with $u = a_1 \beta_1 + ... + a_n \beta_n$.  We then have

$\tau(u) = \tau(a_1 \beta_1 + ... + a_n \beta_n) = a_1 \tau(\beta_1) + ... + a_n \tau(\beta_n)$.

So if we know each $\tau(\beta_i)$, we can figure out where any other vector will be sent by $\tau$.  This does not mean that $\tau(\beta_1), ..., \tau(\beta_n)$ is necessarily a basis for the range, $\tau(V)$.  It could be that $\tau(\beta_1) = \tau(\beta_2)$, in which case these vectors are linearly dependent and can’t both be in the basis.  We do have that $\tau(\beta_1), ..., \tau(\beta_n)$ span $\tau(V)$, however, so as long as they’re linearly independent they’ll form a basis.

The main thing we want to notice about linear transformations for right now is that if both $V$ and $W$ are finite dimensional, then a linear transformation $\tau : V \to W$ can be represented as a matrix.  Suppose that $V$ is $n$-dimensional with the $\beta$ basis mentioned above, and that $W$ is $m$-dimensional with basis $\omega = \{ \omega_1, ..., \omega_m \}$.  Note that the properties of matrix multiplication tell us that any $m \times n$ matrix defines a linear transformation from $V$ to $W$:

$A(\alpha u + \gamma v) = A(\alpha u) + A( \gamma v) = \alpha A u + \gamma A v$

Now suppose $\tau : V \to W$ is any other linear transformation.  Suppose that the coordinate vector of $\tau(\beta_i)$ with respect to the $\omega$ basis is

$\displaystyle \tau(\beta_i) = \left[ \begin{array}{c} a_{1i} \\ a_{2i} \\ \vdots \\ a_{mi} \end{array} \right]_\omega$

Now let $u = b_1 \beta_1 + ... + b_n \beta_n$.  We then have

$\displaystyle \tau(u)$

$\displaystyle \, = b_1 \tau(\beta_1) + b_2 \tau(\beta_2) + ... + b_n \tau(\beta_n)$

$\displaystyle \, = b_1 \left[ \begin{array}{c} a_{11} \\ a_{21} \\ \vdots \\ a_{m1} \end{array} \right] + b_2 \left[ \begin{array}{c} a_{12} \\ a_{22} \\ \vdots \\ a_{m2} \end{array} \right] + ... + b_n \left[ \begin{array}{c} a_{1n} \\ a_{2n} \\ \vdots \\ a_{mn} \end{array} \right]$

$\displaystyle \, = \left[ \begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & & & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{array} \right] \left[ \begin{array}{c} b_1 \\ b_2 \\ \vdots \\ b_n \end{array} \right]$

Thus a linear transformation between finite dimensional vector spaces can be represented as a matrix.  Notice that the entries of our matrix depend on our particular chosen bases: if one basis were altered, the matrix would change, even though the transformation is the same.  We will denote the matrix representing $\tau$ with respect to the $\omega$ and $\beta$ bases as $_\omega[ \tau ]_\beta$