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

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