# Linear Transformation Question

by seansrk
Tags: linear, transformation
 P: 3 A matrix is a linear transformation expressed with respect to a basis for the source space and the target space. Given a linear transformation $T:\mathbb{F}^n \to \mathbb{F}^m$, the corresponding matrix written with respect to a basis $\alpha$ for the source space and a basis $\beta$ for the target space is as follows: $\left[ \begin{array}{cccc} [T(\alpha_1)]_\beta & [T(\alpha_2)]_\beta & ... & [T(\alpha_n)]_\beta \end{array} \right]$ The theory behind this is as follows. Since any vector in a given vector space can be expressed as a linear combination of a set of basis vectors for that vector space, we need only transform an arbitrary basis to capture the transformation. Given some vector $v \in \mathbb{F}^n$ and a basis $\alpha$ we can write $v = a_1\alpha_1 + a_2\alpha_2 + ... + a_n\alpha_n$. Then $v$ transformed is as follows $\begin{eqnarray*} T(v) &=& T(a_1\alpha_1 + a_2\alpha_2 + ... + a_n\alpha_n) \\ &=& T(a_1\alpha_1) + T(a_2\alpha_2) + ... + T(a_n\alpha_n) \\ &=& a_1T(\alpha_1) + a_2T(\alpha_2) + ... + a_nT(\alpha_n) \end{eqnarray*}$
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,339 Linear Transformation Question Any m by n matrix is a linear transformation from $R^m$ to $R^n$. What you have given is a perfectly good linear transformation- although the way you have written it, with the "straight" vertical sides, makes it look more like a determinant than a matrix! The matrix you give represents the linear transformation that maps a vector, $a\vec{i}+ b\vec{j}+ c\vec{k}[itex] into [itex]a(5\vec{i}+ 5\vec{j}+ 9\vec{k})+ b(6\vec{i}- 3\vec{k})+ c(9\vec{i}+ 3\vec{j}- 7\vec{kl})$$= (5a+ 6b+ 9c)\vec{i}+ (5a+ 3b)\vec{i}+ (9a- 3b- 7c)\vec{k}$. I wonder if you aren't confusing "matrix", in general, with "invertible matrix".