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Linear transformations and matrices.

We suppose now that U and V are finite dimensional, and that $\dim U=n$, $\dim V=m$. Let $\phi : U \longrightarrow V$ be a libear mapping.

Choose bases: $e= \{ \overrightarrow{u_1} , \dots , \overrightarrow{u_n}\}$ for U and $v= \{ \overrightarrow{v_1} , \dots , \overrightarrow{v_m}\}$ for V.

Take a vector $\overrightarrow{x} $ in U; there exists a unique n-tuple of scalars $(x_1, \dots , x_n)$ such that $\overrightarrow{x} = \underset{i=1}{\overset{n}{\sum}} x_i \overrightarrow{u_i} $. Then:
\begin{align*}\phi (\overrightarrow{x} ) & = \phi \left( \underset{i=1}{\overset...
...underset{i=1}{\overset{n}{\sum}} x_i \phi ( \overrightarrow{u_i} ).
\end{align*}

Therefore the linear mapping $\phi$ is totally determined by the data of the vectors $\phi ( \overrightarrow{u_k} ),
\; k=1, \dots , n$.

Denote: $\forall k=1, \dots, n, \; \phi ( \overrightarrow{u_i} ) =
\underset{j=1}{\overset{m}{\sum}} a_{ji} \overrightarrow{v_j} $.

We represent this data by the matrix:

$\begin{pmatrix}a_{11} & a_{12} & \dots & a_{1n} \\
a_{21} & a_{22} & \dots & ...
...w \overrightarrow{v_2}\\ \dots \\
\leftarrow \overrightarrow{v_m}\end{matrix}$

$\begin{matrix}\begin{matrix}\uparrow \\ \phi ( \overrightarrow{u_1} ) \end{matr...
...htarrow{u_n} ) \end{matrix}\end{matrix}\begin{matrix}\quad & \quad \end{matrix}$

If the coordinates of $\overrightarrow{x} $ w.r.t. the given basis of U are $\begin{pmatrix}x_1 \\ x_2 \\ \dots \\ x_n \end{pmatrix}$ and if the coordinates of $\overrightarrow{y} $ w.r.t. the given basis of V are $\begin{pmatrix}y_1 \\ y_2 \\ \dots \\ y_m \end{pmatrix}$, they are related by the formulas:


\begin{displaymath}\begin{cases}
y_1= a_{11} x_1 + a_{12} x_2 + \dots + a_{1n} x...
...
y_m= a_{m1} x_1 + a_{m2} x_2 + \dots + a_{mn} x_n
\end{cases}\end{displaymath}

We denote this in the following way:

\fbox{
$\begin{pmatrix}y_1 \\ y_2 \\ \dots \\ y_m \end{pmatrix} =
\begin{pmatri...
... a_{mn}
\end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \\ \dots \\ x_n \end{pmatrix}$ }

Example 6.3.1   $U=V=\mathcal{P}_3=$the vector space of polynomial in one real indeterminate x, with coefficients in $\bbb{R}$ and such that $\deg p(x) \leq 3$.

A basis for U: p1(x)=1, [2(x)=x, p3(x)=x2, p4(x)=x3.

$\phi = \frac {d}{dx}: \mathcal{P}_3 \longrightarrow \mathcal{P}_3$.

We have: $ \frac {d}{dx} (p_1(x))=0$, $ \frac {d}{dx} (p_2(x))= 1 = p_1(x)$, $ \frac {d}{dx} (p_3(x))= 2x =2p_2(x)$, $ \frac {d}{dx} (p_4(x))= 3x^2 = 3p_3(x)$. Thus, the matrix of $\phi$ w.r.t the given basis is: $\begin{pmatrix}0 & 1 & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 3 \\
0 & 0 & 0 & 0
\end{pmatrix}$.


next up previous contents
Next: The algebra of matrices. Up: Linear mappings. Previous: Kernel and Image of
Noah Dana-Picard
2001-02-26