Find the matrix representing the shear transformation \(S\) of \(\RNrSpc{2}\), parallel to the \(x\)-axis which transforms the vector \(\Vect{a} = (0,1)\) into the vector \(S(\Vect{n}) = (1,1)\).
\(\Vect{n}\) is a unit vector. Therefore the shear vector is
\[\Vect{s} = S(\Vect{n}) - \Vect{n} = (1,1) - (0,1) = (1,0)\]
To find the matrix representing \(S\), we determine the effect of \(S\) on the vectors \(\StdBssVec{1}=(1,0)\) and \(\StdBssVec{2}=(0,1)\)
| \(S(1,0)\) | \(=\) | \((1,0) + \left( \DotPr{(1,0)}{(0,1)}\right) \cdot \Vect{s} = (1,0)\) |
| \(S(0,1)\) | \(=\) | \((1,1)\) |
Therefore,
\[ A = \left[ \begin{array}{rr} 1 & 1 \\ 0 & 1 \end{array} \right] \]As in the example above we see that the general shear transformation of \(\RNrSpc{2}\) parallel to the \(x\)-axis is described by
\[ S(x,y) = \left[ \begin{array}{rr} 1 & a \\ 0 & 1 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] \]In particular, \(S\) leaves \(\StdBssVec{1}\) unchanged, and it shears \(\StdBssVec{2}\) into \((0,1) + (a,0) = (a,1)\). So the shear vector is \(\Vect{s} = (a,0)\).
The picture below shows the effect of such a shear transformation on the unit cube of \(\RNrSpc{3}\).
Specifically, let's work out the matrix which represents the shear transformation of \(\RNrSpc{3}\) parallel to the \(xy\)-plane with shear vector \(\Vect{s} = (a,b,0)\). This matrix is of size \((3,3)\), and its columns are given by
| \(S(\StdBssVec{1})\) | \(=\) | \(\StdBssVec{1} = (1,0,0)\) |
| \(S(\StdBssVec{2})\) | \(=\) | \(\StdBssVec{2} = (0,1,0)\) |
| \(S(\StdBssVec{3})\) | \(=\) | \(\StdBssVec{3} + (\DotPr{(0,0,1)}{\StdBssVec{3}})\cdot (a,b,0) = (1,a,b)\) |
Therefore the matrix representing \(S\) is:
\[ S\from \RNrSpc{3} \longrightarrow \RNrSpc{3},\quad S(x,y,z) = \left[ \begin{array}{ccc} 1 & 0 & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{array} \right] \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] \]