Let \(V\) be the orthogonal complement of \(\Vect{n} = (1,1,1)\) in \(\RNrSpc{3}\), and let \(W\) be the subspace of \(\RNrSpc{4}\) spanned by the vectors \(\Vect{b}_1\DefEq (1,0,1,0)\) and \(\Vect{b}_2\DefEq (1,1,1,1)\). Determine whether the linear transformation
\[L\from V \longrightarrow W,\quad L(x,y,z)= (x+y-z)\cdot \Vect{b}_1\ +\ (y+z)\cdot \Vect{b}_2\]is invertible. If it is, find its inverse.
Step 1 Setting up a basis for \(V\). \(V\) has dimension 2 So any basis of \(V\) has two vectors. The vectors
\[\Vect{a}_1\DefEq (1,0,-1) \quad\text{and}\quad \Vect{a}_2\DefEq (1,-2,1)\]are perpendicular to each other, and are perpendicular to \(\Vect{n}\). So \(\EuScript{A}\DefEq (\Vect{a}_1,\Vect{a}_2)\) forms an ordered basis of \(V\).
Step 2 Setting up a basis for \(W\). As \(\Vect{b}_1\) and \(\Vect{b}_2\) are linearly independent, they form an ordered basis \(\EuScript{B}\) for \(W\).
Step 3 Finding the matrix representing \(L\) with respect to the bases \(\EuScript{A}\) and \(\EuScript{B}\). We find
| \(L(\Vect{a}_1)\) | \(=\) | \(L(1,0,-1)\) |
| \(\) | \(=\) | \((1-(-1))\cdot \Vect{b}_1\ +\ (0+(-1))\cdot \Vect{b}_2\) |
| \(\) | \(=\) | \(2\cdot \Vect{b}_1 + (-1)\cdot \Vect{b}_2\) |
| \(L(\Vect{a}_2)\) | \(=\) | \(L(1,-2,1)\) |
| \(\) | \(=\) | \((1+(-2)+(-1))\cdot \Vect{b}_1\ +\ ((-2) + 1)\cdot \Vect{b}_2\) |
| \(\) | \(=\) | \(-2\cdot \Vect{b}_1 + (-1)\cdot \Vect{b}_2\) |
Therefore,
\[ \Mtrx{A}_{\EuScript{B}\EuScript{A}} = \left[ \begin{array}{rr} 2 & -2 \\ -1 & -1 \end{array} \right] \]This matrix has
| \(\det(\Mtrx{A}_{\EuScript{B}\EuScript{A}})\) | \(=\) | \(-4\neq 0\) |
So \(L\) is invertible, and \(L^{-1}\) is represented by
\[ \Mtrx{A}_{\EuScript{A}\EuScript{B}}(L^{-1}) = \left( \Mtrx{A}_{\EuScript{B}\EuScript{A}}\right)^{-1} = -\dfrac{1}{4} \left[ \begin{array}{rr} -1 & 2 \\ 1 & 2 \end{array} \right] \]