Let \(( \mid )\) be the standard inner product on \(\RR^2\).
Let \(\alpha = (1, 2)\), \(\beta = (-1, 1)\). If \(\gamma\) is a vector such that \((\alpha \mid \gamma) = -1\) and \((\beta \mid \gamma) = 3\), find \(\gamma\).
Show that for any \(\alpha\) in \(\RR^2\) we have \(\alpha = (\alpha \mid \epsilon_1)\epsilon_1 + (\alpha \mid \epsilon_2)\epsilon_2\).
Solution
1. The vector in part (a) is \(\gamma=(-7/3,2/3)\).
Write \(\gamma=(x,y)\). Under the standard inner product, \[ (\alpha\mid\gamma)=x+2y=-1, \qquad (\beta\mid\gamma)=-x+y=3. \] Adding these equations gives \(3y=2\), so \(y=2/3\), and then \(x=-1-4/3=-7/3\). Thus \[ \boxed{\gamma=\left(-\frac73,\frac23\right)}. \] Direct substitution checks both prescribed inner products.
2. Every vector expands in the standard orthonormal basis by its inner products.
If \(\alpha=(a,b)\), then for \(\epsilon_1=(1,0)\) and \(\epsilon_2=(0,1)\), \[ (\alpha\mid\epsilon_1)=a, \qquad (\alpha\mid\epsilon_2)=b. \] Therefore \[ (\alpha\mid\epsilon_1)\epsilon_1+(\alpha\mid\epsilon_2)\epsilon_2 =a(1,0)+b(0,1)=(a,b)=\alpha, \] as required [@HK71].