Let \(V\) be a vector space over \(F\). Show that the sum of two inner products on \(V\) is an inner product on \(V\). Is the difference of two inner products an inner product? Show that a positive multiple of an inner product is an inner product.
Solution
Let \((\cdot\mid\cdot)_1\) and \((\cdot\mid\cdot)_2\) be inner products on \(V\).
1. Their sum is an inner product.
Define \[ (\alpha\mid\beta)=(\alpha\mid\beta)_1+(\alpha\mid\beta)_2. \] Linearity in the first variable and conjugate symmetry follow term by term from the corresponding axioms for the two summands. Moreover \[ (\alpha\mid\alpha) =(\alpha\mid\alpha)_1+(\alpha\mid\alpha)_2\ge0, \] and if this sum is zero, both nonnegative summands are zero, so \(\alpha=0\). Thus the sum is positive definite and hence an inner product [@HK71].
2. The difference of two inner products need not be an inner product.
Take the same inner product twice. Their difference is the zero form, and for any nonzero \(\alpha\) it satisfies \[ (\alpha\mid\alpha)_1-(\alpha\mid\alpha)_1=0. \] Hence positive definiteness fails. So the answer is \(\boxed{\text{no}}\).
3. Every positive scalar multiple of an inner product is an inner product.
If \(c>0\), define \[ [\alpha,\beta]=c(\alpha\mid\beta). \] Linearity and conjugate symmetry are immediate, while \[ [\alpha,\alpha]=c(\alpha\mid\alpha)>0 \] for every nonzero \(\alpha\). Thus \([\cdot,\cdot]\) is an inner product.