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  1. If $a \in \mathbb{R}$, let $\beta_a:\mathbb{R}^2 \times \mathbb{R}^2 \rightarrow \mathbb{R}$ be the billinear map defined by

$$\beta_a \left( \left( \begin{array}{ccc} x_1 \\ x_2 \end{array} \right), \left( \begin{array}{ccc} y_1 \\ y_2 \end{array} \right) \right)=4x_1y_1-2x_1y_2-ax_2y_1+3x_2y_2$$

Determine all $a$ such that $\beta_a$ defines an inner product on $\mathbb{R}^2$.

I'm really not sure hpw to move forward with this. The symmetry condition gives us that $$2x_1y_2+ax_2y_1=2x_2y_1+ax_1y_2$$ I think, simply by equating the given orientation to that when the vectors are 'switched round'.

I imagine we have to use linearity in the argument for it to be an IP, but I just can't see where. Help would be much appreciated.

The question (past exam question) goes on...

  1. Let $a \in \mathbb{R}$ s.t. $\beta = \beta_a$ is such an inner product. With respect to the inner product $\beta$, construct an orthonormal basis of $\mathbb{R}^2$ containing the vector $$\left( \begin{array}{ccc} \frac{1}{2} \\ 0 \end{array} \right)$$

Here, I'm assuming Gram-Schmidt needs to be applied, but I'm not entirely sure. I'll work at it and post any progress. Any assistance would be very appreciated.

Thanks in advance.

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up vote 3 down vote accepted

An inner product is a:

  • symmetric
  • positive definite
  • bilinear form.

For symmetry, your equation $2x_1y_2+ax_2y_1=2x_2y_1+ax_1y_2$ is correct, and far simpler than it appears. Remember it must hold for all $x_i$ and $y_j$, so pick convenient ones that make it simpler.

Bilinearity is pretty much automatic, especially if you write the inner product as

$$\left(\begin{array}{cc} x_1 & x_2 \end{array}\right) \left(\begin{array}{cc} 4 & -2 \\ -a & 3 \end{array}\right) \left(\begin{array}{c} y_1 \\ y_2 \end{array}\right)$$. As for positive definiteness, well, just set $x_i = y_i$ and do some algebra (usually, completing a square of some kind). Or if you were really keen, you could calculate the eigenvalues of the above matrix and check they're both positive (or even check they're positive without calculating them, using the trace and determinant).

For the last part, I don't think you use Gram-Schmidt as such. G-S produces an orthonormal basis from an existing basis; you could just throw in another vector and then use G-S on the resulting basis, but it's simpler to just construct an orthonormal one directly. I mean, goodness, you want a basis of 2D space containing a given element, you only need to find one more. Find an orthogonal thing, and then normalise it.

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