Proving something is a real inner product space I have this question here that I'm trying to solve. I need to show that $S$ is equipped with an inner product and is thus an inner product space.
I've started out by trying to prove this with the axiom of symmetry, which it satisfied, but where it gets confusing for me is trying to prove it using the remaining axioms of additivity, homogeneity and positivity.
How would I go about proving $S$ is an inner product space using the remaining axioms, if that is in fact the correct way of going about it?
In order to show that this is an inner product, you do indeed need to check every property from the definition. Let me show an easy way to do it: first, you can write this operation as:
$$\left&lt;u,v\right&gt; = \begin{pmatrix} u_1 &amp; u_2 \end{pmatrix}\begin{pmatrix} 2 &amp; -1 \\ -1 &amp; 1 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix}$$
It now suffices to show that this matrix $A = \begin{pmatrix} 2 &amp; -1 \\ -1 &amp; 1 \end{pmatrix}$, which is the form's associated matrix in the standard basis, is symmetric and positive definite. You can clearly see it's symmetric. For positive-definiteness, you can use Sylvester's Criterion. You can check that the principal minors of the matrix are $2$ and $2 - (-1)(-1) = 1$, both positive, and hence the bilinear form is positive definite, and, with everything we've seen so far, an inner product.
In order to show that this is an inner product, you do indeed need to check every property from the definition. Let me show an easy way to do it: first, you can write this operation as:
$$\left&lt;u,v\right&gt; = \begin{pmatrix} u_1 &amp; u_2 \end{pmatrix}\begin{pmatrix} 2 &amp; -1 \\ -1 &amp; 1 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix}$$
It now suffices to show that this matrix $A = \begin{pmatrix} 2 &amp; -1 \\ -1 &amp; 1 \end{pmatrix}$, which is the form's associated matrix in the standard basis, is symmetric and positive definite. You can clearly see it's symmetric. For positive-definiteness, you can use Sylvester's Criterion. You can check that the principal minors of the matrix are $2$ and $2 - (-1)(-1) = 1$, both positive, and hence the bilinear form is positive definite, and, with everything we've seen so far, an inner product.