# Invariance Of Inner Product and Quadratic Forms

In Riley's Math Methods book, there is a discussion on quadratic forms (see attached). However, I'm generally more lost about the assertion that "In any basis we can write..." the inner product as below. I am wondering why this is true. To be clear, we defined the (standard) inner product where orthogonal vectors $$\mathbf a,\mathbf b$$ have $$\langle \mathbf a,\mathbf b\rangle=0$$. In order to exploit this definition, we spoke earlier in the book about expressing any vector in the vector space as a linear combination of an orthonormal basis set so that we could evaluate $$\langle \mathbf a,\mathbf b\rangle$$ component-wise, as

$$\langle \mathbf a,\mathbf b\rangle=a_1b_1+a_2b_2...$$

If the vectors are not so expressed, then evaluating the inner product is more complex and consists of "cross terms" where we must consider the non-zero inner products of basis vectors (which are not orthogonal). Accordingly, I'm wondering how we can generally say that

$$Q(\mathbf x)=\langle \mathbf x,A\mathbf x \rangle=\mathbf x^TA\mathbf x$$

since this seems to imply that that $$\mathbf x$$ is expressed as a linear combination of orthonormal basis vectors. Now my understanding is that $$Q(\mathbf x)$$ is a (variable) scalar and so invariant under the various bases that we choose, but in order to evaluate it as we did above I assume that $$\mathbf x$$ needs to be expressed as an orthonormal basis right?

Riley

I think what you might be missing is that the matrix $$\mathsf A$$ in $$Q(\mathbf x)=\langle\mathbf x,\mathcal A\mathbf x\rangle=\mathsf x^T\mathsf A\mathsf x$$ depends on the linear operator $$\mathcal A$$ (obviously), the choice of basis, and the choice of inner product. Clearly when $$\mathcal x$$ and $$\mathcal A$$ are respectively the coordinates of $$\mathbf x$$ and matrix of $$\mathcal A$$ relative to some orthonormal basis, this follows from linearity of the inner product. What happens when the basis isn’t orthonormal?
Let $$\mathcal B=(\mathbf v_1,\dots,\mathbf v_n)$$ be an arbitrary ordered basis of $$\mathbb R^n$$ and $$\mathsf B$$ the matrix whose columns are the coordinates of the $$\mathbf v_i$$ in some orthonormal basis $$\mathcal E$$. The matrix $$\mathsf B$$ then converts from $$\mathcal B$$-coordinates to $$\mathcal E$$-coordinates. Then, if $$\mathsf a$$ and $$\mathsf b$$ are the $$\mathcal B$$-coordinate vectors of $$\mathbf a$$ and $$\mathbf b$$, respectively, we have $$\langle\mathbf a,\mathbf b\rangle = (\mathsf B\mathsf x)^T(\mathsf B\mathsf y) = \mathsf x^T(\mathsf B^T\mathsf B)\mathsf y.$$ The matrix $$\mathsf G=\mathsf B^T\mathsf B$$ is known as the Gram matrix of $$\mathsf B$$. Moreover, if $$\mathsf A$$ is the matrix of a linear operator $$\mathcal A$$ with respect to $$\mathcal E$$, then $$\langle\mathbf x,\mathcal A\mathbf x\rangle = (\mathsf B\mathsf x)^T(\mathsf A\mathsf B\mathsf x) = \mathsf x^T(\mathsf B^T\mathsf A\mathsf B)\mathsf x,$$ which is also of the required form. (I could’ve chosen $$\mathsf A$$ to be the matrix of $$\mathcal A$$ relative to $$\mathcal B$$ instead to get $$\mathsf G\mathsf A$$ for the matrix of the quadratic form, but I did it the other way in anticipation of the change-of-basis formula for quadratic forms.)
I’ll leave as an exercise for you to show that this also works for an arbitrary inner product. The key is to show that any scalar product $$(\mathbf a,\mathbf b)$$ can be expressed in coordinates as $$\mathsf a^T\mathsf Q\mathsf b$$ for some fixed symmetric matrix $$\mathsf Q$$. Additionally, for an inner product (a positive-definite scalar product), it’s always possible to find a basis in which it looks like the standard Euclidean inner product. (Use the Gram-Schmidt process.)
• Super answer, thanks for this. So just to confirm, the matrix $A$ that the author has written is, in general, $GA$ as per what you've written above. – 1729_SR Apr 2 at 13:58