Find matrices $S$ and $Q$ which satisfy $B = S^t AS$ and $A=Q^tQ$

Question

For

$ A:= M_{\mathcal{A}}(< \cdot,\cdot>); B:= M_{\mathcal{B}}(< \cdot,\cdot>) $ with $\mathcal{A} = (1, x, x^2,x ^3); \mathcal{B} = (1+x^2, x-2x^2, -2x+x^2+x^3, 1+x^3)$ and $<f; g>:=\int_{-1}^1f(x)g(x)\,dx$

I found

$ A = \begin{pmatrix} 2 & 0 & \frac{2}{3} & 0 \\ 0 & \frac{2}{3} & 0 & \frac{2}{5} \\ \frac{2}{3} & 0 & \frac{2}{5} & 0 \\ 0 & \frac{2}{5} & 0 & \frac{2}{7}\end{pmatrix}, B = \begin{pmatrix} \frac{56}{15} & -\frac{32}{15} & \frac{16}{15} & \frac{8}{3} \\ -\frac{32}{15} & \frac{34}{15} & -\frac{26}{15} & -\frac{14}{15} \\ \frac{16}{15} & -\frac{26}{15} & \frac{184}{105} & \frac{16}{105} \\ \frac{8}{3} & -\frac{14}{15} & \frac{16}{105} & \frac{16}{7}\end{pmatrix}$

I hope these are right. Now I have to find matrices $S \in GL_4(\mathbb{R})$ and $Q \in GL_4(\mathbb{R})$ which satisfy $B = S^t AS$ and $A=Q^tQ$. But I don't have any clue how to find those. Can someone point me into the right direction?

Answer 1

The first one is easy. $S$ is simply the change-of-basis matrix from $\mathcal{B}$ to $\mathcal{A}$. E.g. the third basis vector in $\mathcal{B}$ is $-2x+x^2+x^3$, so the third column of $S$ is $(0,-2,1,1)^T$.

For the second one, there are several ways to find $Q$.

1. As $A$ is generated by real inner products, it is a Gramian matrix, which is positive semidefinite. Therefore it can be orthogonally diagonalised to a nonnegative diagonal matrix, i.e. $A=U^TDU$ where $U$ is orthogonal and $D$ is diagonal. The columns of $U$ are eigenvectors of $A$ and the diagonal entries of $D$ are the eigenvalues of $A$. Take entrywise square root of $D$ and set $Q=\sqrt{D}U$, we have $A=Q^TQ$.
2. You may also view this as a similar question to the $B=S^TAS$ part, but now you have to find an orthonormal basis $\mathcal{C}$ first, so that its Gramian matrix is the identity matrix $I$. Then pick $Q$ as the change-of-basis matrix from $\mathcal{A}$ to $\mathcal{C}$, so that $A=Q^TIQ=Q^TQ$. To find an orthonormal basis, apply Gram-Schmidt orthogonalisation. For instance, consider the pair of basis vectors $1$ and $x^2$. The Gram-Schmidt process then gives a pair of orthogonal vectors \begin{align*} 1,\ x^2-\frac{\langle1,x^2\rangle}{\langle1,1\rangle}1. \end{align*} Normalise them to get unit vectors. Do the similar for the other pair $x,x^3$ to get a full set of orthonormal basis $\mathcal{C}$. Now the change-of-basis matrix $P$ from $\mathcal{C}$ to $\mathcal{A}$ are obtained in the similar manner to the construction of $S$ in the previous part. Hence the the change-of-basis matrix $P$ from $\mathcal{A}$ to $\mathcal{C}$ is given by $Q=P^{-1}$, where $$ P=\pmatrix{ \tfrac1{\sqrt{2}} &0 &-\sqrt{\tfrac58} &0\\ 0 &\sqrt{\tfrac32} &0 &-\sqrt{\tfrac{63}8}\\ 0 &0 &\sqrt{\tfrac{45}8} &0\\ 0 &0 &0 &\sqrt{\tfrac{175}8}} $$
3. A simpler way to solve the problem is to apply simultaneous elementary row and column reductions to $B$. See my answer to another question as an example.