Let $A,B,C$ be two symmetric $n$ x $n$ matrices, where $B$ is also positive definite. Imitate the procedure to obtain a spectral decomposition of $A$ with respect to $B$, by replacing the constraints $\|x\|^2=1$ with $\langle Bx,x\rangle=1$ and the contraints $\langle x,u_j\rangle=0$ with $ \langle Bx,u_j\rangle=0$ Show that this leads to the following results:
- The $n$ x $n$ matrix $U=[u_1,....,u_n]$ satifies the conditions $\langle Bu_i,u_i\rangle=1$, $\langle Bu_i,u_j\rangle=0$, $Au_i=\lambda_i B u_i$ for $i,j=1,...,n$, $i \not= j$
Consider the minimization of the quadratic over $\langle Bx,x\rangle=1$ . $$\min \langle Ax,x \rangle$$ $$s.t \langle Bx,x \rangle =1$$ Thus we have $L(x,\lambda)=\langle Ax,x \rangle + \lambda_1 (1-\langle Bx,x \rangle)$ which implies $$\bigtriangledown_{x}L=2Au_1-2\lambda_1 Bu_1=0, \|u_1\|=1$$ Now consider sequentially the following problems $$\min \langle Ax,x \rangle$$ $$s.t \langle Bx,x \rangle =1$$ $$\langle Bx,u_j \rangle=0 \>j=1,....,k-1$$ So $L(x,\lambda,\delta_1,...\delta_{k-1}=\frac{1}{2} \langle Ax,x \rangle + \frac{\lambda}{2} (1-\langle Bx,x \rangle)+\sum_{j=1}^{k-1} \delta_{j}^{k} \langle Bx,u_j \rangle$
- Define $Λ := diag(λ_1 , . . . , λ_n )$. Prove that (i) implies that we have the simultaneous diagonalizable of $A$ and $B$ $$U^{T}AU=Λ \>, U^{T}BU=I$$ Not sure if I am doing 1 right but also need help with 2
Define $U=[u_1,...u_n]$ and $\Lambda=diag\{\lambda_1,....,\lambda_1\}$. We have $Au_i=\lambda_i B u_i$, $\langle Bu_i,u_i \rangle=1$, $\langle Bu_i,u_j\rangle=0$ Observe: $$U^{T}BU=[u_1,...u_n]^{T}B[u_1,...u_n]=\begin{pmatrix} u_{1}^{T}Bu_1 & \cdots & u_{1}^{T}Bu_n \\ u_{2}^{T}Bu_1 & \cdots & u_{2}^{T}Bu_n \\ \vdots & \ddots & \vdots \\ u_{n}^{T}Bu_1 & \cdots & u_{n}^{T}Bu_n \end{pmatrix}=I$$ and $AU=BU\Lambda \implies U^{T}AU=(U^{T}BU)\Lambda=I \Lambda$