Questions related to Hermitian matrices

Question

Dear Linear Algebra Experts,

(I am preparing for my exams and not able to solve these questions. I apologize in advance if they are straightforward but please help me to grow and learn this field. I would really appreciate your help.)

If $A \in M_n$ and $B \in M_n$ are Hermitian matrices, i.e., $A = A^{\rm H} $ and $B = B^{\rm H}$, where $B$ is positive definite.

(1) Prove that there is a non-singular matrix $Y$ such that $Y^{\rm H}AY = C$ and $Y^{\rm H}BY = D$, where both $C$ and $D$ are diagonal matrices.

(2a) Given a matrix $Y$ such that $Y^{\rm H}AY = C$ and and $Y^{\rm H}BY = D$ where both $C$ and $D$ are diagonal (not necessarily related to the above question 1), prove that the columns of $Y$ are eigenvectors of the following generalized eigenvalue problem $$Ay=\lambda B y$$

(2b) Describe how the corresponding eigenvalues can be obtained from $C$ and $D$.

Answer 1

You transform $B$ to the identity with a matrix $Z$. $$Z^{\star}BZ=I$$ $A$ gets transformed to $Z^{\star}AZ$, still hermitian. Now, we don't want to alter the result from $B$ ( the indentity), but want to transform $Z^{\star}AZ$ to a diagonal. So we find $T$ unitary so that $T^{\star}(Z^{\star}AZ) T= C$. Note that $Z$ does not change the $I$. Summing up $$Z^{\star}T^{\star}ATZ = C \\ Z^{\star}T^{\star}BTZ = I$$ So we take $Y=TZ$.

The other question: if $$Y^{\star}AY=C\\ Y^{\star}BY= D$$ then $$AY =Y^{\star}\ ^{-1}C=Y^{\star}\ ^{-1}D \cdot D^{-1}C =BY \cdot D^{-1}C$$ We see that the column $i$ of $Y$ satisfies $$(Ac_i - \lambda_iB) c_i=0$$ where $D^{-1}C = \text{diag}(\lambda_i)$

Answer 2

Based on the proof given by "orangeskid", below is my take (re-writting in a slightly different way). Please experts correct me if I am wrong.

(1) Since $B$ is a positive definite and Hermitian matrix, then after similarity transformation of $B$ into a diagonal matrix $D$. The matrix $D$ can be factorized $$D = T^{{\rm H}/2} T^{1/2},$$ where $T$ is another diagonal matrix.

Now we plug-in the factorized form of $D$ such that $$ Y^{\rm H} B Y = D = T^{{\rm H}/2} T^{1/2} \\ \Leftrightarrow T^{-{\rm H}/2} Y^{\rm H} B Y T^{-1/2} = I \\ \Leftrightarrow \left(Y T^{-1/2}\right)^{\rm H} B \left(Y T^{-1/2}\right) = I.$$

Now define $Z = Y T^{-1/2} \Leftrightarrow Y = Z T^{1/2} $ such that $$Y^{\rm H} A Y = C \Leftrightarrow \left(Z T^{-1/2}\right)^{\rm H} B \left(Z T^{-1/2}\right) = C$$.

(2) if $$Y^{\rm H}AY=C \\ Y^{\rm H}BY= D \Leftrightarrow BY = Y^{-{\rm H}} D.$$ then (assuming $D$ is invertible as $B$ is Hermitian) $$AY =Y^{-{\rm H}}C= \underbrace{Y^{-{\rm H}} D}_{B Y} D^{-1} C = BY \cdot D^{-1}C$$

We see that the column $i$ of $Y$ satisfies $$(Au_i - \lambda_iB u_i) =0$$ where $D^{-1}C = \text{diag}(\lambda_i)$.