Let $A in M_n $ be rank 1 PSD; $B = Q^{-frac{1}{2}} A Q^{-frac{1}{2}}$, $Q$ is PD. Then can be $B$ be...
Let $A in M_n$ be rank one positive semidefinite (PSD) matrix, and the matrix $A$ can be unitarily diagonalised such that $A = U Lambda U^*$ (eigenvalue decomposition EVD) where $U^*U=UU^* = I$.
For a given $Q$ symmetric positive definite (PD) matrix, and
$$B = Q^{-frac{1}{2}} A Q^{-frac{1}{2}}$$ is also positive semidefinite and rank-one.
Then can $B$ be unitarily diagonlized?
Attempt
begin{align}
B
&= Q^{-frac{1}{2}} A Q^{-frac{1}{2}} \
&= Q^{-frac{1}{2}} U Lambda U^* Q^{-frac{1}{2}} \
end{align}
Then I don't know whether $B = V Sigma V^*$ be unitarily diagonalised? Because $VV^* = Q^{-1} neq V^*V = U^* Q^{-1} U neq I$ unless $Q = I$. I am confused. Please enlighten me.
linear-algebra
asked Nov 25 at 15:20
learning
275
add a comment |
Let $A in M_n$ be rank one positive semidefinite (PSD) matrix, and the matrix $A$ can be unitarily diagonalised such that $A = U Lambda U^*$ (eigenvalue decomposition EVD) where $U^*U=UU^* = I$.
For a given $Q$ symmetric positive definite (PD) matrix, and
$$B = Q^{-frac{1}{2}} A Q^{-frac{1}{2}}$$ is also positive semidefinite and rank-one.
Then can $B$ be unitarily diagonlized?
Attempt
begin{align}
B
&= Q^{-frac{1}{2}} A Q^{-frac{1}{2}} \
&= Q^{-frac{1}{2}} U Lambda U^* Q^{-frac{1}{2}} \
end{align}
Then I don't know whether $B = V Sigma V^*$ be unitarily diagonalised? Because $VV^* = Q^{-1} neq V^*V = U^* Q^{-1} U neq I$ unless $Q = I$. I am confused. Please enlighten me.
linear-algebra
asked Nov 25 at 15:20
learning
275
add a comment |
Let $A in M_n$ be rank one positive semidefinite (PSD) matrix, and the matrix $A$ can be unitarily diagonalised such that $A = U Lambda U^*$ (eigenvalue decomposition EVD) where $U^*U=UU^* = I$.
For a given $Q$ symmetric positive definite (PD) matrix, and
$$B = Q^{-frac{1}{2}} A Q^{-frac{1}{2}}$$ is also positive semidefinite and rank-one.
Then can $B$ be unitarily diagonlized?
Attempt
begin{align}
B
&= Q^{-frac{1}{2}} A Q^{-frac{1}{2}} \
&= Q^{-frac{1}{2}} U Lambda U^* Q^{-frac{1}{2}} \
end{align}
Then I don't know whether $B = V Sigma V^*$ be unitarily diagonalised? Because $VV^* = Q^{-1} neq V^*V = U^* Q^{-1} U neq I$ unless $Q = I$. I am confused. Please enlighten me.
linear-algebra
asked Nov 25 at 15:20
learning
275
Let $A in M_n$ be rank one positive semidefinite (PSD) matrix, and the matrix $A$ can be unitarily diagonalised such that $A = U Lambda U^*$ (eigenvalue decomposition EVD) where $U^*U=UU^* = I$.
For a given $Q$ symmetric positive definite (PD) matrix, and
$$B = Q^{-frac{1}{2}} A Q^{-frac{1}{2}}$$ is also positive semidefinite and rank-one.
Then can $B$ be unitarily diagonlized?
Attempt
begin{align}
B
&= Q^{-frac{1}{2}} A Q^{-frac{1}{2}} \
&= Q^{-frac{1}{2}} U Lambda U^* Q^{-frac{1}{2}} \
end{align}
Then I don't know whether $B = V Sigma V^*$ be unitarily diagonalised? Because $VV^* = Q^{-1} neq V^*V = U^* Q^{-1} U neq I$ unless $Q = I$. I am confused. Please enlighten me.
linear-algebra
linear-algebra
asked Nov 25 at 15:20
learning
275
asked Nov 25 at 15:20
learning
275
edited Nov 25 at 15:25
asked Nov 25 at 15:20
learning
275
asked Nov 25 at 15:20
learning
275
asked Nov 25 at 15:20
learning
275
275
add a comment |
add a comment |
1 Answer
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$B$ is symmetric since
$$
B^*=(Q^{-1/2}AQ^{-1/2})^*=(Q^{-1/2})^*A^*(Q^{-1/2})^*=Q^{-1/2}AQ^{-1/2}
$$
since $Q^{-1/2}$ and $A$ are symmetric. Hence, $B$ is unitarily diagonalisable.
$Q^{-1/2}$ is symmetric since, if $Q=U^*DU$, where $U$ is unitary and $D$ diagonal with positive elements, then $Q^{-1/2}=U^*D^{-1/2}U$.
Finally,
$$
text{rank}(B)=text{rank}(A)=1,
$$
since $Q^{-1/2}$ is invertible.
answered Nov 25 at 15:49
Yiorgos S. Smyrlis
62.4k1383162
add a comment |
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$B$ is symmetric since
$$
B^*=(Q^{-1/2}AQ^{-1/2})^*=(Q^{-1/2})^*A^*(Q^{-1/2})^*=Q^{-1/2}AQ^{-1/2}
$$
since $Q^{-1/2}$ and $A$ are symmetric. Hence, $B$ is unitarily diagonalisable.
$Q^{-1/2}$ is symmetric since, if $Q=U^*DU$, where $U$ is unitary and $D$ diagonal with positive elements, then $Q^{-1/2}=U^*D^{-1/2}U$.
Finally,
$$
text{rank}(B)=text{rank}(A)=1,
$$
since $Q^{-1/2}$ is invertible.
answered Nov 25 at 15:49
Yiorgos S. Smyrlis
62.4k1383162
add a comment |
$B$ is symmetric since
$$
B^*=(Q^{-1/2}AQ^{-1/2})^*=(Q^{-1/2})^*A^*(Q^{-1/2})^*=Q^{-1/2}AQ^{-1/2}
$$
since $Q^{-1/2}$ and $A$ are symmetric. Hence, $B$ is unitarily diagonalisable.
$Q^{-1/2}$ is symmetric since, if $Q=U^*DU$, where $U$ is unitary and $D$ diagonal with positive elements, then $Q^{-1/2}=U^*D^{-1/2}U$.
Finally,
$$
text{rank}(B)=text{rank}(A)=1,
$$
since $Q^{-1/2}$ is invertible.
answered Nov 25 at 15:49
Yiorgos S. Smyrlis
62.4k1383162
add a comment |
$B$ is symmetric since
$$
B^*=(Q^{-1/2}AQ^{-1/2})^*=(Q^{-1/2})^*A^*(Q^{-1/2})^*=Q^{-1/2}AQ^{-1/2}
$$
since $Q^{-1/2}$ and $A$ are symmetric. Hence, $B$ is unitarily diagonalisable.
$Q^{-1/2}$ is symmetric since, if $Q=U^*DU$, where $U$ is unitary and $D$ diagonal with positive elements, then $Q^{-1/2}=U^*D^{-1/2}U$.
Finally,
$$
text{rank}(B)=text{rank}(A)=1,
$$
since $Q^{-1/2}$ is invertible.
answered Nov 25 at 15:49
Yiorgos S. Smyrlis
62.4k1383162
$B$ is symmetric since
$$
B^*=(Q^{-1/2}AQ^{-1/2})^*=(Q^{-1/2})^*A^*(Q^{-1/2})^*=Q^{-1/2}AQ^{-1/2}
$$
since $Q^{-1/2}$ and $A$ are symmetric. Hence, $B$ is unitarily diagonalisable.
$Q^{-1/2}$ is symmetric since, if $Q=U^*DU$, where $U$ is unitary and $D$ diagonal with positive elements, then $Q^{-1/2}=U^*D^{-1/2}U$.
Finally,
$$
text{rank}(B)=text{rank}(A)=1,
$$
since $Q^{-1/2}$ is invertible.
answered Nov 25 at 15:49
Yiorgos S. Smyrlis
62.4k1383162
answered Nov 25 at 15:49
Yiorgos S. Smyrlis
62.4k1383162
answered Nov 25 at 15:49
Yiorgos S. Smyrlis
62.4k1383162
answered Nov 25 at 15:49
Yiorgos S. Smyrlis
62.4k1383162
62.4k1383162
add a comment |
add a comment |
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