Mixing
If $A$ and $Q$ are unitary, then $U = Q^{-1}AQ$ is unitary…
$begingroup$
Here is what I have so far...
Since $A$ and $Q$ are unitary, by definition, we have that $AA^* = A^*A = I$ and $QQ^* = Q^*Q = I$, in other words, $A^* = A^{-1}$ and $Q^* = Q^{-1}$. We can then define $U = Q^*AQ$. We want to show that $U$ is unitary, namely, $UU^* = U^*U = I$. So,
$$UU^* = (Q^*AQ)(Q^*AQ)^* = (Q^*AQ)(QA^*Q^*) = dots$$
This is where I am lost since I have no reason to assume that $Q^2 = I$.
Any help would be great!
linear-algebra matrices
asked Jan 9 at 0:42
Taylor McMillanTaylor McMillan
695
$endgroup$
add a comment |
$begingroup$
Here is what I have so far...
Since $A$ and $Q$ are unitary, by definition, we have that $AA^* = A^*A = I$ and $QQ^* = Q^*Q = I$, in other words, $A^* = A^{-1}$ and $Q^* = Q^{-1}$. We can then define $U = Q^*AQ$. We want to show that $U$ is unitary, namely, $UU^* = U^*U = I$. So,
$$UU^* = (Q^*AQ)(Q^*AQ)^* = (Q^*AQ)(QA^*Q^*) = dots$$
This is where I am lost since I have no reason to assume that $Q^2 = I$.
Any help would be great!
linear-algebra matrices
asked Jan 9 at 0:42
Taylor McMillanTaylor McMillan
695
$endgroup$
add a comment |
$begingroup$
Here is what I have so far...
Since $A$ and $Q$ are unitary, by definition, we have that $AA^* = A^*A = I$ and $QQ^* = Q^*Q = I$, in other words, $A^* = A^{-1}$ and $Q^* = Q^{-1}$. We can then define $U = Q^*AQ$. We want to show that $U$ is unitary, namely, $UU^* = U^*U = I$. So,
$$UU^* = (Q^*AQ)(Q^*AQ)^* = (Q^*AQ)(QA^*Q^*) = dots$$
This is where I am lost since I have no reason to assume that $Q^2 = I$.
Any help would be great!
linear-algebra matrices
asked Jan 9 at 0:42
Taylor McMillanTaylor McMillan
695
$endgroup$
Here is what I have so far...
Since $A$ and $Q$ are unitary, by definition, we have that $AA^* = A^*A = I$ and $QQ^* = Q^*Q = I$, in other words, $A^* = A^{-1}$ and $Q^* = Q^{-1}$. We can then define $U = Q^*AQ$. We want to show that $U$ is unitary, namely, $UU^* = U^*U = I$. So,
$$UU^* = (Q^*AQ)(Q^*AQ)^* = (Q^*AQ)(QA^*Q^*) = dots$$
This is where I am lost since I have no reason to assume that $Q^2 = I$.
Any help would be great!
linear-algebra matrices
linear-algebra matrices
asked Jan 9 at 0:42
Taylor McMillanTaylor McMillan
695
asked Jan 9 at 0:42
Taylor McMillanTaylor McMillan
695
asked Jan 9 at 0:42
Taylor McMillanTaylor McMillan
695
asked Jan 9 at 0:42
Taylor McMillanTaylor McMillan
695
asked Jan 9 at 0:42
Taylor McMillanTaylor McMillan
695
695
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Your mistake is $(AB)^*=B^*A^*$ so $(Q^*AQ)^*=Q^*A^*Q$.
This implies that $UU^*=(Q^*AQ)(Q^*AQ)^*=Q^*AQQ^*A^*Q=I$.
answered Jan 9 at 0:49
Tsemo AristideTsemo Aristide
57.6k11444
$endgroup$
$begingroup$
Of course, socks-shoes property. Thanks!
$endgroup$
– Taylor McMillan
Jan 9 at 18:44
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Your mistake is $(AB)^*=B^*A^*$ so $(Q^*AQ)^*=Q^*A^*Q$.
This implies that $UU^*=(Q^*AQ)(Q^*AQ)^*=Q^*AQQ^*A^*Q=I$.
answered Jan 9 at 0:49
Tsemo AristideTsemo Aristide
57.6k11444
$endgroup$
$begingroup$
Of course, socks-shoes property. Thanks!
$endgroup$
– Taylor McMillan
Jan 9 at 18:44
add a comment |
$begingroup$
Your mistake is $(AB)^*=B^*A^*$ so $(Q^*AQ)^*=Q^*A^*Q$.
This implies that $UU^*=(Q^*AQ)(Q^*AQ)^*=Q^*AQQ^*A^*Q=I$.
answered Jan 9 at 0:49
Tsemo AristideTsemo Aristide
57.6k11444
$endgroup$
$begingroup$
Of course, socks-shoes property. Thanks!
$endgroup$
– Taylor McMillan
Jan 9 at 18:44
add a comment |
$begingroup$
Your mistake is $(AB)^*=B^*A^*$ so $(Q^*AQ)^*=Q^*A^*Q$.
This implies that $UU^*=(Q^*AQ)(Q^*AQ)^*=Q^*AQQ^*A^*Q=I$.
answered Jan 9 at 0:49
Tsemo AristideTsemo Aristide
57.6k11444
$endgroup$
Your mistake is $(AB)^*=B^*A^*$ so $(Q^*AQ)^*=Q^*A^*Q$.
This implies that $UU^*=(Q^*AQ)(Q^*AQ)^*=Q^*AQQ^*A^*Q=I$.
answered Jan 9 at 0:49
Tsemo AristideTsemo Aristide
57.6k11444
answered Jan 9 at 0:49
Tsemo AristideTsemo Aristide
57.6k11444
answered Jan 9 at 0:49
Tsemo AristideTsemo Aristide
57.6k11444
answered Jan 9 at 0:49
Tsemo AristideTsemo Aristide
57.6k11444
57.6k11444
$begingroup$
Of course, socks-shoes property. Thanks!
$endgroup$
– Taylor McMillan
Jan 9 at 18:44
add a comment |
$begingroup$
Of course, socks-shoes property. Thanks!
$endgroup$
– Taylor McMillan
Jan 9 at 18:44
$begingroup$
Of course, socks-shoes property. Thanks!
$endgroup$
– Taylor McMillan
Jan 9 at 18:44
$begingroup$
Of course, socks-shoes property. Thanks!
$endgroup$
– Taylor McMillan
Jan 9 at 18:44
add a comment |
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