Mixing
Does $u=Psi^-1circ Lcirc Phiiff L=Psicirc ucirc Phi^-1$. Same question with matrices.
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Q1) $uin mathcal L(E,F)$ with $E$ and $V$ finite vector spaces over $K$. If $Phi:Eto K^n$ and $Psi:Fto K^p$ are isomorphisms, and if $$u=Psi^-1circ Lcirc Phi,$$
with $L:K^nto K^p$ linear. Does we get that $$Psicirc ucirc Phi^-1=L ?$$
I would say yes since $$u=Psi^-1circ Lcirc Phiiff ucirc Phi=Psi^-1circ Lcirc Phicirc Phi^-1=Psi^-1circ uiff Psicirc ucirc Phi^-1=Psicirc Psi^-1circ L=L,$$
but I have doubt with composition at right...
Q2) If $A=P^-1BQ$ with $P$ ans $Q$ invertible, does we get $$B=P A Q^-1 ??$$
My problem is that $$A=P^-1BQiff Ax=P^-1BQxiff PAx= BQx$$
but I guess that it's wrong that $$PAx=BQxiff PAQ^-1x=BQQ^-1x=Bx,$$
so may be what I'm saying is not correct.
linear-algebra
asked 17 hours ago
Peter
32611
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Q1) $uin mathcal L(E,F)$ with $E$ and $V$ finite vector spaces over $K$. If $Phi:Eto K^n$ and $Psi:Fto K^p$ are isomorphisms, and if $$u=Psi^-1circ Lcirc Phi,$$
with $L:K^nto K^p$ linear. Does we get that $$Psicirc ucirc Phi^-1=L ?$$
I would say yes since $$u=Psi^-1circ Lcirc Phiiff ucirc Phi=Psi^-1circ Lcirc Phicirc Phi^-1=Psi^-1circ uiff Psicirc ucirc Phi^-1=Psicirc Psi^-1circ L=L,$$
but I have doubt with composition at right...
Q2) If $A=P^-1BQ$ with $P$ ans $Q$ invertible, does we get $$B=P A Q^-1 ??$$
My problem is that $$A=P^-1BQiff Ax=P^-1BQxiff PAx= BQx$$
but I guess that it's wrong that $$PAx=BQxiff PAQ^-1x=BQQ^-1x=Bx,$$
so may be what I'm saying is not correct.
linear-algebra
asked 17 hours ago
Peter
32611
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Q1) $uin mathcal L(E,F)$ with $E$ and $V$ finite vector spaces over $K$. If $Phi:Eto K^n$ and $Psi:Fto K^p$ are isomorphisms, and if $$u=Psi^-1circ Lcirc Phi,$$
with $L:K^nto K^p$ linear. Does we get that $$Psicirc ucirc Phi^-1=L ?$$
I would say yes since $$u=Psi^-1circ Lcirc Phiiff ucirc Phi=Psi^-1circ Lcirc Phicirc Phi^-1=Psi^-1circ uiff Psicirc ucirc Phi^-1=Psicirc Psi^-1circ L=L,$$
but I have doubt with composition at right...
Q2) If $A=P^-1BQ$ with $P$ ans $Q$ invertible, does we get $$B=P A Q^-1 ??$$
My problem is that $$A=P^-1BQiff Ax=P^-1BQxiff PAx= BQx$$
but I guess that it's wrong that $$PAx=BQxiff PAQ^-1x=BQQ^-1x=Bx,$$
so may be what I'm saying is not correct.
linear-algebra
asked 17 hours ago
Peter
32611
Q1) $uin mathcal L(E,F)$ with $E$ and $V$ finite vector spaces over $K$. If $Phi:Eto K^n$ and $Psi:Fto K^p$ are isomorphisms, and if $$u=Psi^-1circ Lcirc Phi,$$
with $L:K^nto K^p$ linear. Does we get that $$Psicirc ucirc Phi^-1=L ?$$
I would say yes since $$u=Psi^-1circ Lcirc Phiiff ucirc Phi=Psi^-1circ Lcirc Phicirc Phi^-1=Psi^-1circ uiff Psicirc ucirc Phi^-1=Psicirc Psi^-1circ L=L,$$
but I have doubt with composition at right...
Q2) If $A=P^-1BQ$ with $P$ ans $Q$ invertible, does we get $$B=P A Q^-1 ??$$
My problem is that $$A=P^-1BQiff Ax=P^-1BQxiff PAx= BQx$$
but I guess that it's wrong that $$PAx=BQxiff PAQ^-1x=BQQ^-1x=Bx,$$
so may be what I'm saying is not correct.
linear-algebra
asked 17 hours ago
Peter
32611
asked 17 hours ago
Peter
32611
asked 17 hours ago
Peter
32611
asked 17 hours ago
Peter
32611
32611
add a comment |Â
add a comment |Â
1 Answer
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Yes and yes but not as written.
For Q2), you have to set $u=Qx$ and thus you get $$PAQ^-1u=Bu.$$ But you can't compose by the right in the sense that $$Ax=BxnotRightarrow ACx=BCx.$$
For Q1) it's the same problem. You have to set $u=Phi(x)$ and then get $$Psicirc u (x)=Lcirc Phi(x)iff Phicirc ucirc Phi^-1(u)=L(u),$$
but you can't compose by the right in the sense that $$f(x)=g(x)notRightarrow f(h(x))=g(h(x)).$$
answered 17 hours ago
Surb
36.2k84173
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Yes and yes but not as written.
For Q2), you have to set $u=Qx$ and thus you get $$PAQ^-1u=Bu.$$ But you can't compose by the right in the sense that $$Ax=BxnotRightarrow ACx=BCx.$$
For Q1) it's the same problem. You have to set $u=Phi(x)$ and then get $$Psicirc u (x)=Lcirc Phi(x)iff Phicirc ucirc Phi^-1(u)=L(u),$$
but you can't compose by the right in the sense that $$f(x)=g(x)notRightarrow f(h(x))=g(h(x)).$$
answered 17 hours ago
Surb
36.2k84173
add a comment |Â
up vote
0
down vote
Yes and yes but not as written.
For Q2), you have to set $u=Qx$ and thus you get $$PAQ^-1u=Bu.$$ But you can't compose by the right in the sense that $$Ax=BxnotRightarrow ACx=BCx.$$
For Q1) it's the same problem. You have to set $u=Phi(x)$ and then get $$Psicirc u (x)=Lcirc Phi(x)iff Phicirc ucirc Phi^-1(u)=L(u),$$
but you can't compose by the right in the sense that $$f(x)=g(x)notRightarrow f(h(x))=g(h(x)).$$
answered 17 hours ago
Surb
36.2k84173
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Yes and yes but not as written.
For Q2), you have to set $u=Qx$ and thus you get $$PAQ^-1u=Bu.$$ But you can't compose by the right in the sense that $$Ax=BxnotRightarrow ACx=BCx.$$
For Q1) it's the same problem. You have to set $u=Phi(x)$ and then get $$Psicirc u (x)=Lcirc Phi(x)iff Phicirc ucirc Phi^-1(u)=L(u),$$
but you can't compose by the right in the sense that $$f(x)=g(x)notRightarrow f(h(x))=g(h(x)).$$
answered 17 hours ago
Surb
36.2k84173
Yes and yes but not as written.
For Q2), you have to set $u=Qx$ and thus you get $$PAQ^-1u=Bu.$$ But you can't compose by the right in the sense that $$Ax=BxnotRightarrow ACx=BCx.$$
For Q1) it's the same problem. You have to set $u=Phi(x)$ and then get $$Psicirc u (x)=Lcirc Phi(x)iff Phicirc ucirc Phi^-1(u)=L(u),$$
but you can't compose by the right in the sense that $$f(x)=g(x)notRightarrow f(h(x))=g(h(x)).$$
answered 17 hours ago
Surb
36.2k84173
answered 17 hours ago
Surb
36.2k84173
answered 17 hours ago
Surb
36.2k84173
answered 17 hours ago
Surb
36.2k84173
36.2k84173
add a comment |Â
add a comment |Â
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