Mixing
Simultaneous Diagonalisation [duplicate]
$begingroup$
This question already has an answer here:
Find a matrix that simultaneously diagonalizes to matrices
1 answer
Im stuck at this problem
Find an invertible Real Matrix $P$ such that $P^{-1}AP$ and $P^{-1}BP$ are both diagonal where $A$ and $B$ are real matrices.
a) $A=begin{bmatrix}
1&2\
0&2\
end{bmatrix}$ and
$B=begin{bmatrix}
3&-8\
0&-1\
end{bmatrix}$
b) $A=begin{bmatrix}
1&1\
1&1\
end{bmatrix}$ and $B=begin{bmatrix}
1&a\
a&1\
end{bmatrix}$
I know that diagonalisable commuting matrices can be simultaneously diagonalised but I'm not able to proceed.Kindly help.
linear-algebra diagonalization
asked Jan 1 at 10:38
SundarNarasimhanSundarNarasimhan
233
$endgroup$
marked as duplicate by Dietrich Burde, Saad, metamorphy, Xander Henderson, José Carlos Santos linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 1 at 15:38
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Find a matrix that simultaneously diagonalizes to matrices
1 answer
Im stuck at this problem
Find an invertible Real Matrix $P$ such that $P^{-1}AP$ and $P^{-1}BP$ are both diagonal where $A$ and $B$ are real matrices.
a) $A=begin{bmatrix}
1&2\
0&2\
end{bmatrix}$ and
$B=begin{bmatrix}
3&-8\
0&-1\
end{bmatrix}$
b) $A=begin{bmatrix}
1&1\
1&1\
end{bmatrix}$ and $B=begin{bmatrix}
1&a\
a&1\
end{bmatrix}$
I know that diagonalisable commuting matrices can be simultaneously diagonalised but I'm not able to proceed.Kindly help.
linear-algebra diagonalization
asked Jan 1 at 10:38
SundarNarasimhanSundarNarasimhan
233
$endgroup$
marked as duplicate by Dietrich Burde, Saad, metamorphy, Xander Henderson, José Carlos Santos linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 1 at 15:38
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Could you find common eigenvectors of $A, B$?
$endgroup$
– xbh
Jan 1 at 10:40
$begingroup$
Dietrich Burde Hi I did see that in both a and b cases the matrices commute
$endgroup$
– SundarNarasimhan
Jan 1 at 10:45
$begingroup$
Thank you xbh I was able to find common Eigenvectors of $A$ and $B$ ,thereby it becomes easy
$endgroup$
– SundarNarasimhan
Jan 1 at 11:01
add a comment |
$begingroup$
This question already has an answer here:
Find a matrix that simultaneously diagonalizes to matrices
1 answer
Im stuck at this problem
Find an invertible Real Matrix $P$ such that $P^{-1}AP$ and $P^{-1}BP$ are both diagonal where $A$ and $B$ are real matrices.
a) $A=begin{bmatrix}
1&2\
0&2\
end{bmatrix}$ and
$B=begin{bmatrix}
3&-8\
0&-1\
end{bmatrix}$
b) $A=begin{bmatrix}
1&1\
1&1\
end{bmatrix}$ and $B=begin{bmatrix}
1&a\
a&1\
end{bmatrix}$
I know that diagonalisable commuting matrices can be simultaneously diagonalised but I'm not able to proceed.Kindly help.
linear-algebra diagonalization
asked Jan 1 at 10:38
SundarNarasimhanSundarNarasimhan
233
$endgroup$
This question already has an answer here:
Find a matrix that simultaneously diagonalizes to matrices
1 answer
Im stuck at this problem
Find an invertible Real Matrix $P$ such that $P^{-1}AP$ and $P^{-1}BP$ are both diagonal where $A$ and $B$ are real matrices.
a) $A=begin{bmatrix}
1&2\
0&2\
end{bmatrix}$ and
$B=begin{bmatrix}
3&-8\
0&-1\
end{bmatrix}$
b) $A=begin{bmatrix}
1&1\
1&1\
end{bmatrix}$ and $B=begin{bmatrix}
1&a\
a&1\
end{bmatrix}$
I know that diagonalisable commuting matrices can be simultaneously diagonalised but I'm not able to proceed.Kindly help.
This question already has an answer here:
Find a matrix that simultaneously diagonalizes to matrices
1 answer
linear-algebra diagonalization
linear-algebra diagonalization
asked Jan 1 at 10:38
SundarNarasimhanSundarNarasimhan
233
asked Jan 1 at 10:38
SundarNarasimhanSundarNarasimhan
233
asked Jan 1 at 10:38
SundarNarasimhanSundarNarasimhan
233
asked Jan 1 at 10:38
SundarNarasimhanSundarNarasimhan
233
asked Jan 1 at 10:38
SundarNarasimhanSundarNarasimhan
233
233
marked as duplicate by Dietrich Burde, Saad, metamorphy, Xander Henderson, José Carlos Santos linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 1 at 15:38
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Dietrich Burde, Saad, metamorphy, Xander Henderson, José Carlos Santos linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 1 at 15:38
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Could you find common eigenvectors of $A, B$?
$endgroup$
– xbh
Jan 1 at 10:40
$begingroup$
Dietrich Burde Hi I did see that in both a and b cases the matrices commute
$endgroup$
– SundarNarasimhan
Jan 1 at 10:45
$begingroup$
Thank you xbh I was able to find common Eigenvectors of $A$ and $B$ ,thereby it becomes easy
$endgroup$
– SundarNarasimhan
Jan 1 at 11:01
add a comment |
$begingroup$
Could you find common eigenvectors of $A, B$?
$endgroup$
– xbh
Jan 1 at 10:40
$begingroup$
Dietrich Burde Hi I did see that in both a and b cases the matrices commute
$endgroup$
– SundarNarasimhan
Jan 1 at 10:45
$begingroup$
Thank you xbh I was able to find common Eigenvectors of $A$ and $B$ ,thereby it becomes easy
$endgroup$
– SundarNarasimhan
Jan 1 at 11:01
$begingroup$
Could you find common eigenvectors of $A, B$?
$endgroup$
– xbh
Jan 1 at 10:40
$begingroup$
Could you find common eigenvectors of $A, B$?
$endgroup$
– xbh
Jan 1 at 10:40
$begingroup$
Dietrich Burde Hi I did see that in both a and b cases the matrices commute
$endgroup$
– SundarNarasimhan
Jan 1 at 10:45
$begingroup$
Dietrich Burde Hi I did see that in both a and b cases the matrices commute
$endgroup$
– SundarNarasimhan
Jan 1 at 10:45
$begingroup$
Thank you xbh I was able to find common Eigenvectors of $A$ and $B$ ,thereby it becomes easy
$endgroup$
– SundarNarasimhan
Jan 1 at 11:01
$begingroup$
Thank you xbh I was able to find common Eigenvectors of $A$ and $B$ ,thereby it becomes easy
$endgroup$
– SundarNarasimhan
Jan 1 at 11:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Concerning the first pair of matrices, it turns out that the vectors $(1,0)$ and $(2,1)$ are eigenvectors of each of them. Therefore, take$$P=begin{bmatrix}1&2\0&1end{bmatrix}.$$Can you solve the other problem now?
answered Jan 1 at 10:42
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
$begingroup$
Thank you very much I was able to do the other problem also
$endgroup$
– SundarNarasimhan
Jan 1 at 11:04
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Concerning the first pair of matrices, it turns out that the vectors $(1,0)$ and $(2,1)$ are eigenvectors of each of them. Therefore, take$$P=begin{bmatrix}1&2\0&1end{bmatrix}.$$Can you solve the other problem now?
answered Jan 1 at 10:42
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
$begingroup$
Thank you very much I was able to do the other problem also
$endgroup$
– SundarNarasimhan
Jan 1 at 11:04
add a comment |
$begingroup$
Concerning the first pair of matrices, it turns out that the vectors $(1,0)$ and $(2,1)$ are eigenvectors of each of them. Therefore, take$$P=begin{bmatrix}1&2\0&1end{bmatrix}.$$Can you solve the other problem now?
answered Jan 1 at 10:42
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
$begingroup$
Thank you very much I was able to do the other problem also
$endgroup$
– SundarNarasimhan
Jan 1 at 11:04
add a comment |
$begingroup$
Concerning the first pair of matrices, it turns out that the vectors $(1,0)$ and $(2,1)$ are eigenvectors of each of them. Therefore, take$$P=begin{bmatrix}1&2\0&1end{bmatrix}.$$Can you solve the other problem now?
answered Jan 1 at 10:42
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
Concerning the first pair of matrices, it turns out that the vectors $(1,0)$ and $(2,1)$ are eigenvectors of each of them. Therefore, take$$P=begin{bmatrix}1&2\0&1end{bmatrix}.$$Can you solve the other problem now?
answered Jan 1 at 10:42
José Carlos SantosJosé Carlos Santos
153k22123225
answered Jan 1 at 10:42
José Carlos SantosJosé Carlos Santos
153k22123225
answered Jan 1 at 10:42
José Carlos SantosJosé Carlos Santos
153k22123225
answered Jan 1 at 10:42
José Carlos SantosJosé Carlos Santos
153k22123225
153k22123225
$begingroup$
Thank you very much I was able to do the other problem also
$endgroup$
– SundarNarasimhan
Jan 1 at 11:04
add a comment |
$begingroup$
Thank you very much I was able to do the other problem also
$endgroup$
– SundarNarasimhan
Jan 1 at 11:04
$begingroup$
Thank you very much I was able to do the other problem also
$endgroup$
– SundarNarasimhan
Jan 1 at 11:04
$begingroup$
Thank you very much I was able to do the other problem also
$endgroup$
– SundarNarasimhan
Jan 1 at 11:04
add a comment |
$begingroup$
Could you find common eigenvectors of $A, B$?
$endgroup$
– xbh
Jan 1 at 10:40
$begingroup$
Dietrich Burde Hi I did see that in both a and b cases the matrices commute
$endgroup$
– SundarNarasimhan
Jan 1 at 10:45
$begingroup$
Thank you xbh I was able to find common Eigenvectors of $A$ and $B$ ,thereby it becomes easy
$endgroup$
– SundarNarasimhan
Jan 1 at 11:01