How to rewrite matrix formula for Diagonalizable matrix $A=PDP^{-1}$
$begingroup$
I am working on an old exam containing a question about Diagonalizable matrix, I am quite confident about the subject overall but there is one simple thing that bothers me, a lot!
We are given the formula $A=PDP^{-1}$ I know from my memory that this can be rewritten as $D=P^{-1}AP$ to solve for D, but I cannot find out how to do it. I have watched a few examples on Wikipedia and in some PDF file that hinted me in the correct direction.
What I believe,
For example:
$A=PD$ can be rewritten as $AP^{-1}=D$ if this was regular variables, but they are not, they are matrices, and with matrices, you can only put the "latest" number in front of the equation like this:
$A=PD$ can be rewritten as $P^{-1}A=P^{-1}PD=D$ this works fine but with the formula $A=PDP^{-1}$ I get stuck in an infinite loop of moving things in front of the equation and everything becomes a mess. There is clearly something easy I have missed out. Here are my calculations:
$A=PDP^{-1}$
$P^{-1}A=P^{-1}PDP^{-1}$
$P^{-1}A=DP^{-1}$
Now I want to get rid of $P^{-1}$ from the right-hand side but since I can only but things in front of D, everything gets messy and I get stuck in a never-ending loop of making things more complicated and adding $PP^{-1}=I$ to the equation hoping it would help but it doesn't :(
linear-algebra matrices matrix-equations diagonalization
edited Jan 10 at 10:03
Bernard
122k740116
asked Jan 10 at 9:59
J. DoeJ. Doe
627
$endgroup$
add a comment |
$begingroup$
I am working on an old exam containing a question about Diagonalizable matrix, I am quite confident about the subject overall but there is one simple thing that bothers me, a lot!
We are given the formula $A=PDP^{-1}$ I know from my memory that this can be rewritten as $D=P^{-1}AP$ to solve for D, but I cannot find out how to do it. I have watched a few examples on Wikipedia and in some PDF file that hinted me in the correct direction.
What I believe,
For example:
$A=PD$ can be rewritten as $AP^{-1}=D$ if this was regular variables, but they are not, they are matrices, and with matrices, you can only put the "latest" number in front of the equation like this:
$A=PD$ can be rewritten as $P^{-1}A=P^{-1}PD=D$ this works fine but with the formula $A=PDP^{-1}$ I get stuck in an infinite loop of moving things in front of the equation and everything becomes a mess. There is clearly something easy I have missed out. Here are my calculations:
$A=PDP^{-1}$
$P^{-1}A=P^{-1}PDP^{-1}$
$P^{-1}A=DP^{-1}$
Now I want to get rid of $P^{-1}$ from the right-hand side but since I can only but things in front of D, everything gets messy and I get stuck in a never-ending loop of making things more complicated and adding $PP^{-1}=I$ to the equation hoping it would help but it doesn't :(
linear-algebra matrices matrix-equations diagonalization
edited Jan 10 at 10:03
Bernard
122k740116
asked Jan 10 at 9:59
J. DoeJ. Doe
627
$endgroup$
1
$begingroup$
Right-multiply the equation $P^{-1}A=DP^{-1}$ by $P$,$$P^{-1}AP=DP^{-1}P=D$$Since matrix products don't commute in general, you can either pre-multiply or left-multiply a matrix $X$ with another matrix $Y$, as in $XY$, or you can do post-multiplication or right-multiplication, as in $YX$.
$endgroup$
– Shubham Johri
Jan 10 at 10:02
$begingroup$
So you actually are looking for $P$ which will transform $A$ into a diagonal form, right?
$endgroup$
– denklo
Jan 10 at 10:07
$begingroup$
@denklo Looking for D, but exact values are not needed, I know how to do that, I do not know how to rewrite matrix equations
$endgroup$
– J. Doe
Jan 10 at 10:10
$begingroup$
@ShubhamJohri Okay so the trick is that I can multiply in front as I have done, or at the end of the equation?
$endgroup$
– J. Doe
Jan 10 at 10:11
1
$begingroup$
That's correct, @J.Doe.
$endgroup$
– Shubham Johri
Jan 10 at 10:13
add a comment |
$begingroup$
I am working on an old exam containing a question about Diagonalizable matrix, I am quite confident about the subject overall but there is one simple thing that bothers me, a lot!
We are given the formula $A=PDP^{-1}$ I know from my memory that this can be rewritten as $D=P^{-1}AP$ to solve for D, but I cannot find out how to do it. I have watched a few examples on Wikipedia and in some PDF file that hinted me in the correct direction.
What I believe,
For example:
$A=PD$ can be rewritten as $AP^{-1}=D$ if this was regular variables, but they are not, they are matrices, and with matrices, you can only put the "latest" number in front of the equation like this:
$A=PD$ can be rewritten as $P^{-1}A=P^{-1}PD=D$ this works fine but with the formula $A=PDP^{-1}$ I get stuck in an infinite loop of moving things in front of the equation and everything becomes a mess. There is clearly something easy I have missed out. Here are my calculations:
$A=PDP^{-1}$
$P^{-1}A=P^{-1}PDP^{-1}$
$P^{-1}A=DP^{-1}$
Now I want to get rid of $P^{-1}$ from the right-hand side but since I can only but things in front of D, everything gets messy and I get stuck in a never-ending loop of making things more complicated and adding $PP^{-1}=I$ to the equation hoping it would help but it doesn't :(
linear-algebra matrices matrix-equations diagonalization
edited Jan 10 at 10:03
Bernard
122k740116
asked Jan 10 at 9:59
J. DoeJ. Doe
627
$endgroup$
I am working on an old exam containing a question about Diagonalizable matrix, I am quite confident about the subject overall but there is one simple thing that bothers me, a lot!
We are given the formula $A=PDP^{-1}$ I know from my memory that this can be rewritten as $D=P^{-1}AP$ to solve for D, but I cannot find out how to do it. I have watched a few examples on Wikipedia and in some PDF file that hinted me in the correct direction.
What I believe,
For example:
$A=PD$ can be rewritten as $AP^{-1}=D$ if this was regular variables, but they are not, they are matrices, and with matrices, you can only put the "latest" number in front of the equation like this:
$A=PD$ can be rewritten as $P^{-1}A=P^{-1}PD=D$ this works fine but with the formula $A=PDP^{-1}$ I get stuck in an infinite loop of moving things in front of the equation and everything becomes a mess. There is clearly something easy I have missed out. Here are my calculations:
$A=PDP^{-1}$
$P^{-1}A=P^{-1}PDP^{-1}$
$P^{-1}A=DP^{-1}$
Now I want to get rid of $P^{-1}$ from the right-hand side but since I can only but things in front of D, everything gets messy and I get stuck in a never-ending loop of making things more complicated and adding $PP^{-1}=I$ to the equation hoping it would help but it doesn't :(
linear-algebra matrices matrix-equations diagonalization
linear-algebra matrices matrix-equations diagonalization
edited Jan 10 at 10:03
Bernard
122k740116
asked Jan 10 at 9:59
J. DoeJ. Doe
627
edited Jan 10 at 10:03
Bernard
122k740116
asked Jan 10 at 9:59
J. DoeJ. Doe
627
edited Jan 10 at 10:03
Bernard
122k740116
edited Jan 10 at 10:03
Bernard
122k740116
edited Jan 10 at 10:03
Bernard
122k740116
122k740116
asked Jan 10 at 9:59
J. DoeJ. Doe
627
asked Jan 10 at 9:59
J. DoeJ. Doe
627
asked Jan 10 at 9:59
J. DoeJ. Doe
627
627
1
$begingroup$
Right-multiply the equation $P^{-1}A=DP^{-1}$ by $P$,$$P^{-1}AP=DP^{-1}P=D$$Since matrix products don't commute in general, you can either pre-multiply or left-multiply a matrix $X$ with another matrix $Y$, as in $XY$, or you can do post-multiplication or right-multiplication, as in $YX$.
$endgroup$
– Shubham Johri
Jan 10 at 10:02
$begingroup$
So you actually are looking for $P$ which will transform $A$ into a diagonal form, right?
$endgroup$
– denklo
Jan 10 at 10:07
$begingroup$
@denklo Looking for D, but exact values are not needed, I know how to do that, I do not know how to rewrite matrix equations
$endgroup$
– J. Doe
Jan 10 at 10:10
$begingroup$
@ShubhamJohri Okay so the trick is that I can multiply in front as I have done, or at the end of the equation?
$endgroup$
– J. Doe
Jan 10 at 10:11
1
$begingroup$
That's correct, @J.Doe.
$endgroup$
– Shubham Johri
Jan 10 at 10:13
add a comment |
1
$begingroup$
Right-multiply the equation $P^{-1}A=DP^{-1}$ by $P$,$$P^{-1}AP=DP^{-1}P=D$$Since matrix products don't commute in general, you can either pre-multiply or left-multiply a matrix $X$ with another matrix $Y$, as in $XY$, or you can do post-multiplication or right-multiplication, as in $YX$.
$endgroup$
– Shubham Johri
Jan 10 at 10:02
$begingroup$
So you actually are looking for $P$ which will transform $A$ into a diagonal form, right?
$endgroup$
– denklo
Jan 10 at 10:07
$begingroup$
@denklo Looking for D, but exact values are not needed, I know how to do that, I do not know how to rewrite matrix equations
$endgroup$
– J. Doe
Jan 10 at 10:10
$begingroup$
@ShubhamJohri Okay so the trick is that I can multiply in front as I have done, or at the end of the equation?
$endgroup$
– J. Doe
Jan 10 at 10:11
1
$begingroup$
That's correct, @J.Doe.
$endgroup$
– Shubham Johri
Jan 10 at 10:13
1
1
$begingroup$
Right-multiply the equation $P^{-1}A=DP^{-1}$ by $P$,$$P^{-1}AP=DP^{-1}P=D$$Since matrix products don't commute in general, you can either pre-multiply or left-multiply a matrix $X$ with another matrix $Y$, as in $XY$, or you can do post-multiplication or right-multiplication, as in $YX$.
$endgroup$
– Shubham Johri
Jan 10 at 10:02
$begingroup$
Right-multiply the equation $P^{-1}A=DP^{-1}$ by $P$,$$P^{-1}AP=DP^{-1}P=D$$Since matrix products don't commute in general, you can either pre-multiply or left-multiply a matrix $X$ with another matrix $Y$, as in $XY$, or you can do post-multiplication or right-multiplication, as in $YX$.
$endgroup$
– Shubham Johri
Jan 10 at 10:02
$begingroup$
So you actually are looking for $P$ which will transform $A$ into a diagonal form, right?
$endgroup$
– denklo
Jan 10 at 10:07
$begingroup$
So you actually are looking for $P$ which will transform $A$ into a diagonal form, right?
$endgroup$
– denklo
Jan 10 at 10:07
$begingroup$
@denklo Looking for D, but exact values are not needed, I know how to do that, I do not know how to rewrite matrix equations
$endgroup$
– J. Doe
Jan 10 at 10:10
$begingroup$
@denklo Looking for D, but exact values are not needed, I know how to do that, I do not know how to rewrite matrix equations
$endgroup$
– J. Doe
Jan 10 at 10:10
$begingroup$
@ShubhamJohri Okay so the trick is that I can multiply in front as I have done, or at the end of the equation?
$endgroup$
– J. Doe
Jan 10 at 10:11
$begingroup$
@ShubhamJohri Okay so the trick is that I can multiply in front as I have done, or at the end of the equation?
$endgroup$
– J. Doe
Jan 10 at 10:11
1
1
$begingroup$
That's correct, @J.Doe.
$endgroup$
– Shubham Johri
Jan 10 at 10:13
$begingroup$
That's correct, @J.Doe.
$endgroup$
– Shubham Johri
Jan 10 at 10:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You just have to take into account that matrix multiplication is not commutative.
So from $A=PDP^{-1}$, just multiply with $P^{-1}$ on the left, and with $P$ on the right. As matrix multiplication is associative, you obtain
$$P^{-1}AP=P^{-1}(PDP^{-1})P=(P^{-1}P)D(P^{-1}P)=D.$$
answered Jan 10 at 10:10
BernardBernard
122k740116
$endgroup$
add a comment |
$begingroup$
I know from my memory that this can be rewritten as $D=P^{-1}AP$ to solve for D, but I cannot find out how to do it.
We start with the identity $A=PDP^{-1}$.
Begin by multiplying both sides on the left by $P^{-1}$ and then on the right by $P$, then make use of the properties that
- For any invertible matrix $A$, $AA^{-1} = A^{-1}A = I$.
- For all square matrices $A$, we have $AI=IA=A$.
- Matrix multiplication is associative, i.e. $ABC = (AB)C = A(BC)$, where the multiplicative is assumed to be defined for the matrices $A,B,C$.
So, we obtain,
$$begin{align}
A=PDP^{-1} &iff P^{-1}AP=P^{-1}PDP^{-1}P \
&iff P^{-1}AP=(P^{-1}P)D(P^{-1}P) = IDI = D
end{align}$$
Thus,
$$P^{-1}AP = D$$
answered Jan 10 at 10:10
Eevee TrainerEevee Trainer
7,03321337
$endgroup$
$begingroup$
Nice thing about this result is that it's not exclusive to matrices at all. As invertible matrices form a Abelian Group over addition and Group over multiplication we see this results holds for all structures that have those properties!
$endgroup$
– DavidG
Jan 12 at 9:19
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You just have to take into account that matrix multiplication is not commutative.
So from $A=PDP^{-1}$, just multiply with $P^{-1}$ on the left, and with $P$ on the right. As matrix multiplication is associative, you obtain
$$P^{-1}AP=P^{-1}(PDP^{-1})P=(P^{-1}P)D(P^{-1}P)=D.$$
answered Jan 10 at 10:10
BernardBernard
122k740116
$endgroup$
add a comment |
$begingroup$
You just have to take into account that matrix multiplication is not commutative.
So from $A=PDP^{-1}$, just multiply with $P^{-1}$ on the left, and with $P$ on the right. As matrix multiplication is associative, you obtain
$$P^{-1}AP=P^{-1}(PDP^{-1})P=(P^{-1}P)D(P^{-1}P)=D.$$
answered Jan 10 at 10:10
BernardBernard
122k740116
$endgroup$
add a comment |
$begingroup$
You just have to take into account that matrix multiplication is not commutative.
So from $A=PDP^{-1}$, just multiply with $P^{-1}$ on the left, and with $P$ on the right. As matrix multiplication is associative, you obtain
$$P^{-1}AP=P^{-1}(PDP^{-1})P=(P^{-1}P)D(P^{-1}P)=D.$$
answered Jan 10 at 10:10
BernardBernard
122k740116
$endgroup$
You just have to take into account that matrix multiplication is not commutative.
So from $A=PDP^{-1}$, just multiply with $P^{-1}$ on the left, and with $P$ on the right. As matrix multiplication is associative, you obtain
$$P^{-1}AP=P^{-1}(PDP^{-1})P=(P^{-1}P)D(P^{-1}P)=D.$$
answered Jan 10 at 10:10
BernardBernard
122k740116
answered Jan 10 at 10:10
BernardBernard
122k740116
answered Jan 10 at 10:10
BernardBernard
122k740116
answered Jan 10 at 10:10
BernardBernard
122k740116
122k740116
add a comment |
add a comment |
$begingroup$
I know from my memory that this can be rewritten as $D=P^{-1}AP$ to solve for D, but I cannot find out how to do it.
We start with the identity $A=PDP^{-1}$.
Begin by multiplying both sides on the left by $P^{-1}$ and then on the right by $P$, then make use of the properties that
- For any invertible matrix $A$, $AA^{-1} = A^{-1}A = I$.
- For all square matrices $A$, we have $AI=IA=A$.
- Matrix multiplication is associative, i.e. $ABC = (AB)C = A(BC)$, where the multiplicative is assumed to be defined for the matrices $A,B,C$.
So, we obtain,
$$begin{align}
A=PDP^{-1} &iff P^{-1}AP=P^{-1}PDP^{-1}P \
&iff P^{-1}AP=(P^{-1}P)D(P^{-1}P) = IDI = D
end{align}$$
Thus,
$$P^{-1}AP = D$$
answered Jan 10 at 10:10
Eevee TrainerEevee Trainer
7,03321337
$endgroup$
$begingroup$
Nice thing about this result is that it's not exclusive to matrices at all. As invertible matrices form a Abelian Group over addition and Group over multiplication we see this results holds for all structures that have those properties!
$endgroup$
– DavidG
Jan 12 at 9:19
add a comment |
$begingroup$
I know from my memory that this can be rewritten as $D=P^{-1}AP$ to solve for D, but I cannot find out how to do it.
We start with the identity $A=PDP^{-1}$.
Begin by multiplying both sides on the left by $P^{-1}$ and then on the right by $P$, then make use of the properties that
- For any invertible matrix $A$, $AA^{-1} = A^{-1}A = I$.
- For all square matrices $A$, we have $AI=IA=A$.
- Matrix multiplication is associative, i.e. $ABC = (AB)C = A(BC)$, where the multiplicative is assumed to be defined for the matrices $A,B,C$.
So, we obtain,
$$begin{align}
A=PDP^{-1} &iff P^{-1}AP=P^{-1}PDP^{-1}P \
&iff P^{-1}AP=(P^{-1}P)D(P^{-1}P) = IDI = D
end{align}$$
Thus,
$$P^{-1}AP = D$$
answered Jan 10 at 10:10
Eevee TrainerEevee Trainer
7,03321337
$endgroup$
$begingroup$
Nice thing about this result is that it's not exclusive to matrices at all. As invertible matrices form a Abelian Group over addition and Group over multiplication we see this results holds for all structures that have those properties!
$endgroup$
– DavidG
Jan 12 at 9:19
add a comment |
$begingroup$
I know from my memory that this can be rewritten as $D=P^{-1}AP$ to solve for D, but I cannot find out how to do it.
We start with the identity $A=PDP^{-1}$.
Begin by multiplying both sides on the left by $P^{-1}$ and then on the right by $P$, then make use of the properties that
- For any invertible matrix $A$, $AA^{-1} = A^{-1}A = I$.
- For all square matrices $A$, we have $AI=IA=A$.
- Matrix multiplication is associative, i.e. $ABC = (AB)C = A(BC)$, where the multiplicative is assumed to be defined for the matrices $A,B,C$.
So, we obtain,
$$begin{align}
A=PDP^{-1} &iff P^{-1}AP=P^{-1}PDP^{-1}P \
&iff P^{-1}AP=(P^{-1}P)D(P^{-1}P) = IDI = D
end{align}$$
Thus,
$$P^{-1}AP = D$$
answered Jan 10 at 10:10
Eevee TrainerEevee Trainer
7,03321337
$endgroup$
I know from my memory that this can be rewritten as $D=P^{-1}AP$ to solve for D, but I cannot find out how to do it.
We start with the identity $A=PDP^{-1}$.
Begin by multiplying both sides on the left by $P^{-1}$ and then on the right by $P$, then make use of the properties that
- For any invertible matrix $A$, $AA^{-1} = A^{-1}A = I$.
- For all square matrices $A$, we have $AI=IA=A$.
- Matrix multiplication is associative, i.e. $ABC = (AB)C = A(BC)$, where the multiplicative is assumed to be defined for the matrices $A,B,C$.
So, we obtain,
$$begin{align}
A=PDP^{-1} &iff P^{-1}AP=P^{-1}PDP^{-1}P \
&iff P^{-1}AP=(P^{-1}P)D(P^{-1}P) = IDI = D
end{align}$$
Thus,
$$P^{-1}AP = D$$
answered Jan 10 at 10:10
Eevee TrainerEevee Trainer
7,03321337
answered Jan 10 at 10:10
Eevee TrainerEevee Trainer
7,03321337
answered Jan 10 at 10:10
Eevee TrainerEevee Trainer
7,03321337
answered Jan 10 at 10:10
Eevee TrainerEevee Trainer
7,03321337
7,03321337
$begingroup$
Nice thing about this result is that it's not exclusive to matrices at all. As invertible matrices form a Abelian Group over addition and Group over multiplication we see this results holds for all structures that have those properties!
$endgroup$
– DavidG
Jan 12 at 9:19
add a comment |
$begingroup$
Nice thing about this result is that it's not exclusive to matrices at all. As invertible matrices form a Abelian Group over addition and Group over multiplication we see this results holds for all structures that have those properties!
$endgroup$
– DavidG
Jan 12 at 9:19
$begingroup$
Nice thing about this result is that it's not exclusive to matrices at all. As invertible matrices form a Abelian Group over addition and Group over multiplication we see this results holds for all structures that have those properties!
$endgroup$
– DavidG
Jan 12 at 9:19
$begingroup$
Nice thing about this result is that it's not exclusive to matrices at all. As invertible matrices form a Abelian Group over addition and Group over multiplication we see this results holds for all structures that have those properties!
$endgroup$
– DavidG
Jan 12 at 9:19
add a comment |
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1
$begingroup$
Right-multiply the equation $P^{-1}A=DP^{-1}$ by $P$,$$P^{-1}AP=DP^{-1}P=D$$Since matrix products don't commute in general, you can either pre-multiply or left-multiply a matrix $X$ with another matrix $Y$, as in $XY$, or you can do post-multiplication or right-multiplication, as in $YX$.
$endgroup$
– Shubham Johri
Jan 10 at 10:02
$begingroup$
So you actually are looking for $P$ which will transform $A$ into a diagonal form, right?
$endgroup$
– denklo
Jan 10 at 10:07
$begingroup$
@denklo Looking for D, but exact values are not needed, I know how to do that, I do not know how to rewrite matrix equations
$endgroup$
– J. Doe
Jan 10 at 10:10
$begingroup$
@ShubhamJohri Okay so the trick is that I can multiply in front as I have done, or at the end of the equation?
$endgroup$
– J. Doe
Jan 10 at 10:11
1
$begingroup$
That's correct, @J.Doe.
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– Shubham Johri
Jan 10 at 10:13