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Understanding the formula for a diagonal matrix $A = P^-1DP$.
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I'm trying to understand the procedure for finding the powers of a matrix using the diagonal relation $A^n = P^-1D^nP$. Here's what I understand so far.
- We find eigenvalues of A. The matrix D is formed with eigenvalues in
the diagonal line and zeros everywhere else. The order of entering
diagonal values doesn't matter. - The matrix $P$ is a matrix that contains eigenvectors of $A$. Again, the order does not matter.
Is this right? I'm assuming matrix is nice (invertible etc). Am I right in thinking that the diagonal matrix itself isn't useful (i.e. doesn't give you $A^2$ unless you find $P$ and $P^-1$ too).
linear-algebra matrices
asked Jul 23 at 23:17
Nebulae
525
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up vote
1
down vote
favorite
I'm trying to understand the procedure for finding the powers of a matrix using the diagonal relation $A^n = P^-1D^nP$. Here's what I understand so far.
- We find eigenvalues of A. The matrix D is formed with eigenvalues in
the diagonal line and zeros everywhere else. The order of entering
diagonal values doesn't matter. - The matrix $P$ is a matrix that contains eigenvectors of $A$. Again, the order does not matter.
Is this right? I'm assuming matrix is nice (invertible etc). Am I right in thinking that the diagonal matrix itself isn't useful (i.e. doesn't give you $A^2$ unless you find $P$ and $P^-1$ too).
linear-algebra matrices
asked Jul 23 at 23:17
Nebulae
525
The order of the eigenvectors in $P$ has to match the order of the eigenvalues in $D$
– Omnomnomnom
Jul 23 at 23:19
@Omnomnomnom Could you elaborate please? Say I've my eigenvalue entries as $a_11 = lambda_1, a_22 = lambda_2, ldots$.
– Nebulae
Jul 23 at 23:23
You're assuming the matrix is diagonalizable, which requires that for each eigenvalue the algebraic multiplicity equals the geometric multiplicity. You can have singular matrices which are diagonalizable, e.g. the zero matrix. You cannot find $A$ with just $D$, since distinct matrices can have the same eigenvalues.
– Thoth
Jul 23 at 23:23
@Thoth Thanks. Makes sense.
– Nebulae
Jul 23 at 23:35
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to understand the procedure for finding the powers of a matrix using the diagonal relation $A^n = P^-1D^nP$. Here's what I understand so far.
- We find eigenvalues of A. The matrix D is formed with eigenvalues in
the diagonal line and zeros everywhere else. The order of entering
diagonal values doesn't matter. - The matrix $P$ is a matrix that contains eigenvectors of $A$. Again, the order does not matter.
Is this right? I'm assuming matrix is nice (invertible etc). Am I right in thinking that the diagonal matrix itself isn't useful (i.e. doesn't give you $A^2$ unless you find $P$ and $P^-1$ too).
linear-algebra matrices
asked Jul 23 at 23:17
Nebulae
525
I'm trying to understand the procedure for finding the powers of a matrix using the diagonal relation $A^n = P^-1D^nP$. Here's what I understand so far.
- We find eigenvalues of A. The matrix D is formed with eigenvalues in
the diagonal line and zeros everywhere else. The order of entering
diagonal values doesn't matter. - The matrix $P$ is a matrix that contains eigenvectors of $A$. Again, the order does not matter.
Is this right? I'm assuming matrix is nice (invertible etc). Am I right in thinking that the diagonal matrix itself isn't useful (i.e. doesn't give you $A^2$ unless you find $P$ and $P^-1$ too).
linear-algebra matrices
asked Jul 23 at 23:17
Nebulae
525
asked Jul 23 at 23:17
Nebulae
525
asked Jul 23 at 23:17
Nebulae
525
asked Jul 23 at 23:17
Nebulae
525
525
The order of the eigenvectors in $P$ has to match the order of the eigenvalues in $D$
– Omnomnomnom
Jul 23 at 23:19
@Omnomnomnom Could you elaborate please? Say I've my eigenvalue entries as $a_11 = lambda_1, a_22 = lambda_2, ldots$.
– Nebulae
Jul 23 at 23:23
You're assuming the matrix is diagonalizable, which requires that for each eigenvalue the algebraic multiplicity equals the geometric multiplicity. You can have singular matrices which are diagonalizable, e.g. the zero matrix. You cannot find $A$ with just $D$, since distinct matrices can have the same eigenvalues.
– Thoth
Jul 23 at 23:23
@Thoth Thanks. Makes sense.
– Nebulae
Jul 23 at 23:35
add a comment |Â
The order of the eigenvectors in $P$ has to match the order of the eigenvalues in $D$
– Omnomnomnom
Jul 23 at 23:19
@Omnomnomnom Could you elaborate please? Say I've my eigenvalue entries as $a_11 = lambda_1, a_22 = lambda_2, ldots$.
– Nebulae
Jul 23 at 23:23
You're assuming the matrix is diagonalizable, which requires that for each eigenvalue the algebraic multiplicity equals the geometric multiplicity. You can have singular matrices which are diagonalizable, e.g. the zero matrix. You cannot find $A$ with just $D$, since distinct matrices can have the same eigenvalues.
– Thoth
Jul 23 at 23:23
@Thoth Thanks. Makes sense.
– Nebulae
Jul 23 at 23:35
The order of the eigenvectors in $P$ has to match the order of the eigenvalues in $D$
– Omnomnomnom
Jul 23 at 23:19
The order of the eigenvectors in $P$ has to match the order of the eigenvalues in $D$
– Omnomnomnom
Jul 23 at 23:19
@Omnomnomnom Could you elaborate please? Say I've my eigenvalue entries as $a_11 = lambda_1, a_22 = lambda_2, ldots$.
– Nebulae
Jul 23 at 23:23
@Omnomnomnom Could you elaborate please? Say I've my eigenvalue entries as $a_11 = lambda_1, a_22 = lambda_2, ldots$.
– Nebulae
Jul 23 at 23:23
You're assuming the matrix is diagonalizable, which requires that for each eigenvalue the algebraic multiplicity equals the geometric multiplicity. You can have singular matrices which are diagonalizable, e.g. the zero matrix. You cannot find $A$ with just $D$, since distinct matrices can have the same eigenvalues.
– Thoth
Jul 23 at 23:23
You're assuming the matrix is diagonalizable, which requires that for each eigenvalue the algebraic multiplicity equals the geometric multiplicity. You can have singular matrices which are diagonalizable, e.g. the zero matrix. You cannot find $A$ with just $D$, since distinct matrices can have the same eigenvalues.
– Thoth
Jul 23 at 23:23
@Thoth Thanks. Makes sense.
– Nebulae
Jul 23 at 23:35
@Thoth Thanks. Makes sense.
– Nebulae
Jul 23 at 23:35
add a comment |Â
1 Answer
1
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oldest
votes
up vote
1
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accepted
Indicating the eigenvalues along the diagonal $D$ with $lambda_i=D_ii$ we need that the corresponding eigenvectors $vec v_i$ are placed as the i-th column of the matrix $P$.
Therefore the order of the eigenvalues in $D$ doesn't matter but the corresponding eigenvectors must be placed in $P$ accordingly and viceversa.
answered Jul 23 at 23:27
gimusi
65.2k73583
Thanks. When does $A^n = P^-1A^nP$? I had a question which asked me to find the diagonalisation of a matrix $M^2$ which it said can be obtained as $P^-1M^2P$. I was expecting $P^-1D^2P$, which threw me off.
– Nebulae
Jul 23 at 23:38
@Farthing Once we have the diagonalization $M=P^-1DP$ we are done indeed $M^2=P^-1DPP^-1DP=P^-1D^2P$.
– gimusi
Jul 23 at 23:44
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Indicating the eigenvalues along the diagonal $D$ with $lambda_i=D_ii$ we need that the corresponding eigenvectors $vec v_i$ are placed as the i-th column of the matrix $P$.
Therefore the order of the eigenvalues in $D$ doesn't matter but the corresponding eigenvectors must be placed in $P$ accordingly and viceversa.
answered Jul 23 at 23:27
gimusi
65.2k73583
Thanks. When does $A^n = P^-1A^nP$? I had a question which asked me to find the diagonalisation of a matrix $M^2$ which it said can be obtained as $P^-1M^2P$. I was expecting $P^-1D^2P$, which threw me off.
– Nebulae
Jul 23 at 23:38
@Farthing Once we have the diagonalization $M=P^-1DP$ we are done indeed $M^2=P^-1DPP^-1DP=P^-1D^2P$.
– gimusi
Jul 23 at 23:44
add a comment |Â
up vote
1
down vote
accepted
Indicating the eigenvalues along the diagonal $D$ with $lambda_i=D_ii$ we need that the corresponding eigenvectors $vec v_i$ are placed as the i-th column of the matrix $P$.
Therefore the order of the eigenvalues in $D$ doesn't matter but the corresponding eigenvectors must be placed in $P$ accordingly and viceversa.
answered Jul 23 at 23:27
gimusi
65.2k73583
Thanks. When does $A^n = P^-1A^nP$? I had a question which asked me to find the diagonalisation of a matrix $M^2$ which it said can be obtained as $P^-1M^2P$. I was expecting $P^-1D^2P$, which threw me off.
– Nebulae
Jul 23 at 23:38
@Farthing Once we have the diagonalization $M=P^-1DP$ we are done indeed $M^2=P^-1DPP^-1DP=P^-1D^2P$.
– gimusi
Jul 23 at 23:44
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Indicating the eigenvalues along the diagonal $D$ with $lambda_i=D_ii$ we need that the corresponding eigenvectors $vec v_i$ are placed as the i-th column of the matrix $P$.
Therefore the order of the eigenvalues in $D$ doesn't matter but the corresponding eigenvectors must be placed in $P$ accordingly and viceversa.
answered Jul 23 at 23:27
gimusi
65.2k73583
Indicating the eigenvalues along the diagonal $D$ with $lambda_i=D_ii$ we need that the corresponding eigenvectors $vec v_i$ are placed as the i-th column of the matrix $P$.
Therefore the order of the eigenvalues in $D$ doesn't matter but the corresponding eigenvectors must be placed in $P$ accordingly and viceversa.
answered Jul 23 at 23:27
gimusi
65.2k73583
answered Jul 23 at 23:27
gimusi
65.2k73583
answered Jul 23 at 23:27
gimusi
65.2k73583
answered Jul 23 at 23:27
gimusi
65.2k73583
65.2k73583
Thanks. When does $A^n = P^-1A^nP$? I had a question which asked me to find the diagonalisation of a matrix $M^2$ which it said can be obtained as $P^-1M^2P$. I was expecting $P^-1D^2P$, which threw me off.
– Nebulae
Jul 23 at 23:38
@Farthing Once we have the diagonalization $M=P^-1DP$ we are done indeed $M^2=P^-1DPP^-1DP=P^-1D^2P$.
– gimusi
Jul 23 at 23:44
add a comment |Â
Thanks. When does $A^n = P^-1A^nP$? I had a question which asked me to find the diagonalisation of a matrix $M^2$ which it said can be obtained as $P^-1M^2P$. I was expecting $P^-1D^2P$, which threw me off.
– Nebulae
Jul 23 at 23:38
@Farthing Once we have the diagonalization $M=P^-1DP$ we are done indeed $M^2=P^-1DPP^-1DP=P^-1D^2P$.
– gimusi
Jul 23 at 23:44
Thanks. When does $A^n = P^-1A^nP$? I had a question which asked me to find the diagonalisation of a matrix $M^2$ which it said can be obtained as $P^-1M^2P$. I was expecting $P^-1D^2P$, which threw me off.
– Nebulae
Jul 23 at 23:38
Thanks. When does $A^n = P^-1A^nP$? I had a question which asked me to find the diagonalisation of a matrix $M^2$ which it said can be obtained as $P^-1M^2P$. I was expecting $P^-1D^2P$, which threw me off.
– Nebulae
Jul 23 at 23:38
@Farthing Once we have the diagonalization $M=P^-1DP$ we are done indeed $M^2=P^-1DPP^-1DP=P^-1D^2P$.
– gimusi
Jul 23 at 23:44
@Farthing Once we have the diagonalization $M=P^-1DP$ we are done indeed $M^2=P^-1DPP^-1DP=P^-1D^2P$.
– gimusi
Jul 23 at 23:44
add a comment |Â
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The order of the eigenvectors in $P$ has to match the order of the eigenvalues in $D$
– Omnomnomnom
Jul 23 at 23:19
@Omnomnomnom Could you elaborate please? Say I've my eigenvalue entries as $a_11 = lambda_1, a_22 = lambda_2, ldots$.
– Nebulae
Jul 23 at 23:23
You're assuming the matrix is diagonalizable, which requires that for each eigenvalue the algebraic multiplicity equals the geometric multiplicity. You can have singular matrices which are diagonalizable, e.g. the zero matrix. You cannot find $A$ with just $D$, since distinct matrices can have the same eigenvalues.
– Thoth
Jul 23 at 23:23
@Thoth Thanks. Makes sense.
– Nebulae
Jul 23 at 23:35