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Find $A^n , n ge 1$
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Let $u = (1, 2, 3)$ and $v = (1, frac12,frac13)$ · Let $A= u^tv.$ Find $A^n , n ge 1$
My attempts :$ A = u^tv$ = $beginbmatrix 1 \ 2 \ 3 endbmatrix[ 1 frac12 frac13 ]$.
$A = beginbmatrix 1 &1& frac23 \ 2& 1& frac23\3& frac32 & 1 endbmatrix$.
i don't know How to find $A^n $?
linear-algebra
edited Jul 26 at 18:20
mvw
30.2k22250
asked Jul 26 at 17:39
stupid
53518
add a comment |Â
up vote
1
down vote
favorite
Let $u = (1, 2, 3)$ and $v = (1, frac12,frac13)$ · Let $A= u^tv.$ Find $A^n , n ge 1$
My attempts :$ A = u^tv$ = $beginbmatrix 1 \ 2 \ 3 endbmatrix[ 1 frac12 frac13 ]$.
$A = beginbmatrix 1 &1& frac23 \ 2& 1& frac23\3& frac32 & 1 endbmatrix$.
i don't know How to find $A^n $?
linear-algebra
edited Jul 26 at 18:20
mvw
30.2k22250
asked Jul 26 at 17:39
stupid
53518
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $u = (1, 2, 3)$ and $v = (1, frac12,frac13)$ · Let $A= u^tv.$ Find $A^n , n ge 1$
My attempts :$ A = u^tv$ = $beginbmatrix 1 \ 2 \ 3 endbmatrix[ 1 frac12 frac13 ]$.
$A = beginbmatrix 1 &1& frac23 \ 2& 1& frac23\3& frac32 & 1 endbmatrix$.
i don't know How to find $A^n $?
linear-algebra
edited Jul 26 at 18:20
mvw
30.2k22250
asked Jul 26 at 17:39
stupid
53518
Let $u = (1, 2, 3)$ and $v = (1, frac12,frac13)$ · Let $A= u^tv.$ Find $A^n , n ge 1$
My attempts :$ A = u^tv$ = $beginbmatrix 1 \ 2 \ 3 endbmatrix[ 1 frac12 frac13 ]$.
$A = beginbmatrix 1 &1& frac23 \ 2& 1& frac23\3& frac32 & 1 endbmatrix$.
i don't know How to find $A^n $?
linear-algebra
edited Jul 26 at 18:20
mvw
30.2k22250
asked Jul 26 at 17:39
stupid
53518
edited Jul 26 at 18:20
mvw
30.2k22250
edited Jul 26 at 18:20
mvw
30.2k22250
edited Jul 26 at 18:20
mvw
30.2k22250
30.2k22250
asked Jul 26 at 17:39
stupid
53518
asked Jul 26 at 17:39
stupid
53518
asked Jul 26 at 17:39
stupid
53518
53518
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
6
down vote
accepted
$A^2 = u^tvu^tv= u^t (vu^t)v\
A^n = u^tvu^tv= u^t (vu^t)^n-1v\
(vu^t) = 3\
A^n = 3^n-1 A$
answered Jul 26 at 17:57
Doug M
39k31749
superb,,,,,can u tell me,, how this thinking came in ur mind ??
– stupid
Jul 26 at 20:35
1
Always difficult to say exactly where an idea comes from.... Diagonlization...This is a standard bit of logic. $A = PDP^-1, A^2 = PDP^-1PDP^-1 = PD^2P^-1$ by induction $A^n = PD^nP^-1.$ With that in the back of my mind, it seemed that it was unnecessary to go to the step of actually diagonlizing, and proceed to $AA$ and see what simplified.
– Doug M
Jul 26 at 21:46
okks thanks u @Doug M
– stupid
Jul 27 at 1:50
add a comment |Â
up vote
2
down vote
The value for $A=u^tv$ you got is not correct.
The correct value is
$$A=u^tv=beginbmatrix 1 \ 2 \ 3 endbmatrixbeginbmatrix 1 & dfrac 1 2 & dfrac 1 3 endbmatrix=beginbmatrix 1 & dfrac 1 2 & dfrac 1 3 \ 2 & 1 & dfrac 2 3 \ 3 & dfrac 3 2 & 1 endbmatrix$$
Now you could use diagonalization.
Let $A$ be a matrix then if you diagonalize $A$ then you get $A=PDP^-1$ with $B$ as a diagonal matrix and $P^-1$ an invertible matrix. If you try for $A^2$, then it is equal to $A^2=PD^2P^-1$. Similarly for $A^n=PD^nP^-1$
$1.$Find the characteristic polynomial $p(t)$ of $A$.
$2.$Find eigenvalues $λ$ of the matrix $A$ and their algebraic multiplicities from the characteristic polynomial $p(t)$.
$3.$For each eigenvalue $λ$ of $A$, find a basis of the eigenspace $E_λ$.
If there is an eigenvalue $λ$ such that the geometric multiplicity of $λ$, dim$(E_λ)$, is less than the algebraic multiplicity of $λ$, then the matrix $A$ is not diagonalizable. If not, $A$ is diagonalizable, and proceed to the next step.
$4.$If we combine all basis vectors for all eigenspaces, we obtained n linearly independent eigenvectors $v1,v2,…,vn$.
$5.$Define the nonsingular matrix $P=[v_1v_2…v_n]$.
$6.$Define the diagonal matrix $D$, whose $(i,i)$-entry is the eigenvalue λ such that the $i^th$ column vector $v_i$ is in the eigenspace $E_λ$
$7.$Then the matrix $A$ is diagonalized as $P^-1AP=D$
answered Jul 26 at 17:42
Key Flex
4,047423
@ Key Flexhow can i find the VAlue of P and B ? can elaborate that more
– stupid
Jul 26 at 17:43
1
@stupid I added step by step procedure how to find it.
– Key Flex
Jul 26 at 17:49
The matrix $A$ has rank 1, so two of the eigenvalues are automatically equal to 0. The other eigenvalue turns out to be 3.
– Joe Silverman
Jul 26 at 17:54
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
$A^2 = u^tvu^tv= u^t (vu^t)v\
A^n = u^tvu^tv= u^t (vu^t)^n-1v\
(vu^t) = 3\
A^n = 3^n-1 A$
answered Jul 26 at 17:57
Doug M
39k31749
superb,,,,,can u tell me,, how this thinking came in ur mind ??
– stupid
Jul 26 at 20:35
1
Always difficult to say exactly where an idea comes from.... Diagonlization...This is a standard bit of logic. $A = PDP^-1, A^2 = PDP^-1PDP^-1 = PD^2P^-1$ by induction $A^n = PD^nP^-1.$ With that in the back of my mind, it seemed that it was unnecessary to go to the step of actually diagonlizing, and proceed to $AA$ and see what simplified.
– Doug M
Jul 26 at 21:46
okks thanks u @Doug M
– stupid
Jul 27 at 1:50
add a comment |Â
up vote
6
down vote
accepted
$A^2 = u^tvu^tv= u^t (vu^t)v\
A^n = u^tvu^tv= u^t (vu^t)^n-1v\
(vu^t) = 3\
A^n = 3^n-1 A$
answered Jul 26 at 17:57
Doug M
39k31749
superb,,,,,can u tell me,, how this thinking came in ur mind ??
– stupid
Jul 26 at 20:35
1
Always difficult to say exactly where an idea comes from.... Diagonlization...This is a standard bit of logic. $A = PDP^-1, A^2 = PDP^-1PDP^-1 = PD^2P^-1$ by induction $A^n = PD^nP^-1.$ With that in the back of my mind, it seemed that it was unnecessary to go to the step of actually diagonlizing, and proceed to $AA$ and see what simplified.
– Doug M
Jul 26 at 21:46
okks thanks u @Doug M
– stupid
Jul 27 at 1:50
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
$A^2 = u^tvu^tv= u^t (vu^t)v\
A^n = u^tvu^tv= u^t (vu^t)^n-1v\
(vu^t) = 3\
A^n = 3^n-1 A$
answered Jul 26 at 17:57
Doug M
39k31749
$A^2 = u^tvu^tv= u^t (vu^t)v\
A^n = u^tvu^tv= u^t (vu^t)^n-1v\
(vu^t) = 3\
A^n = 3^n-1 A$
answered Jul 26 at 17:57
Doug M
39k31749
answered Jul 26 at 17:57
Doug M
39k31749
answered Jul 26 at 17:57
Doug M
39k31749
answered Jul 26 at 17:57
Doug M
39k31749
39k31749
superb,,,,,can u tell me,, how this thinking came in ur mind ??
– stupid
Jul 26 at 20:35
1
Always difficult to say exactly where an idea comes from.... Diagonlization...This is a standard bit of logic. $A = PDP^-1, A^2 = PDP^-1PDP^-1 = PD^2P^-1$ by induction $A^n = PD^nP^-1.$ With that in the back of my mind, it seemed that it was unnecessary to go to the step of actually diagonlizing, and proceed to $AA$ and see what simplified.
– Doug M
Jul 26 at 21:46
okks thanks u @Doug M
– stupid
Jul 27 at 1:50
add a comment |Â
superb,,,,,can u tell me,, how this thinking came in ur mind ??
– stupid
Jul 26 at 20:35
1
Always difficult to say exactly where an idea comes from.... Diagonlization...This is a standard bit of logic. $A = PDP^-1, A^2 = PDP^-1PDP^-1 = PD^2P^-1$ by induction $A^n = PD^nP^-1.$ With that in the back of my mind, it seemed that it was unnecessary to go to the step of actually diagonlizing, and proceed to $AA$ and see what simplified.
– Doug M
Jul 26 at 21:46
okks thanks u @Doug M
– stupid
Jul 27 at 1:50
superb,,,,,can u tell me,, how this thinking came in ur mind ??
– stupid
Jul 26 at 20:35
superb,,,,,can u tell me,, how this thinking came in ur mind ??
– stupid
Jul 26 at 20:35
1
1
Always difficult to say exactly where an idea comes from.... Diagonlization...This is a standard bit of logic. $A = PDP^-1, A^2 = PDP^-1PDP^-1 = PD^2P^-1$ by induction $A^n = PD^nP^-1.$ With that in the back of my mind, it seemed that it was unnecessary to go to the step of actually diagonlizing, and proceed to $AA$ and see what simplified.
– Doug M
Jul 26 at 21:46
Always difficult to say exactly where an idea comes from.... Diagonlization...This is a standard bit of logic. $A = PDP^-1, A^2 = PDP^-1PDP^-1 = PD^2P^-1$ by induction $A^n = PD^nP^-1.$ With that in the back of my mind, it seemed that it was unnecessary to go to the step of actually diagonlizing, and proceed to $AA$ and see what simplified.
– Doug M
Jul 26 at 21:46
okks thanks u @Doug M
– stupid
Jul 27 at 1:50
okks thanks u @Doug M
– stupid
Jul 27 at 1:50
add a comment |Â
up vote
2
down vote
The value for $A=u^tv$ you got is not correct.
The correct value is
$$A=u^tv=beginbmatrix 1 \ 2 \ 3 endbmatrixbeginbmatrix 1 & dfrac 1 2 & dfrac 1 3 endbmatrix=beginbmatrix 1 & dfrac 1 2 & dfrac 1 3 \ 2 & 1 & dfrac 2 3 \ 3 & dfrac 3 2 & 1 endbmatrix$$
Now you could use diagonalization.
Let $A$ be a matrix then if you diagonalize $A$ then you get $A=PDP^-1$ with $B$ as a diagonal matrix and $P^-1$ an invertible matrix. If you try for $A^2$, then it is equal to $A^2=PD^2P^-1$. Similarly for $A^n=PD^nP^-1$
$1.$Find the characteristic polynomial $p(t)$ of $A$.
$2.$Find eigenvalues $λ$ of the matrix $A$ and their algebraic multiplicities from the characteristic polynomial $p(t)$.
$3.$For each eigenvalue $λ$ of $A$, find a basis of the eigenspace $E_λ$.
If there is an eigenvalue $λ$ such that the geometric multiplicity of $λ$, dim$(E_λ)$, is less than the algebraic multiplicity of $λ$, then the matrix $A$ is not diagonalizable. If not, $A$ is diagonalizable, and proceed to the next step.
$4.$If we combine all basis vectors for all eigenspaces, we obtained n linearly independent eigenvectors $v1,v2,…,vn$.
$5.$Define the nonsingular matrix $P=[v_1v_2…v_n]$.
$6.$Define the diagonal matrix $D$, whose $(i,i)$-entry is the eigenvalue λ such that the $i^th$ column vector $v_i$ is in the eigenspace $E_λ$
$7.$Then the matrix $A$ is diagonalized as $P^-1AP=D$
answered Jul 26 at 17:42
Key Flex
4,047423
@ Key Flexhow can i find the VAlue of P and B ? can elaborate that more
– stupid
Jul 26 at 17:43
1
@stupid I added step by step procedure how to find it.
– Key Flex
Jul 26 at 17:49
The matrix $A$ has rank 1, so two of the eigenvalues are automatically equal to 0. The other eigenvalue turns out to be 3.
– Joe Silverman
Jul 26 at 17:54
add a comment |Â
up vote
2
down vote
The value for $A=u^tv$ you got is not correct.
The correct value is
$$A=u^tv=beginbmatrix 1 \ 2 \ 3 endbmatrixbeginbmatrix 1 & dfrac 1 2 & dfrac 1 3 endbmatrix=beginbmatrix 1 & dfrac 1 2 & dfrac 1 3 \ 2 & 1 & dfrac 2 3 \ 3 & dfrac 3 2 & 1 endbmatrix$$
Now you could use diagonalization.
Let $A$ be a matrix then if you diagonalize $A$ then you get $A=PDP^-1$ with $B$ as a diagonal matrix and $P^-1$ an invertible matrix. If you try for $A^2$, then it is equal to $A^2=PD^2P^-1$. Similarly for $A^n=PD^nP^-1$
$1.$Find the characteristic polynomial $p(t)$ of $A$.
$2.$Find eigenvalues $λ$ of the matrix $A$ and their algebraic multiplicities from the characteristic polynomial $p(t)$.
$3.$For each eigenvalue $λ$ of $A$, find a basis of the eigenspace $E_λ$.
If there is an eigenvalue $λ$ such that the geometric multiplicity of $λ$, dim$(E_λ)$, is less than the algebraic multiplicity of $λ$, then the matrix $A$ is not diagonalizable. If not, $A$ is diagonalizable, and proceed to the next step.
$4.$If we combine all basis vectors for all eigenspaces, we obtained n linearly independent eigenvectors $v1,v2,…,vn$.
$5.$Define the nonsingular matrix $P=[v_1v_2…v_n]$.
$6.$Define the diagonal matrix $D$, whose $(i,i)$-entry is the eigenvalue λ such that the $i^th$ column vector $v_i$ is in the eigenspace $E_λ$
$7.$Then the matrix $A$ is diagonalized as $P^-1AP=D$
answered Jul 26 at 17:42
Key Flex
4,047423
@ Key Flexhow can i find the VAlue of P and B ? can elaborate that more
– stupid
Jul 26 at 17:43
1
@stupid I added step by step procedure how to find it.
– Key Flex
Jul 26 at 17:49
The matrix $A$ has rank 1, so two of the eigenvalues are automatically equal to 0. The other eigenvalue turns out to be 3.
– Joe Silverman
Jul 26 at 17:54
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The value for $A=u^tv$ you got is not correct.
The correct value is
$$A=u^tv=beginbmatrix 1 \ 2 \ 3 endbmatrixbeginbmatrix 1 & dfrac 1 2 & dfrac 1 3 endbmatrix=beginbmatrix 1 & dfrac 1 2 & dfrac 1 3 \ 2 & 1 & dfrac 2 3 \ 3 & dfrac 3 2 & 1 endbmatrix$$
Now you could use diagonalization.
Let $A$ be a matrix then if you diagonalize $A$ then you get $A=PDP^-1$ with $B$ as a diagonal matrix and $P^-1$ an invertible matrix. If you try for $A^2$, then it is equal to $A^2=PD^2P^-1$. Similarly for $A^n=PD^nP^-1$
$1.$Find the characteristic polynomial $p(t)$ of $A$.
$2.$Find eigenvalues $λ$ of the matrix $A$ and their algebraic multiplicities from the characteristic polynomial $p(t)$.
$3.$For each eigenvalue $λ$ of $A$, find a basis of the eigenspace $E_λ$.
If there is an eigenvalue $λ$ such that the geometric multiplicity of $λ$, dim$(E_λ)$, is less than the algebraic multiplicity of $λ$, then the matrix $A$ is not diagonalizable. If not, $A$ is diagonalizable, and proceed to the next step.
$4.$If we combine all basis vectors for all eigenspaces, we obtained n linearly independent eigenvectors $v1,v2,…,vn$.
$5.$Define the nonsingular matrix $P=[v_1v_2…v_n]$.
$6.$Define the diagonal matrix $D$, whose $(i,i)$-entry is the eigenvalue λ such that the $i^th$ column vector $v_i$ is in the eigenspace $E_λ$
$7.$Then the matrix $A$ is diagonalized as $P^-1AP=D$
answered Jul 26 at 17:42
Key Flex
4,047423
The value for $A=u^tv$ you got is not correct.
The correct value is
$$A=u^tv=beginbmatrix 1 \ 2 \ 3 endbmatrixbeginbmatrix 1 & dfrac 1 2 & dfrac 1 3 endbmatrix=beginbmatrix 1 & dfrac 1 2 & dfrac 1 3 \ 2 & 1 & dfrac 2 3 \ 3 & dfrac 3 2 & 1 endbmatrix$$
Now you could use diagonalization.
Let $A$ be a matrix then if you diagonalize $A$ then you get $A=PDP^-1$ with $B$ as a diagonal matrix and $P^-1$ an invertible matrix. If you try for $A^2$, then it is equal to $A^2=PD^2P^-1$. Similarly for $A^n=PD^nP^-1$
$1.$Find the characteristic polynomial $p(t)$ of $A$.
$2.$Find eigenvalues $λ$ of the matrix $A$ and their algebraic multiplicities from the characteristic polynomial $p(t)$.
$3.$For each eigenvalue $λ$ of $A$, find a basis of the eigenspace $E_λ$.
If there is an eigenvalue $λ$ such that the geometric multiplicity of $λ$, dim$(E_λ)$, is less than the algebraic multiplicity of $λ$, then the matrix $A$ is not diagonalizable. If not, $A$ is diagonalizable, and proceed to the next step.
$4.$If we combine all basis vectors for all eigenspaces, we obtained n linearly independent eigenvectors $v1,v2,…,vn$.
$5.$Define the nonsingular matrix $P=[v_1v_2…v_n]$.
$6.$Define the diagonal matrix $D$, whose $(i,i)$-entry is the eigenvalue λ such that the $i^th$ column vector $v_i$ is in the eigenspace $E_λ$
$7.$Then the matrix $A$ is diagonalized as $P^-1AP=D$
answered Jul 26 at 17:42
Key Flex
4,047423
edited Jul 26 at 18:03
answered Jul 26 at 17:42
Key Flex
4,047423
answered Jul 26 at 17:42
Key Flex
4,047423
answered Jul 26 at 17:42
Key Flex
4,047423
4,047423
@ Key Flexhow can i find the VAlue of P and B ? can elaborate that more
– stupid
Jul 26 at 17:43
1
@stupid I added step by step procedure how to find it.
– Key Flex
Jul 26 at 17:49
The matrix $A$ has rank 1, so two of the eigenvalues are automatically equal to 0. The other eigenvalue turns out to be 3.
– Joe Silverman
Jul 26 at 17:54
add a comment |Â
@ Key Flexhow can i find the VAlue of P and B ? can elaborate that more
– stupid
Jul 26 at 17:43
1
@stupid I added step by step procedure how to find it.
– Key Flex
Jul 26 at 17:49
The matrix $A$ has rank 1, so two of the eigenvalues are automatically equal to 0. The other eigenvalue turns out to be 3.
– Joe Silverman
Jul 26 at 17:54
@ Key Flexhow can i find the VAlue of P and B ? can elaborate that more
– stupid
Jul 26 at 17:43
@ Key Flexhow can i find the VAlue of P and B ? can elaborate that more
– stupid
Jul 26 at 17:43
1
1
@stupid I added step by step procedure how to find it.
– Key Flex
Jul 26 at 17:49
@stupid I added step by step procedure how to find it.
– Key Flex
Jul 26 at 17:49
The matrix $A$ has rank 1, so two of the eigenvalues are automatically equal to 0. The other eigenvalue turns out to be 3.
– Joe Silverman
Jul 26 at 17:54
The matrix $A$ has rank 1, so two of the eigenvalues are automatically equal to 0. The other eigenvalue turns out to be 3.
– Joe Silverman
Jul 26 at 17:54
add a comment |Â
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