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Determining a matrix from its characteristic equation
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Given the characteristic equation of a $2 times 2$ matrix: $lambda^2-10 lambda+12=0$. How do you find the two different matrices that both include the equation?
linear-algebra matrices
edited 43 mins ago
Robert Howard
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Given the characteristic equation of a $2 times 2$ matrix: $lambda^2-10 lambda+12=0$. How do you find the two different matrices that both include the equation?
linear-algebra matrices
edited 43 mins ago
Robert Howard
1,399622
asked 3 hours ago
Madison
137
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Madison is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
2
down vote
favorite
up vote
2
down vote
favorite
Given the characteristic equation of a $2 times 2$ matrix: $lambda^2-10 lambda+12=0$. How do you find the two different matrices that both include the equation?
linear-algebra matrices
edited 43 mins ago
Robert Howard
1,399622
asked 3 hours ago
Madison
137
New contributor
Madison is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Given the characteristic equation of a $2 times 2$ matrix: $lambda^2-10 lambda+12=0$. How do you find the two different matrices that both include the equation?
linear-algebra matrices
linear-algebra matrices
edited 43 mins ago
Robert Howard
1,399622
asked 3 hours ago
Madison
137
New contributor
Madison is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 43 mins ago
Robert Howard
1,399622
asked 3 hours ago
Madison
137
New contributor
Madison is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 43 mins ago
Robert Howard
1,399622
edited 43 mins ago
Robert Howard
1,399622
edited 43 mins ago
Robert Howard
1,399622
1,399622
asked 3 hours ago
Madison
137
New contributor
Madison is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Madison
137
asked 3 hours ago
Madison
137
137
New contributor
Madison is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Madison is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Madison is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
For convenience of referencing, let's say our $2times2$ matrix is $M=beginbmatrixa&b\c&dendbmatrix$.
Then its characteristic polynomial is
$$p(lambda)=detbeginbmatrixa-lambda&b\c&d-lambdaendbmatrix=(a-lambda)(d-lambda)-bc=lambda^2-(operatornametrM)lambda+(det M),$$
where $operatornametrM=a+d$ is the trace and $det M=ad-bc$ is the determinant of the matrix $M$.
From here you can create lots of matrices with the desired characteristic equation. First, you want $operatornametrM=a+d=10$, so pick any two numbers $a$ and $d$ that add up to $10$. Once you've placed those, pick $c$ and $d$ to make $det M=ad-bc=12$.
answered 2 hours ago
zipirovich
10.5k11630
add a comment |Â
up vote
1
down vote
Given any polynomial
$p(x) = displaystyle sum_0^n p_i x^i in Bbb F[x] tag 1$
which is satisfied by a matrix $A in M_n(Bbb F)$, that is, such that
$p(A) = displaystyle sum_0^n p_i A^i = 0, tag 2$
then any matrix $B$ similar to $A$,
$B = QAQ^-1 tag 3$
also satisfies $p(x)$ since
$p(B) = displaystyle sum_0^n p_i B^i = sum_0^n p_i (QAQ^-1)^i = sum_0^n p_i QA^iQ^-1 = Q left ( sum_0^n p_i A^i right ) Q^-1, tag 4$
where we have used the fact that $(QAQ^-1)^i = QA^iQ^-1$.
Thus there are in general very many matrices which satisfy $p(x)$, given that at least one does. The exceptional case is $A = mu I$ for some $mu in F$; then $B = QAQ^-1 = Q mu I Q^-1 = mu I$ and we don't get anything new. But if we add a strictly upper triangular matrix to $A$, the resulting matrix $A'$ has the same characteristic polynomial but $QA'Q^-1 ne A'$ for at least some $Q$; in this way we may treat this exceptional case.
Can we then find a matrix which satisfies the polynomial $p(x)$? With $deg p = n$, we will look for matrices in $M_n(Bbb F)$. Suppose for the moment that $p(x)$ is monic, that is, that $p_n = 1$; it is then well-known that the companion matrix of $p(x)$, viz.
$C(p) = beginbmatrix 0 & 0 & ldots & 0 & -p_0 \ 1 & 0 & ldots & 0 & -p_1 \ 0 & 1 & ldots & 0 & -p_2 \ vdots & vdots & ddots & vdots & vdots \ 0 & 0 & ldots & 1 & -p_n - 1 endbmatrix, tag 5$
has characteristic polynomial $p(x)$; therefore,
$p(C(p)) = 0; tag 6$
we may clearly remove the restriction $p_n = 1$ from (6) by simply multiplying by $p_n$; therefore we have shown that any polynomial of degree $n$ is satisfied by some matrix in $M_n(Bbb F)$; and via the similarity operation (4) or its variant discussed above, we may construct other matrices satisfying $p(x)$.
Of course we can apply all of this to the polynomial $x^2 - 10x + 12$, and see that there are many $2 times 2$ matrices such that $A^2 - 10 A + 12 = 0$.
answered 1 hour ago
Robert Lewis
41.1k22659
add a comment |Â
up vote
0
down vote
Note that $lambda^2 - 10lambda +12 =(lambda^2-10lambda + 9)+3 = (lambda-9)(lambda-1)+3$
So then one such matrix is
$$beginbmatrix 9&1\ -3&1endbmatrix$$
Do you see a quick way to permute the elements of this matrix and then end up with the same polynomial?
answered 2 hours ago
Theo C.
13418
2
Other option is $lambda(lambda-10)+12=0$. Or many other choices
– Andrei
2 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
For convenience of referencing, let's say our $2times2$ matrix is $M=beginbmatrixa&b\c&dendbmatrix$.
Then its characteristic polynomial is
$$p(lambda)=detbeginbmatrixa-lambda&b\c&d-lambdaendbmatrix=(a-lambda)(d-lambda)-bc=lambda^2-(operatornametrM)lambda+(det M),$$
where $operatornametrM=a+d$ is the trace and $det M=ad-bc$ is the determinant of the matrix $M$.
From here you can create lots of matrices with the desired characteristic equation. First, you want $operatornametrM=a+d=10$, so pick any two numbers $a$ and $d$ that add up to $10$. Once you've placed those, pick $c$ and $d$ to make $det M=ad-bc=12$.
answered 2 hours ago
zipirovich
10.5k11630
add a comment |Â
up vote
4
down vote
For convenience of referencing, let's say our $2times2$ matrix is $M=beginbmatrixa&b\c&dendbmatrix$.
Then its characteristic polynomial is
$$p(lambda)=detbeginbmatrixa-lambda&b\c&d-lambdaendbmatrix=(a-lambda)(d-lambda)-bc=lambda^2-(operatornametrM)lambda+(det M),$$
where $operatornametrM=a+d$ is the trace and $det M=ad-bc$ is the determinant of the matrix $M$.
From here you can create lots of matrices with the desired characteristic equation. First, you want $operatornametrM=a+d=10$, so pick any two numbers $a$ and $d$ that add up to $10$. Once you've placed those, pick $c$ and $d$ to make $det M=ad-bc=12$.
answered 2 hours ago
zipirovich
10.5k11630
add a comment |Â
up vote
4
down vote
up vote
4
down vote
For convenience of referencing, let's say our $2times2$ matrix is $M=beginbmatrixa&b\c&dendbmatrix$.
Then its characteristic polynomial is
$$p(lambda)=detbeginbmatrixa-lambda&b\c&d-lambdaendbmatrix=(a-lambda)(d-lambda)-bc=lambda^2-(operatornametrM)lambda+(det M),$$
where $operatornametrM=a+d$ is the trace and $det M=ad-bc$ is the determinant of the matrix $M$.
From here you can create lots of matrices with the desired characteristic equation. First, you want $operatornametrM=a+d=10$, so pick any two numbers $a$ and $d$ that add up to $10$. Once you've placed those, pick $c$ and $d$ to make $det M=ad-bc=12$.
answered 2 hours ago
zipirovich
10.5k11630
For convenience of referencing, let's say our $2times2$ matrix is $M=beginbmatrixa&b\c&dendbmatrix$.
Then its characteristic polynomial is
$$p(lambda)=detbeginbmatrixa-lambda&b\c&d-lambdaendbmatrix=(a-lambda)(d-lambda)-bc=lambda^2-(operatornametrM)lambda+(det M),$$
where $operatornametrM=a+d$ is the trace and $det M=ad-bc$ is the determinant of the matrix $M$.
From here you can create lots of matrices with the desired characteristic equation. First, you want $operatornametrM=a+d=10$, so pick any two numbers $a$ and $d$ that add up to $10$. Once you've placed those, pick $c$ and $d$ to make $det M=ad-bc=12$.
answered 2 hours ago
zipirovich
10.5k11630
answered 2 hours ago
zipirovich
10.5k11630
answered 2 hours ago
zipirovich
10.5k11630
answered 2 hours ago
zipirovich
10.5k11630
10.5k11630
add a comment |Â
add a comment |Â
up vote
1
down vote
Given any polynomial
$p(x) = displaystyle sum_0^n p_i x^i in Bbb F[x] tag 1$
which is satisfied by a matrix $A in M_n(Bbb F)$, that is, such that
$p(A) = displaystyle sum_0^n p_i A^i = 0, tag 2$
then any matrix $B$ similar to $A$,
$B = QAQ^-1 tag 3$
also satisfies $p(x)$ since
$p(B) = displaystyle sum_0^n p_i B^i = sum_0^n p_i (QAQ^-1)^i = sum_0^n p_i QA^iQ^-1 = Q left ( sum_0^n p_i A^i right ) Q^-1, tag 4$
where we have used the fact that $(QAQ^-1)^i = QA^iQ^-1$.
Thus there are in general very many matrices which satisfy $p(x)$, given that at least one does. The exceptional case is $A = mu I$ for some $mu in F$; then $B = QAQ^-1 = Q mu I Q^-1 = mu I$ and we don't get anything new. But if we add a strictly upper triangular matrix to $A$, the resulting matrix $A'$ has the same characteristic polynomial but $QA'Q^-1 ne A'$ for at least some $Q$; in this way we may treat this exceptional case.
Can we then find a matrix which satisfies the polynomial $p(x)$? With $deg p = n$, we will look for matrices in $M_n(Bbb F)$. Suppose for the moment that $p(x)$ is monic, that is, that $p_n = 1$; it is then well-known that the companion matrix of $p(x)$, viz.
$C(p) = beginbmatrix 0 & 0 & ldots & 0 & -p_0 \ 1 & 0 & ldots & 0 & -p_1 \ 0 & 1 & ldots & 0 & -p_2 \ vdots & vdots & ddots & vdots & vdots \ 0 & 0 & ldots & 1 & -p_n - 1 endbmatrix, tag 5$
has characteristic polynomial $p(x)$; therefore,
$p(C(p)) = 0; tag 6$
we may clearly remove the restriction $p_n = 1$ from (6) by simply multiplying by $p_n$; therefore we have shown that any polynomial of degree $n$ is satisfied by some matrix in $M_n(Bbb F)$; and via the similarity operation (4) or its variant discussed above, we may construct other matrices satisfying $p(x)$.
Of course we can apply all of this to the polynomial $x^2 - 10x + 12$, and see that there are many $2 times 2$ matrices such that $A^2 - 10 A + 12 = 0$.
answered 1 hour ago
Robert Lewis
41.1k22659
add a comment |Â
up vote
1
down vote
Given any polynomial
$p(x) = displaystyle sum_0^n p_i x^i in Bbb F[x] tag 1$
which is satisfied by a matrix $A in M_n(Bbb F)$, that is, such that
$p(A) = displaystyle sum_0^n p_i A^i = 0, tag 2$
then any matrix $B$ similar to $A$,
$B = QAQ^-1 tag 3$
also satisfies $p(x)$ since
$p(B) = displaystyle sum_0^n p_i B^i = sum_0^n p_i (QAQ^-1)^i = sum_0^n p_i QA^iQ^-1 = Q left ( sum_0^n p_i A^i right ) Q^-1, tag 4$
where we have used the fact that $(QAQ^-1)^i = QA^iQ^-1$.
Thus there are in general very many matrices which satisfy $p(x)$, given that at least one does. The exceptional case is $A = mu I$ for some $mu in F$; then $B = QAQ^-1 = Q mu I Q^-1 = mu I$ and we don't get anything new. But if we add a strictly upper triangular matrix to $A$, the resulting matrix $A'$ has the same characteristic polynomial but $QA'Q^-1 ne A'$ for at least some $Q$; in this way we may treat this exceptional case.
Can we then find a matrix which satisfies the polynomial $p(x)$? With $deg p = n$, we will look for matrices in $M_n(Bbb F)$. Suppose for the moment that $p(x)$ is monic, that is, that $p_n = 1$; it is then well-known that the companion matrix of $p(x)$, viz.
$C(p) = beginbmatrix 0 & 0 & ldots & 0 & -p_0 \ 1 & 0 & ldots & 0 & -p_1 \ 0 & 1 & ldots & 0 & -p_2 \ vdots & vdots & ddots & vdots & vdots \ 0 & 0 & ldots & 1 & -p_n - 1 endbmatrix, tag 5$
has characteristic polynomial $p(x)$; therefore,
$p(C(p)) = 0; tag 6$
we may clearly remove the restriction $p_n = 1$ from (6) by simply multiplying by $p_n$; therefore we have shown that any polynomial of degree $n$ is satisfied by some matrix in $M_n(Bbb F)$; and via the similarity operation (4) or its variant discussed above, we may construct other matrices satisfying $p(x)$.
Of course we can apply all of this to the polynomial $x^2 - 10x + 12$, and see that there are many $2 times 2$ matrices such that $A^2 - 10 A + 12 = 0$.
answered 1 hour ago
Robert Lewis
41.1k22659
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Given any polynomial
$p(x) = displaystyle sum_0^n p_i x^i in Bbb F[x] tag 1$
which is satisfied by a matrix $A in M_n(Bbb F)$, that is, such that
$p(A) = displaystyle sum_0^n p_i A^i = 0, tag 2$
then any matrix $B$ similar to $A$,
$B = QAQ^-1 tag 3$
also satisfies $p(x)$ since
$p(B) = displaystyle sum_0^n p_i B^i = sum_0^n p_i (QAQ^-1)^i = sum_0^n p_i QA^iQ^-1 = Q left ( sum_0^n p_i A^i right ) Q^-1, tag 4$
where we have used the fact that $(QAQ^-1)^i = QA^iQ^-1$.
Thus there are in general very many matrices which satisfy $p(x)$, given that at least one does. The exceptional case is $A = mu I$ for some $mu in F$; then $B = QAQ^-1 = Q mu I Q^-1 = mu I$ and we don't get anything new. But if we add a strictly upper triangular matrix to $A$, the resulting matrix $A'$ has the same characteristic polynomial but $QA'Q^-1 ne A'$ for at least some $Q$; in this way we may treat this exceptional case.
Can we then find a matrix which satisfies the polynomial $p(x)$? With $deg p = n$, we will look for matrices in $M_n(Bbb F)$. Suppose for the moment that $p(x)$ is monic, that is, that $p_n = 1$; it is then well-known that the companion matrix of $p(x)$, viz.
$C(p) = beginbmatrix 0 & 0 & ldots & 0 & -p_0 \ 1 & 0 & ldots & 0 & -p_1 \ 0 & 1 & ldots & 0 & -p_2 \ vdots & vdots & ddots & vdots & vdots \ 0 & 0 & ldots & 1 & -p_n - 1 endbmatrix, tag 5$
has characteristic polynomial $p(x)$; therefore,
$p(C(p)) = 0; tag 6$
we may clearly remove the restriction $p_n = 1$ from (6) by simply multiplying by $p_n$; therefore we have shown that any polynomial of degree $n$ is satisfied by some matrix in $M_n(Bbb F)$; and via the similarity operation (4) or its variant discussed above, we may construct other matrices satisfying $p(x)$.
Of course we can apply all of this to the polynomial $x^2 - 10x + 12$, and see that there are many $2 times 2$ matrices such that $A^2 - 10 A + 12 = 0$.
answered 1 hour ago
Robert Lewis
41.1k22659
Given any polynomial
$p(x) = displaystyle sum_0^n p_i x^i in Bbb F[x] tag 1$
which is satisfied by a matrix $A in M_n(Bbb F)$, that is, such that
$p(A) = displaystyle sum_0^n p_i A^i = 0, tag 2$
then any matrix $B$ similar to $A$,
$B = QAQ^-1 tag 3$
also satisfies $p(x)$ since
$p(B) = displaystyle sum_0^n p_i B^i = sum_0^n p_i (QAQ^-1)^i = sum_0^n p_i QA^iQ^-1 = Q left ( sum_0^n p_i A^i right ) Q^-1, tag 4$
where we have used the fact that $(QAQ^-1)^i = QA^iQ^-1$.
Thus there are in general very many matrices which satisfy $p(x)$, given that at least one does. The exceptional case is $A = mu I$ for some $mu in F$; then $B = QAQ^-1 = Q mu I Q^-1 = mu I$ and we don't get anything new. But if we add a strictly upper triangular matrix to $A$, the resulting matrix $A'$ has the same characteristic polynomial but $QA'Q^-1 ne A'$ for at least some $Q$; in this way we may treat this exceptional case.
Can we then find a matrix which satisfies the polynomial $p(x)$? With $deg p = n$, we will look for matrices in $M_n(Bbb F)$. Suppose for the moment that $p(x)$ is monic, that is, that $p_n = 1$; it is then well-known that the companion matrix of $p(x)$, viz.
$C(p) = beginbmatrix 0 & 0 & ldots & 0 & -p_0 \ 1 & 0 & ldots & 0 & -p_1 \ 0 & 1 & ldots & 0 & -p_2 \ vdots & vdots & ddots & vdots & vdots \ 0 & 0 & ldots & 1 & -p_n - 1 endbmatrix, tag 5$
has characteristic polynomial $p(x)$; therefore,
$p(C(p)) = 0; tag 6$
we may clearly remove the restriction $p_n = 1$ from (6) by simply multiplying by $p_n$; therefore we have shown that any polynomial of degree $n$ is satisfied by some matrix in $M_n(Bbb F)$; and via the similarity operation (4) or its variant discussed above, we may construct other matrices satisfying $p(x)$.
Of course we can apply all of this to the polynomial $x^2 - 10x + 12$, and see that there are many $2 times 2$ matrices such that $A^2 - 10 A + 12 = 0$.
answered 1 hour ago
Robert Lewis
41.1k22659
edited 1 hour ago
answered 1 hour ago
Robert Lewis
41.1k22659
answered 1 hour ago
Robert Lewis
41.1k22659
answered 1 hour ago
Robert Lewis
41.1k22659
41.1k22659
add a comment |Â
add a comment |Â
up vote
0
down vote
Note that $lambda^2 - 10lambda +12 =(lambda^2-10lambda + 9)+3 = (lambda-9)(lambda-1)+3$
So then one such matrix is
$$beginbmatrix 9&1\ -3&1endbmatrix$$
Do you see a quick way to permute the elements of this matrix and then end up with the same polynomial?
answered 2 hours ago
Theo C.
13418
2
Other option is $lambda(lambda-10)+12=0$. Or many other choices
– Andrei
2 hours ago
add a comment |Â
up vote
0
down vote
Note that $lambda^2 - 10lambda +12 =(lambda^2-10lambda + 9)+3 = (lambda-9)(lambda-1)+3$
So then one such matrix is
$$beginbmatrix 9&1\ -3&1endbmatrix$$
Do you see a quick way to permute the elements of this matrix and then end up with the same polynomial?
answered 2 hours ago
Theo C.
13418
2
Other option is $lambda(lambda-10)+12=0$. Or many other choices
– Andrei
2 hours ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Note that $lambda^2 - 10lambda +12 =(lambda^2-10lambda + 9)+3 = (lambda-9)(lambda-1)+3$
So then one such matrix is
$$beginbmatrix 9&1\ -3&1endbmatrix$$
Do you see a quick way to permute the elements of this matrix and then end up with the same polynomial?
answered 2 hours ago
Theo C.
13418
Note that $lambda^2 - 10lambda +12 =(lambda^2-10lambda + 9)+3 = (lambda-9)(lambda-1)+3$
So then one such matrix is
$$beginbmatrix 9&1\ -3&1endbmatrix$$
Do you see a quick way to permute the elements of this matrix and then end up with the same polynomial?
answered 2 hours ago
Theo C.
13418
answered 2 hours ago
Theo C.
13418
answered 2 hours ago
Theo C.
13418
answered 2 hours ago
Theo C.
13418
13418
2
Other option is $lambda(lambda-10)+12=0$. Or many other choices
– Andrei
2 hours ago
add a comment |Â
2
Other option is $lambda(lambda-10)+12=0$. Or many other choices
– Andrei
2 hours ago
2
2
Other option is $lambda(lambda-10)+12=0$. Or many other choices
– Andrei
2 hours ago
Other option is $lambda(lambda-10)+12=0$. Or many other choices
– Andrei
2 hours ago
add a comment |Â
Madison is a new contributor. Be nice, and check out our Code of Conduct.
Madison is a new contributor. Be nice, and check out our Code of Conduct.
Madison is a new contributor. Be nice, and check out our Code of Conduct.
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