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Finding a matrix given it's characteristic polynomial
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I am asked to find a $2 times 2$ matrix with real and whole entries given it's characteristic polynomial:
$$p^2 -5p +1.$$
This is what I have done thus far:
I equated the polynomial to zero, and the roots (eigenvalues) were found to be $2.5 +/- sqrt(21/2$
I named the matrix to be solved $C$,
so $det(C) =$ product of eigenvalues $= 1$
$trace(C) =$ sum of eigenvalues $=5$
I then tried to find C by solving $T^-1 times D times T$, where $D$ is a matrix whose diagonal entries are the eigenvalues solved above, and $T$ is any matrix who's determinant is non zero.
I used $T$ as a $2 times 2$ being
$$beginbmatrix 1 & 1 \ 0 & 1 endbmatrix.$$
I solved $T^-1 times D times T$, and threw it in a calculator to make sure I made no algebraic mistakes, but the answer I received is wrong.
I appreciate any help, thank you.
linear-algebra matrices polynomials
edited 3 hours ago
Robert Lewis
40.1k22459
asked 3 hours ago
Michel
724
add a comment |Â
up vote
3
down vote
favorite
I am asked to find a $2 times 2$ matrix with real and whole entries given it's characteristic polynomial:
$$p^2 -5p +1.$$
This is what I have done thus far:
I equated the polynomial to zero, and the roots (eigenvalues) were found to be $2.5 +/- sqrt(21/2$
I named the matrix to be solved $C$,
so $det(C) =$ product of eigenvalues $= 1$
$trace(C) =$ sum of eigenvalues $=5$
I then tried to find C by solving $T^-1 times D times T$, where $D$ is a matrix whose diagonal entries are the eigenvalues solved above, and $T$ is any matrix who's determinant is non zero.
I used $T$ as a $2 times 2$ being
$$beginbmatrix 1 & 1 \ 0 & 1 endbmatrix.$$
I solved $T^-1 times D times T$, and threw it in a calculator to make sure I made no algebraic mistakes, but the answer I received is wrong.
I appreciate any help, thank you.
linear-algebra matrices polynomials
edited 3 hours ago
Robert Lewis
40.1k22459
asked 3 hours ago
Michel
724
2
Characteristic polynomial is invariant under similarity, so you made a computational mistake.
– RghtHndSd
3 hours ago
2
it is pretty easy to guess a 2 by 2 matrix of integers with trace 5 and determinant 1
– Will Jagy
3 hours ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I am asked to find a $2 times 2$ matrix with real and whole entries given it's characteristic polynomial:
$$p^2 -5p +1.$$
This is what I have done thus far:
I equated the polynomial to zero, and the roots (eigenvalues) were found to be $2.5 +/- sqrt(21/2$
I named the matrix to be solved $C$,
so $det(C) =$ product of eigenvalues $= 1$
$trace(C) =$ sum of eigenvalues $=5$
I then tried to find C by solving $T^-1 times D times T$, where $D$ is a matrix whose diagonal entries are the eigenvalues solved above, and $T$ is any matrix who's determinant is non zero.
I used $T$ as a $2 times 2$ being
$$beginbmatrix 1 & 1 \ 0 & 1 endbmatrix.$$
I solved $T^-1 times D times T$, and threw it in a calculator to make sure I made no algebraic mistakes, but the answer I received is wrong.
I appreciate any help, thank you.
linear-algebra matrices polynomials
edited 3 hours ago
Robert Lewis
40.1k22459
asked 3 hours ago
Michel
724
I am asked to find a $2 times 2$ matrix with real and whole entries given it's characteristic polynomial:
$$p^2 -5p +1.$$
This is what I have done thus far:
I equated the polynomial to zero, and the roots (eigenvalues) were found to be $2.5 +/- sqrt(21/2$
I named the matrix to be solved $C$,
so $det(C) =$ product of eigenvalues $= 1$
$trace(C) =$ sum of eigenvalues $=5$
I then tried to find C by solving $T^-1 times D times T$, where $D$ is a matrix whose diagonal entries are the eigenvalues solved above, and $T$ is any matrix who's determinant is non zero.
I used $T$ as a $2 times 2$ being
$$beginbmatrix 1 & 1 \ 0 & 1 endbmatrix.$$
I solved $T^-1 times D times T$, and threw it in a calculator to make sure I made no algebraic mistakes, but the answer I received is wrong.
I appreciate any help, thank you.
linear-algebra matrices polynomials
linear-algebra matrices polynomials
edited 3 hours ago
Robert Lewis
40.1k22459
asked 3 hours ago
Michel
724
edited 3 hours ago
Robert Lewis
40.1k22459
asked 3 hours ago
Michel
724
edited 3 hours ago
Robert Lewis
40.1k22459
edited 3 hours ago
Robert Lewis
40.1k22459
edited 3 hours ago
Robert Lewis
40.1k22459
40.1k22459
asked 3 hours ago
Michel
724
asked 3 hours ago
Michel
724
asked 3 hours ago
Michel
724
724
2
Characteristic polynomial is invariant under similarity, so you made a computational mistake.
– RghtHndSd
3 hours ago
2
it is pretty easy to guess a 2 by 2 matrix of integers with trace 5 and determinant 1
– Will Jagy
3 hours ago
add a comment |Â
2
Characteristic polynomial is invariant under similarity, so you made a computational mistake.
– RghtHndSd
3 hours ago
2
it is pretty easy to guess a 2 by 2 matrix of integers with trace 5 and determinant 1
– Will Jagy
3 hours ago
2
2
Characteristic polynomial is invariant under similarity, so you made a computational mistake.
– RghtHndSd
3 hours ago
Characteristic polynomial is invariant under similarity, so you made a computational mistake.
– RghtHndSd
3 hours ago
2
2
it is pretty easy to guess a 2 by 2 matrix of integers with trace 5 and determinant 1
– Will Jagy
3 hours ago
it is pretty easy to guess a 2 by 2 matrix of integers with trace 5 and determinant 1
– Will Jagy
3 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
Your choice of $T$ would not lead to whole entries.
Guide:
Let $C=beginbmatrixa & b \ c & d endbmatrix$.
We need $a+d=5$ and $ad-bc=1$
You can let $a=0$, then we end up having $-bc=1$. Can you choose some possible values of $b$ and $c$ to let it satisfy the equation?
answered 3 hours ago
Siong Thye Goh
85.9k1458107
add a comment |Â
up vote
2
down vote
Let's start off by looking at the characteristic polynomial of the $2 times 2$ matrix
$A = beginbmatrix 0 & 1 \ -a & -b endbmatrix: tag 1$
$det (A - lambda I) = det left (beginbmatrix -lambda & 1 \ -a & -b - lambda endbmatrix right ) = lambda^2 + blambda + a; tag 2$
we see from (2) that we may always present a $2 times 2$ matrix with given characteristic polynomial $lambda^2 + blambda + a$ in the form $A$; for example, if he quadratic is $lambda^2 + 5lambda + 1$, as in the present problem (tho' I have replaced $p$ with $lambda$), we may take
$P = beginbmatrix 0 & 1 \ -1 & -5 endbmatrix, tag 3$
which may be easily checked:
$det(P - lambda I) = -lambda(-5 - lambda) - 1 ( -1) = lambda^2 + 5lambda + 1. tag 4$
The matrices $A$ and $P$ above are part of a general paradigm for constructing matrices with a given characteristic polynomial, and it extends to higher dimensions. Now, there are a very many matrices possessed of a given characteristic polynomial, since it is a similarity invariant; that is, the characteristic polynomials of $X$ and $S^-1XS$ are always the same; thus it behooves us to find a matrix of particularly simple, general form for a given polynomial. If
$q(lambda) = displaystyle sum_1^n q_i lambda^i, ; q_n = 1, tag5$
we define $C(q(lambda))$ to be the $n times n$ matrix
$C(q(lambda)) = beginbmatrix 0 & 1 & 0 & ldots & 0 \ 0 & 0 & 1 & ldots & 0 \
vdots & vdots & ldots & vdots & vdots \ -q_0 & -q_1 & -q_2 & ldots & -q_n - 1 endbmatrix; tag 6$
that is, $C(q(lambda))$ has all $1$s on the superdiagonal, the negatives of the coefficients of $q(lambda)$ on the $n$-th row, and $0$s everywhere else. We have
$C(q(lambda) - lambda I = beginbmatrix -lambda & 1 & 0 & ldots & 0 \ 0 & -lambda & 1 & ldots & 0 \
vdots & vdots & ldots & vdots & vdots \ -q_0 & -q_1 & -q_2 & ldots & -q_n - 1 - lambda endbmatrix; tag 7$
it is easy to see, by expanding in minors along the $n$-th row, that
$det(C(q(lambda)) - lambda I) = q(lambda); tag 8$
also, since a matrix and its transpose have equal determinants, the transposed form $C(q(lambda))$, $C^T(q(lambda))$, also gives rise to the same characteristic polynomial. The matrices $C(q(lambda))$ and $C^T(q(lambda))$ are known as companion matrices for the polynomial $q(lambda)$; the linked article has more of the story.
answered 1 hour ago
Robert Lewis
40.1k22459
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Your choice of $T$ would not lead to whole entries.
Guide:
Let $C=beginbmatrixa & b \ c & d endbmatrix$.
We need $a+d=5$ and $ad-bc=1$
You can let $a=0$, then we end up having $-bc=1$. Can you choose some possible values of $b$ and $c$ to let it satisfy the equation?
answered 3 hours ago
Siong Thye Goh
85.9k1458107
add a comment |Â
up vote
2
down vote
Your choice of $T$ would not lead to whole entries.
Guide:
Let $C=beginbmatrixa & b \ c & d endbmatrix$.
We need $a+d=5$ and $ad-bc=1$
You can let $a=0$, then we end up having $-bc=1$. Can you choose some possible values of $b$ and $c$ to let it satisfy the equation?
answered 3 hours ago
Siong Thye Goh
85.9k1458107
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Your choice of $T$ would not lead to whole entries.
Guide:
Let $C=beginbmatrixa & b \ c & d endbmatrix$.
We need $a+d=5$ and $ad-bc=1$
You can let $a=0$, then we end up having $-bc=1$. Can you choose some possible values of $b$ and $c$ to let it satisfy the equation?
answered 3 hours ago
Siong Thye Goh
85.9k1458107
Your choice of $T$ would not lead to whole entries.
Guide:
Let $C=beginbmatrixa & b \ c & d endbmatrix$.
We need $a+d=5$ and $ad-bc=1$
You can let $a=0$, then we end up having $-bc=1$. Can you choose some possible values of $b$ and $c$ to let it satisfy the equation?
answered 3 hours ago
Siong Thye Goh
85.9k1458107
edited 3 hours ago
answered 3 hours ago
Siong Thye Goh
85.9k1458107
answered 3 hours ago
Siong Thye Goh
85.9k1458107
answered 3 hours ago
Siong Thye Goh
85.9k1458107
85.9k1458107
add a comment |Â
add a comment |Â
up vote
2
down vote
Let's start off by looking at the characteristic polynomial of the $2 times 2$ matrix
$A = beginbmatrix 0 & 1 \ -a & -b endbmatrix: tag 1$
$det (A - lambda I) = det left (beginbmatrix -lambda & 1 \ -a & -b - lambda endbmatrix right ) = lambda^2 + blambda + a; tag 2$
we see from (2) that we may always present a $2 times 2$ matrix with given characteristic polynomial $lambda^2 + blambda + a$ in the form $A$; for example, if he quadratic is $lambda^2 + 5lambda + 1$, as in the present problem (tho' I have replaced $p$ with $lambda$), we may take
$P = beginbmatrix 0 & 1 \ -1 & -5 endbmatrix, tag 3$
which may be easily checked:
$det(P - lambda I) = -lambda(-5 - lambda) - 1 ( -1) = lambda^2 + 5lambda + 1. tag 4$
The matrices $A$ and $P$ above are part of a general paradigm for constructing matrices with a given characteristic polynomial, and it extends to higher dimensions. Now, there are a very many matrices possessed of a given characteristic polynomial, since it is a similarity invariant; that is, the characteristic polynomials of $X$ and $S^-1XS$ are always the same; thus it behooves us to find a matrix of particularly simple, general form for a given polynomial. If
$q(lambda) = displaystyle sum_1^n q_i lambda^i, ; q_n = 1, tag5$
we define $C(q(lambda))$ to be the $n times n$ matrix
$C(q(lambda)) = beginbmatrix 0 & 1 & 0 & ldots & 0 \ 0 & 0 & 1 & ldots & 0 \
vdots & vdots & ldots & vdots & vdots \ -q_0 & -q_1 & -q_2 & ldots & -q_n - 1 endbmatrix; tag 6$
that is, $C(q(lambda))$ has all $1$s on the superdiagonal, the negatives of the coefficients of $q(lambda)$ on the $n$-th row, and $0$s everywhere else. We have
$C(q(lambda) - lambda I = beginbmatrix -lambda & 1 & 0 & ldots & 0 \ 0 & -lambda & 1 & ldots & 0 \
vdots & vdots & ldots & vdots & vdots \ -q_0 & -q_1 & -q_2 & ldots & -q_n - 1 - lambda endbmatrix; tag 7$
it is easy to see, by expanding in minors along the $n$-th row, that
$det(C(q(lambda)) - lambda I) = q(lambda); tag 8$
also, since a matrix and its transpose have equal determinants, the transposed form $C(q(lambda))$, $C^T(q(lambda))$, also gives rise to the same characteristic polynomial. The matrices $C(q(lambda))$ and $C^T(q(lambda))$ are known as companion matrices for the polynomial $q(lambda)$; the linked article has more of the story.
answered 1 hour ago
Robert Lewis
40.1k22459
add a comment |Â
up vote
2
down vote
Let's start off by looking at the characteristic polynomial of the $2 times 2$ matrix
$A = beginbmatrix 0 & 1 \ -a & -b endbmatrix: tag 1$
$det (A - lambda I) = det left (beginbmatrix -lambda & 1 \ -a & -b - lambda endbmatrix right ) = lambda^2 + blambda + a; tag 2$
we see from (2) that we may always present a $2 times 2$ matrix with given characteristic polynomial $lambda^2 + blambda + a$ in the form $A$; for example, if he quadratic is $lambda^2 + 5lambda + 1$, as in the present problem (tho' I have replaced $p$ with $lambda$), we may take
$P = beginbmatrix 0 & 1 \ -1 & -5 endbmatrix, tag 3$
which may be easily checked:
$det(P - lambda I) = -lambda(-5 - lambda) - 1 ( -1) = lambda^2 + 5lambda + 1. tag 4$
The matrices $A$ and $P$ above are part of a general paradigm for constructing matrices with a given characteristic polynomial, and it extends to higher dimensions. Now, there are a very many matrices possessed of a given characteristic polynomial, since it is a similarity invariant; that is, the characteristic polynomials of $X$ and $S^-1XS$ are always the same; thus it behooves us to find a matrix of particularly simple, general form for a given polynomial. If
$q(lambda) = displaystyle sum_1^n q_i lambda^i, ; q_n = 1, tag5$
we define $C(q(lambda))$ to be the $n times n$ matrix
$C(q(lambda)) = beginbmatrix 0 & 1 & 0 & ldots & 0 \ 0 & 0 & 1 & ldots & 0 \
vdots & vdots & ldots & vdots & vdots \ -q_0 & -q_1 & -q_2 & ldots & -q_n - 1 endbmatrix; tag 6$
that is, $C(q(lambda))$ has all $1$s on the superdiagonal, the negatives of the coefficients of $q(lambda)$ on the $n$-th row, and $0$s everywhere else. We have
$C(q(lambda) - lambda I = beginbmatrix -lambda & 1 & 0 & ldots & 0 \ 0 & -lambda & 1 & ldots & 0 \
vdots & vdots & ldots & vdots & vdots \ -q_0 & -q_1 & -q_2 & ldots & -q_n - 1 - lambda endbmatrix; tag 7$
it is easy to see, by expanding in minors along the $n$-th row, that
$det(C(q(lambda)) - lambda I) = q(lambda); tag 8$
also, since a matrix and its transpose have equal determinants, the transposed form $C(q(lambda))$, $C^T(q(lambda))$, also gives rise to the same characteristic polynomial. The matrices $C(q(lambda))$ and $C^T(q(lambda))$ are known as companion matrices for the polynomial $q(lambda)$; the linked article has more of the story.
answered 1 hour ago
Robert Lewis
40.1k22459
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let's start off by looking at the characteristic polynomial of the $2 times 2$ matrix
$A = beginbmatrix 0 & 1 \ -a & -b endbmatrix: tag 1$
$det (A - lambda I) = det left (beginbmatrix -lambda & 1 \ -a & -b - lambda endbmatrix right ) = lambda^2 + blambda + a; tag 2$
we see from (2) that we may always present a $2 times 2$ matrix with given characteristic polynomial $lambda^2 + blambda + a$ in the form $A$; for example, if he quadratic is $lambda^2 + 5lambda + 1$, as in the present problem (tho' I have replaced $p$ with $lambda$), we may take
$P = beginbmatrix 0 & 1 \ -1 & -5 endbmatrix, tag 3$
which may be easily checked:
$det(P - lambda I) = -lambda(-5 - lambda) - 1 ( -1) = lambda^2 + 5lambda + 1. tag 4$
The matrices $A$ and $P$ above are part of a general paradigm for constructing matrices with a given characteristic polynomial, and it extends to higher dimensions. Now, there are a very many matrices possessed of a given characteristic polynomial, since it is a similarity invariant; that is, the characteristic polynomials of $X$ and $S^-1XS$ are always the same; thus it behooves us to find a matrix of particularly simple, general form for a given polynomial. If
$q(lambda) = displaystyle sum_1^n q_i lambda^i, ; q_n = 1, tag5$
we define $C(q(lambda))$ to be the $n times n$ matrix
$C(q(lambda)) = beginbmatrix 0 & 1 & 0 & ldots & 0 \ 0 & 0 & 1 & ldots & 0 \
vdots & vdots & ldots & vdots & vdots \ -q_0 & -q_1 & -q_2 & ldots & -q_n - 1 endbmatrix; tag 6$
that is, $C(q(lambda))$ has all $1$s on the superdiagonal, the negatives of the coefficients of $q(lambda)$ on the $n$-th row, and $0$s everywhere else. We have
$C(q(lambda) - lambda I = beginbmatrix -lambda & 1 & 0 & ldots & 0 \ 0 & -lambda & 1 & ldots & 0 \
vdots & vdots & ldots & vdots & vdots \ -q_0 & -q_1 & -q_2 & ldots & -q_n - 1 - lambda endbmatrix; tag 7$
it is easy to see, by expanding in minors along the $n$-th row, that
$det(C(q(lambda)) - lambda I) = q(lambda); tag 8$
also, since a matrix and its transpose have equal determinants, the transposed form $C(q(lambda))$, $C^T(q(lambda))$, also gives rise to the same characteristic polynomial. The matrices $C(q(lambda))$ and $C^T(q(lambda))$ are known as companion matrices for the polynomial $q(lambda)$; the linked article has more of the story.
answered 1 hour ago
Robert Lewis
40.1k22459
Let's start off by looking at the characteristic polynomial of the $2 times 2$ matrix
$A = beginbmatrix 0 & 1 \ -a & -b endbmatrix: tag 1$
$det (A - lambda I) = det left (beginbmatrix -lambda & 1 \ -a & -b - lambda endbmatrix right ) = lambda^2 + blambda + a; tag 2$
we see from (2) that we may always present a $2 times 2$ matrix with given characteristic polynomial $lambda^2 + blambda + a$ in the form $A$; for example, if he quadratic is $lambda^2 + 5lambda + 1$, as in the present problem (tho' I have replaced $p$ with $lambda$), we may take
$P = beginbmatrix 0 & 1 \ -1 & -5 endbmatrix, tag 3$
which may be easily checked:
$det(P - lambda I) = -lambda(-5 - lambda) - 1 ( -1) = lambda^2 + 5lambda + 1. tag 4$
The matrices $A$ and $P$ above are part of a general paradigm for constructing matrices with a given characteristic polynomial, and it extends to higher dimensions. Now, there are a very many matrices possessed of a given characteristic polynomial, since it is a similarity invariant; that is, the characteristic polynomials of $X$ and $S^-1XS$ are always the same; thus it behooves us to find a matrix of particularly simple, general form for a given polynomial. If
$q(lambda) = displaystyle sum_1^n q_i lambda^i, ; q_n = 1, tag5$
we define $C(q(lambda))$ to be the $n times n$ matrix
$C(q(lambda)) = beginbmatrix 0 & 1 & 0 & ldots & 0 \ 0 & 0 & 1 & ldots & 0 \
vdots & vdots & ldots & vdots & vdots \ -q_0 & -q_1 & -q_2 & ldots & -q_n - 1 endbmatrix; tag 6$
that is, $C(q(lambda))$ has all $1$s on the superdiagonal, the negatives of the coefficients of $q(lambda)$ on the $n$-th row, and $0$s everywhere else. We have
$C(q(lambda) - lambda I = beginbmatrix -lambda & 1 & 0 & ldots & 0 \ 0 & -lambda & 1 & ldots & 0 \
vdots & vdots & ldots & vdots & vdots \ -q_0 & -q_1 & -q_2 & ldots & -q_n - 1 - lambda endbmatrix; tag 7$
it is easy to see, by expanding in minors along the $n$-th row, that
$det(C(q(lambda)) - lambda I) = q(lambda); tag 8$
also, since a matrix and its transpose have equal determinants, the transposed form $C(q(lambda))$, $C^T(q(lambda))$, also gives rise to the same characteristic polynomial. The matrices $C(q(lambda))$ and $C^T(q(lambda))$ are known as companion matrices for the polynomial $q(lambda)$; the linked article has more of the story.
answered 1 hour ago
Robert Lewis
40.1k22459
answered 1 hour ago
Robert Lewis
40.1k22459
answered 1 hour ago
Robert Lewis
40.1k22459
answered 1 hour ago
Robert Lewis
40.1k22459
40.1k22459
add a comment |Â
add a comment |Â
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2
Characteristic polynomial is invariant under similarity, so you made a computational mistake.
– RghtHndSd
3 hours ago
2
it is pretty easy to guess a 2 by 2 matrix of integers with trace 5 and determinant 1
– Will Jagy
3 hours ago