Given a Characteristic Polynomial of a Matrix…
$begingroup$
This question contains three parts. I have already answered the first two. The last part is confusing me.
Suppose $A$ is a $4 times 4$ matrix whose characteristic polynomial is $p(x) = (x - 1)(x + 2)^2(x - 3)$.
Part (a): Show that $A$ is invertible. Find the characteristic polynomial of $A^{-1}$.
We have that the roots of a characteristic polynomial are the eigenvalues of $A$. That is, $lambda = -2, -2, 1, 3$ are our eigenvalues. The determinant of an $n times n$ matrix is the product of its eigenvalues. Hence, det$A = 12$. An $n times n$ matrix is invertible if and only if its determinant is nonzero. Therefore, $A$ is invertible.
Since none of the eigenvalues are zero, we have that $lambda$ is an eigenvalue of $A$ if and only if $frac{1}{lambda}$ is an eigenvalue of $A^{-1}$. Then, the characteristic polynomial for $A^{-1}$ is $q(x) = (x - 1)(x + 1/2)^2(x - 1/3)$.
Part (b): Find the determinant and trace of $A$ and $A^{-1}$.
This is easy since the determinant of an $n times n$ matrix is the product of its eigenvalues and the trace of an $n times n$ matrix is the sum of its eigenvalues.
Part (c): Express $A^{-1}$ as a polynomial in $A$. Explain your answer.
Not really sure what part (c) is getting at.
linear-algebra matrices polynomials determinant
asked Jan 6 at 0:36
Taylor McMillanTaylor McMillan
695
$endgroup$
add a comment |
$begingroup$
This question contains three parts. I have already answered the first two. The last part is confusing me.
Suppose $A$ is a $4 times 4$ matrix whose characteristic polynomial is $p(x) = (x - 1)(x + 2)^2(x - 3)$.
Part (a): Show that $A$ is invertible. Find the characteristic polynomial of $A^{-1}$.
We have that the roots of a characteristic polynomial are the eigenvalues of $A$. That is, $lambda = -2, -2, 1, 3$ are our eigenvalues. The determinant of an $n times n$ matrix is the product of its eigenvalues. Hence, det$A = 12$. An $n times n$ matrix is invertible if and only if its determinant is nonzero. Therefore, $A$ is invertible.
Since none of the eigenvalues are zero, we have that $lambda$ is an eigenvalue of $A$ if and only if $frac{1}{lambda}$ is an eigenvalue of $A^{-1}$. Then, the characteristic polynomial for $A^{-1}$ is $q(x) = (x - 1)(x + 1/2)^2(x - 1/3)$.
Part (b): Find the determinant and trace of $A$ and $A^{-1}$.
This is easy since the determinant of an $n times n$ matrix is the product of its eigenvalues and the trace of an $n times n$ matrix is the sum of its eigenvalues.
Part (c): Express $A^{-1}$ as a polynomial in $A$. Explain your answer.
Not really sure what part (c) is getting at.
linear-algebra matrices polynomials determinant
asked Jan 6 at 0:36
Taylor McMillanTaylor McMillan
695
$endgroup$
1
$begingroup$
For the part $a$, you did not address the issue of multiplicities... I suggest instead to note that $det(xI-A)=det(A)det(xA^{-1}-I)=det(A)*(-1/x)^4det(x^{-1}I-A)$, which does give your result.
$endgroup$
– Mindlack
Jan 6 at 0:49
add a comment |
$begingroup$
This question contains three parts. I have already answered the first two. The last part is confusing me.
Suppose $A$ is a $4 times 4$ matrix whose characteristic polynomial is $p(x) = (x - 1)(x + 2)^2(x - 3)$.
Part (a): Show that $A$ is invertible. Find the characteristic polynomial of $A^{-1}$.
We have that the roots of a characteristic polynomial are the eigenvalues of $A$. That is, $lambda = -2, -2, 1, 3$ are our eigenvalues. The determinant of an $n times n$ matrix is the product of its eigenvalues. Hence, det$A = 12$. An $n times n$ matrix is invertible if and only if its determinant is nonzero. Therefore, $A$ is invertible.
Since none of the eigenvalues are zero, we have that $lambda$ is an eigenvalue of $A$ if and only if $frac{1}{lambda}$ is an eigenvalue of $A^{-1}$. Then, the characteristic polynomial for $A^{-1}$ is $q(x) = (x - 1)(x + 1/2)^2(x - 1/3)$.
Part (b): Find the determinant and trace of $A$ and $A^{-1}$.
This is easy since the determinant of an $n times n$ matrix is the product of its eigenvalues and the trace of an $n times n$ matrix is the sum of its eigenvalues.
Part (c): Express $A^{-1}$ as a polynomial in $A$. Explain your answer.
Not really sure what part (c) is getting at.
linear-algebra matrices polynomials determinant
asked Jan 6 at 0:36
Taylor McMillanTaylor McMillan
695
$endgroup$
This question contains three parts. I have already answered the first two. The last part is confusing me.
Suppose $A$ is a $4 times 4$ matrix whose characteristic polynomial is $p(x) = (x - 1)(x + 2)^2(x - 3)$.
Part (a): Show that $A$ is invertible. Find the characteristic polynomial of $A^{-1}$.
We have that the roots of a characteristic polynomial are the eigenvalues of $A$. That is, $lambda = -2, -2, 1, 3$ are our eigenvalues. The determinant of an $n times n$ matrix is the product of its eigenvalues. Hence, det$A = 12$. An $n times n$ matrix is invertible if and only if its determinant is nonzero. Therefore, $A$ is invertible.
Since none of the eigenvalues are zero, we have that $lambda$ is an eigenvalue of $A$ if and only if $frac{1}{lambda}$ is an eigenvalue of $A^{-1}$. Then, the characteristic polynomial for $A^{-1}$ is $q(x) = (x - 1)(x + 1/2)^2(x - 1/3)$.
Part (b): Find the determinant and trace of $A$ and $A^{-1}$.
This is easy since the determinant of an $n times n$ matrix is the product of its eigenvalues and the trace of an $n times n$ matrix is the sum of its eigenvalues.
Part (c): Express $A^{-1}$ as a polynomial in $A$. Explain your answer.
Not really sure what part (c) is getting at.
linear-algebra matrices polynomials determinant
linear-algebra matrices polynomials determinant
asked Jan 6 at 0:36
Taylor McMillanTaylor McMillan
695
asked Jan 6 at 0:36
Taylor McMillanTaylor McMillan
695
asked Jan 6 at 0:36
Taylor McMillanTaylor McMillan
695
asked Jan 6 at 0:36
Taylor McMillanTaylor McMillan
695
asked Jan 6 at 0:36
Taylor McMillanTaylor McMillan
695
695
1
$begingroup$
For the part $a$, you did not address the issue of multiplicities... I suggest instead to note that $det(xI-A)=det(A)det(xA^{-1}-I)=det(A)*(-1/x)^4det(x^{-1}I-A)$, which does give your result.
$endgroup$
– Mindlack
Jan 6 at 0:49
add a comment |
1
$begingroup$
For the part $a$, you did not address the issue of multiplicities... I suggest instead to note that $det(xI-A)=det(A)det(xA^{-1}-I)=det(A)*(-1/x)^4det(x^{-1}I-A)$, which does give your result.
$endgroup$
– Mindlack
Jan 6 at 0:49
1
1
$begingroup$
For the part $a$, you did not address the issue of multiplicities... I suggest instead to note that $det(xI-A)=det(A)det(xA^{-1}-I)=det(A)*(-1/x)^4det(x^{-1}I-A)$, which does give your result.
$endgroup$
– Mindlack
Jan 6 at 0:49
$begingroup$
For the part $a$, you did not address the issue of multiplicities... I suggest instead to note that $det(xI-A)=det(A)det(xA^{-1}-I)=det(A)*(-1/x)^4det(x^{-1}I-A)$, which does give your result.
$endgroup$
– Mindlack
Jan 6 at 0:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
By the Cayley-Hamilton theorem, we have $(A-1)(A+2)^2(A-3)=0$, that is,
$A^4-9A^2-4A+12I=0$.
Multiply both sides by $A^{-1}$, and be amazed!
answered Jan 6 at 0:43
A. PongráczA. Pongrácz
6,0831929
$endgroup$
1
$begingroup$
suggest you type in the $I$ for $12 I$ I see, you also wrote the factored version without any $I$
$endgroup$
– Will Jagy
Jan 6 at 0:49
1
$begingroup$
You are right, of course. Thanks.
$endgroup$
– A. Pongrácz
Jan 6 at 1:00
$begingroup$
Much appreciated. Is there an approach that does not use Cayley-Hamilton? Or is this the "obvious" way to go about this problem?
$endgroup$
– Taylor McMillan
Jan 6 at 1:07
1
$begingroup$
Not sure about obvious, but certainly the simplest.
$endgroup$
– A. Pongrácz
Jan 6 at 1:08
add a comment |
$begingroup$
Hint for part (c): Cayley-Hamilton. Then multiply by $A^{-1}$ and solve for the inverse.
answered Jan 6 at 0:42
jmerryjmerry
17k11633
$endgroup$
$begingroup$
What tells you to use Cayley-Hamilton? Just the fact that it gives us a polynomial for $A$, and since $A$ is invertible, we should then be able to find a polynomial $A^{-1}$ in terms of $A$?
$endgroup$
– Taylor McMillan
Jan 6 at 1:06
add a comment |
Your Answer
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2 Answers
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2 Answers
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$begingroup$
By the Cayley-Hamilton theorem, we have $(A-1)(A+2)^2(A-3)=0$, that is,
$A^4-9A^2-4A+12I=0$.
Multiply both sides by $A^{-1}$, and be amazed!
answered Jan 6 at 0:43
A. PongráczA. Pongrácz
6,0831929
$endgroup$
1
$begingroup$
suggest you type in the $I$ for $12 I$ I see, you also wrote the factored version without any $I$
$endgroup$
– Will Jagy
Jan 6 at 0:49
1
$begingroup$
You are right, of course. Thanks.
$endgroup$
– A. Pongrácz
Jan 6 at 1:00
$begingroup$
Much appreciated. Is there an approach that does not use Cayley-Hamilton? Or is this the "obvious" way to go about this problem?
$endgroup$
– Taylor McMillan
Jan 6 at 1:07
1
$begingroup$
Not sure about obvious, but certainly the simplest.
$endgroup$
– A. Pongrácz
Jan 6 at 1:08
add a comment |
$begingroup$
By the Cayley-Hamilton theorem, we have $(A-1)(A+2)^2(A-3)=0$, that is,
$A^4-9A^2-4A+12I=0$.
Multiply both sides by $A^{-1}$, and be amazed!
answered Jan 6 at 0:43
A. PongráczA. Pongrácz
6,0831929
$endgroup$
1
$begingroup$
suggest you type in the $I$ for $12 I$ I see, you also wrote the factored version without any $I$
$endgroup$
– Will Jagy
Jan 6 at 0:49
1
$begingroup$
You are right, of course. Thanks.
$endgroup$
– A. Pongrácz
Jan 6 at 1:00
$begingroup$
Much appreciated. Is there an approach that does not use Cayley-Hamilton? Or is this the "obvious" way to go about this problem?
$endgroup$
– Taylor McMillan
Jan 6 at 1:07
1
$begingroup$
Not sure about obvious, but certainly the simplest.
$endgroup$
– A. Pongrácz
Jan 6 at 1:08
add a comment |
$begingroup$
By the Cayley-Hamilton theorem, we have $(A-1)(A+2)^2(A-3)=0$, that is,
$A^4-9A^2-4A+12I=0$.
Multiply both sides by $A^{-1}$, and be amazed!
answered Jan 6 at 0:43
A. PongráczA. Pongrácz
6,0831929
$endgroup$
By the Cayley-Hamilton theorem, we have $(A-1)(A+2)^2(A-3)=0$, that is,
$A^4-9A^2-4A+12I=0$.
Multiply both sides by $A^{-1}$, and be amazed!
answered Jan 6 at 0:43
A. PongráczA. Pongrácz
6,0831929
edited Jan 6 at 1:00
answered Jan 6 at 0:43
A. PongráczA. Pongrácz
6,0831929
answered Jan 6 at 0:43
A. PongráczA. Pongrácz
6,0831929
answered Jan 6 at 0:43
A. PongráczA. Pongrácz
6,0831929
6,0831929
1
$begingroup$
suggest you type in the $I$ for $12 I$ I see, you also wrote the factored version without any $I$
$endgroup$
– Will Jagy
Jan 6 at 0:49
1
$begingroup$
You are right, of course. Thanks.
$endgroup$
– A. Pongrácz
Jan 6 at 1:00
$begingroup$
Much appreciated. Is there an approach that does not use Cayley-Hamilton? Or is this the "obvious" way to go about this problem?
$endgroup$
– Taylor McMillan
Jan 6 at 1:07
1
$begingroup$
Not sure about obvious, but certainly the simplest.
$endgroup$
– A. Pongrácz
Jan 6 at 1:08
add a comment |
1
$begingroup$
suggest you type in the $I$ for $12 I$ I see, you also wrote the factored version without any $I$
$endgroup$
– Will Jagy
Jan 6 at 0:49
1
$begingroup$
You are right, of course. Thanks.
$endgroup$
– A. Pongrácz
Jan 6 at 1:00
$begingroup$
Much appreciated. Is there an approach that does not use Cayley-Hamilton? Or is this the "obvious" way to go about this problem?
$endgroup$
– Taylor McMillan
Jan 6 at 1:07
1
$begingroup$
Not sure about obvious, but certainly the simplest.
$endgroup$
– A. Pongrácz
Jan 6 at 1:08
1
1
$begingroup$
suggest you type in the $I$ for $12 I$ I see, you also wrote the factored version without any $I$
$endgroup$
– Will Jagy
Jan 6 at 0:49
$begingroup$
suggest you type in the $I$ for $12 I$ I see, you also wrote the factored version without any $I$
$endgroup$
– Will Jagy
Jan 6 at 0:49
1
1
$begingroup$
You are right, of course. Thanks.
$endgroup$
– A. Pongrácz
Jan 6 at 1:00
$begingroup$
You are right, of course. Thanks.
$endgroup$
– A. Pongrácz
Jan 6 at 1:00
$begingroup$
Much appreciated. Is there an approach that does not use Cayley-Hamilton? Or is this the "obvious" way to go about this problem?
$endgroup$
– Taylor McMillan
Jan 6 at 1:07
$begingroup$
Much appreciated. Is there an approach that does not use Cayley-Hamilton? Or is this the "obvious" way to go about this problem?
$endgroup$
– Taylor McMillan
Jan 6 at 1:07
1
1
$begingroup$
Not sure about obvious, but certainly the simplest.
$endgroup$
– A. Pongrácz
Jan 6 at 1:08
$begingroup$
Not sure about obvious, but certainly the simplest.
$endgroup$
– A. Pongrácz
Jan 6 at 1:08
add a comment |
$begingroup$
Hint for part (c): Cayley-Hamilton. Then multiply by $A^{-1}$ and solve for the inverse.
answered Jan 6 at 0:42
jmerryjmerry
17k11633
$endgroup$
$begingroup$
What tells you to use Cayley-Hamilton? Just the fact that it gives us a polynomial for $A$, and since $A$ is invertible, we should then be able to find a polynomial $A^{-1}$ in terms of $A$?
$endgroup$
– Taylor McMillan
Jan 6 at 1:06
add a comment |
$begingroup$
Hint for part (c): Cayley-Hamilton. Then multiply by $A^{-1}$ and solve for the inverse.
answered Jan 6 at 0:42
jmerryjmerry
17k11633
$endgroup$
$begingroup$
What tells you to use Cayley-Hamilton? Just the fact that it gives us a polynomial for $A$, and since $A$ is invertible, we should then be able to find a polynomial $A^{-1}$ in terms of $A$?
$endgroup$
– Taylor McMillan
Jan 6 at 1:06
add a comment |
$begingroup$
Hint for part (c): Cayley-Hamilton. Then multiply by $A^{-1}$ and solve for the inverse.
answered Jan 6 at 0:42
jmerryjmerry
17k11633
$endgroup$
Hint for part (c): Cayley-Hamilton. Then multiply by $A^{-1}$ and solve for the inverse.
answered Jan 6 at 0:42
jmerryjmerry
17k11633
answered Jan 6 at 0:42
jmerryjmerry
17k11633
answered Jan 6 at 0:42
jmerryjmerry
17k11633
answered Jan 6 at 0:42
jmerryjmerry
17k11633
17k11633
$begingroup$
What tells you to use Cayley-Hamilton? Just the fact that it gives us a polynomial for $A$, and since $A$ is invertible, we should then be able to find a polynomial $A^{-1}$ in terms of $A$?
$endgroup$
– Taylor McMillan
Jan 6 at 1:06
add a comment |
$begingroup$
What tells you to use Cayley-Hamilton? Just the fact that it gives us a polynomial for $A$, and since $A$ is invertible, we should then be able to find a polynomial $A^{-1}$ in terms of $A$?
$endgroup$
– Taylor McMillan
Jan 6 at 1:06
$begingroup$
What tells you to use Cayley-Hamilton? Just the fact that it gives us a polynomial for $A$, and since $A$ is invertible, we should then be able to find a polynomial $A^{-1}$ in terms of $A$?
$endgroup$
– Taylor McMillan
Jan 6 at 1:06
$begingroup$
What tells you to use Cayley-Hamilton? Just the fact that it gives us a polynomial for $A$, and since $A$ is invertible, we should then be able to find a polynomial $A^{-1}$ in terms of $A$?
$endgroup$
– Taylor McMillan
Jan 6 at 1:06
add a comment |
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$begingroup$
For the part $a$, you did not address the issue of multiplicities... I suggest instead to note that $det(xI-A)=det(A)det(xA^{-1}-I)=det(A)*(-1/x)^4det(x^{-1}I-A)$, which does give your result.
$endgroup$
– Mindlack
Jan 6 at 0:49