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If $lambda$ is a characteristic root of a non-singular matrix $A$, then prove that $det(A)/lambda$ is a characteristic root of $Adj A$.
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Let $A$ be a $ntimes n$ matrix. Then the characteristic equation is $mathrmdet(A-lambda×I)=0$, with possibly characteristic roots of $A$ $lambda_1,dots,lambda_n$. Let $A$ be regular, i.e. $mathrmdet(A)neq0$. Now $mathrmAdj(A)=mathrmdet(A)mathrmInv(A)$. I don't know how to find its characteristic roots.
linear-algebra
edited Aug 1 at 11:11
zzuussee
1,152419
asked Aug 1 at 10:56
kanchan b
23
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Let $A$ be a $ntimes n$ matrix. Then the characteristic equation is $mathrmdet(A-lambda×I)=0$, with possibly characteristic roots of $A$ $lambda_1,dots,lambda_n$. Let $A$ be regular, i.e. $mathrmdet(A)neq0$. Now $mathrmAdj(A)=mathrmdet(A)mathrmInv(A)$. I don't know how to find its characteristic roots.
linear-algebra
edited Aug 1 at 11:11
zzuussee
1,152419
asked Aug 1 at 10:56
kanchan b
23
$detleft(det(A)A^-1-fracdet(A)lambdaIright)=det(A)det(A^-1-frac1lambdaI)=det(A)det(A^-1)det(lambda I-A)=(-1)^ndet(A)det(A^-1)det(A-lambda I)=0$. The last equality being because $det(A-lambda I)=0$. The assumption of $A$ invertible allows to know that $lambda neq 0$ and that $A^-1$ exists.
– JessicaMcRae
Aug 1 at 11:07
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Let $A$ be a $ntimes n$ matrix. Then the characteristic equation is $mathrmdet(A-lambda×I)=0$, with possibly characteristic roots of $A$ $lambda_1,dots,lambda_n$. Let $A$ be regular, i.e. $mathrmdet(A)neq0$. Now $mathrmAdj(A)=mathrmdet(A)mathrmInv(A)$. I don't know how to find its characteristic roots.
linear-algebra
edited Aug 1 at 11:11
zzuussee
1,152419
asked Aug 1 at 10:56
kanchan b
23
Let $A$ be a $ntimes n$ matrix. Then the characteristic equation is $mathrmdet(A-lambda×I)=0$, with possibly characteristic roots of $A$ $lambda_1,dots,lambda_n$. Let $A$ be regular, i.e. $mathrmdet(A)neq0$. Now $mathrmAdj(A)=mathrmdet(A)mathrmInv(A)$. I don't know how to find its characteristic roots.
linear-algebra
edited Aug 1 at 11:11
zzuussee
1,152419
asked Aug 1 at 10:56
kanchan b
23
edited Aug 1 at 11:11
zzuussee
1,152419
edited Aug 1 at 11:11
zzuussee
1,152419
edited Aug 1 at 11:11
zzuussee
1,152419
1,152419
asked Aug 1 at 10:56
kanchan b
23
asked Aug 1 at 10:56
kanchan b
23
asked Aug 1 at 10:56
kanchan b
23
23
$detleft(det(A)A^-1-fracdet(A)lambdaIright)=det(A)det(A^-1-frac1lambdaI)=det(A)det(A^-1)det(lambda I-A)=(-1)^ndet(A)det(A^-1)det(A-lambda I)=0$. The last equality being because $det(A-lambda I)=0$. The assumption of $A$ invertible allows to know that $lambda neq 0$ and that $A^-1$ exists.
– JessicaMcRae
Aug 1 at 11:07
add a comment |Â
$detleft(det(A)A^-1-fracdet(A)lambdaIright)=det(A)det(A^-1-frac1lambdaI)=det(A)det(A^-1)det(lambda I-A)=(-1)^ndet(A)det(A^-1)det(A-lambda I)=0$. The last equality being because $det(A-lambda I)=0$. The assumption of $A$ invertible allows to know that $lambda neq 0$ and that $A^-1$ exists.
– JessicaMcRae
Aug 1 at 11:07
$detleft(det(A)A^-1-fracdet(A)lambdaIright)=det(A)det(A^-1-frac1lambdaI)=det(A)det(A^-1)det(lambda I-A)=(-1)^ndet(A)det(A^-1)det(A-lambda I)=0$. The last equality being because $det(A-lambda I)=0$. The assumption of $A$ invertible allows to know that $lambda neq 0$ and that $A^-1$ exists.
– JessicaMcRae
Aug 1 at 11:07
$detleft(det(A)A^-1-fracdet(A)lambdaIright)=det(A)det(A^-1-frac1lambdaI)=det(A)det(A^-1)det(lambda I-A)=(-1)^ndet(A)det(A^-1)det(A-lambda I)=0$. The last equality being because $det(A-lambda I)=0$. The assumption of $A$ invertible allows to know that $lambda neq 0$ and that $A^-1$ exists.
– JessicaMcRae
Aug 1 at 11:07
add a comment |Â
2 Answers
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Let $A$ be invertible. As you rightly remarked, then $mathrmadj(A)=mathrmdet(A)A^-1$. We can even prove the following stronger proposition:
Proposition: For $A$ invertible, $lambdaneq 0$ is an eigenvalue of $A$ iff $mathrmdet(A)lambda^-1$ is an eigenvalue of $mathrmadj(A)$:
Proof:
$$
Av=lambda vLeftrightarrow mathrmadj(A)(Av)=mathrmadj(A)(lambda v)Leftrightarrow mathrmdet(A)A^-1(Av)=mathrmdet(A)A^-1(lambda v)Leftrightarrow mathrmdet(A)v=mathrmdet(A)lambda A^-1(v)Leftrightarrowmathrmdet(A)lambda^-1v=mathrmdet(A)A^-1(v)Leftrightarrowmathrmdet(A)lambda^-1v=mathrmadj(A)v
$$
$Box$
You may rephrase this result in terms of roots of the characteristic polynomial, as you naturally have for $lambdaneq 0$ that $mathrmdet(A-lambda E_n)=0$ iff $lambda$ is an eigenvalue of $A$ iff $mathrmdet(A)lambda^-1$ is an eigenvalue of $mathrmadj(A)$ iff $mathrmdet(mathrmadj(A)-mathrmdet(A)lambda^-1E_n)=0$.
Obviously, the condition $lambdaneq 0$ is w.l.o.g. as $0$ is a eigenvalue of $A$ iff $det(A-0E_n)=det(A)=0$ iff $A$ is singular.
answered Aug 1 at 11:15
zzuussee
1,152419
add a comment |Â
up vote
0
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Let $B:=det(A) A^-1$. For a scalar $mu ne 0 $ and a vector $x ne 0$ we have
$Bx= mu x iff det(A)A^-1x= mu x iff det(A)x= mu Ax iff Ax= fracdet(A)mux.$
answered Aug 1 at 11:16
Fred
37k1237
I think you mean $mathrmdet(A)$ instead of $mathrmdet(a)$.
– zzuussee
Aug 1 at 13:50
Oops, you are right !
– Fred
Aug 2 at 7:59
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let $A$ be invertible. As you rightly remarked, then $mathrmadj(A)=mathrmdet(A)A^-1$. We can even prove the following stronger proposition:
Proposition: For $A$ invertible, $lambdaneq 0$ is an eigenvalue of $A$ iff $mathrmdet(A)lambda^-1$ is an eigenvalue of $mathrmadj(A)$:
Proof:
$$
Av=lambda vLeftrightarrow mathrmadj(A)(Av)=mathrmadj(A)(lambda v)Leftrightarrow mathrmdet(A)A^-1(Av)=mathrmdet(A)A^-1(lambda v)Leftrightarrow mathrmdet(A)v=mathrmdet(A)lambda A^-1(v)Leftrightarrowmathrmdet(A)lambda^-1v=mathrmdet(A)A^-1(v)Leftrightarrowmathrmdet(A)lambda^-1v=mathrmadj(A)v
$$
$Box$
You may rephrase this result in terms of roots of the characteristic polynomial, as you naturally have for $lambdaneq 0$ that $mathrmdet(A-lambda E_n)=0$ iff $lambda$ is an eigenvalue of $A$ iff $mathrmdet(A)lambda^-1$ is an eigenvalue of $mathrmadj(A)$ iff $mathrmdet(mathrmadj(A)-mathrmdet(A)lambda^-1E_n)=0$.
Obviously, the condition $lambdaneq 0$ is w.l.o.g. as $0$ is a eigenvalue of $A$ iff $det(A-0E_n)=det(A)=0$ iff $A$ is singular.
answered Aug 1 at 11:15
zzuussee
1,152419
add a comment |Â
up vote
0
down vote
Let $A$ be invertible. As you rightly remarked, then $mathrmadj(A)=mathrmdet(A)A^-1$. We can even prove the following stronger proposition:
Proposition: For $A$ invertible, $lambdaneq 0$ is an eigenvalue of $A$ iff $mathrmdet(A)lambda^-1$ is an eigenvalue of $mathrmadj(A)$:
Proof:
$$
Av=lambda vLeftrightarrow mathrmadj(A)(Av)=mathrmadj(A)(lambda v)Leftrightarrow mathrmdet(A)A^-1(Av)=mathrmdet(A)A^-1(lambda v)Leftrightarrow mathrmdet(A)v=mathrmdet(A)lambda A^-1(v)Leftrightarrowmathrmdet(A)lambda^-1v=mathrmdet(A)A^-1(v)Leftrightarrowmathrmdet(A)lambda^-1v=mathrmadj(A)v
$$
$Box$
You may rephrase this result in terms of roots of the characteristic polynomial, as you naturally have for $lambdaneq 0$ that $mathrmdet(A-lambda E_n)=0$ iff $lambda$ is an eigenvalue of $A$ iff $mathrmdet(A)lambda^-1$ is an eigenvalue of $mathrmadj(A)$ iff $mathrmdet(mathrmadj(A)-mathrmdet(A)lambda^-1E_n)=0$.
Obviously, the condition $lambdaneq 0$ is w.l.o.g. as $0$ is a eigenvalue of $A$ iff $det(A-0E_n)=det(A)=0$ iff $A$ is singular.
answered Aug 1 at 11:15
zzuussee
1,152419
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $A$ be invertible. As you rightly remarked, then $mathrmadj(A)=mathrmdet(A)A^-1$. We can even prove the following stronger proposition:
Proposition: For $A$ invertible, $lambdaneq 0$ is an eigenvalue of $A$ iff $mathrmdet(A)lambda^-1$ is an eigenvalue of $mathrmadj(A)$:
Proof:
$$
Av=lambda vLeftrightarrow mathrmadj(A)(Av)=mathrmadj(A)(lambda v)Leftrightarrow mathrmdet(A)A^-1(Av)=mathrmdet(A)A^-1(lambda v)Leftrightarrow mathrmdet(A)v=mathrmdet(A)lambda A^-1(v)Leftrightarrowmathrmdet(A)lambda^-1v=mathrmdet(A)A^-1(v)Leftrightarrowmathrmdet(A)lambda^-1v=mathrmadj(A)v
$$
$Box$
You may rephrase this result in terms of roots of the characteristic polynomial, as you naturally have for $lambdaneq 0$ that $mathrmdet(A-lambda E_n)=0$ iff $lambda$ is an eigenvalue of $A$ iff $mathrmdet(A)lambda^-1$ is an eigenvalue of $mathrmadj(A)$ iff $mathrmdet(mathrmadj(A)-mathrmdet(A)lambda^-1E_n)=0$.
Obviously, the condition $lambdaneq 0$ is w.l.o.g. as $0$ is a eigenvalue of $A$ iff $det(A-0E_n)=det(A)=0$ iff $A$ is singular.
answered Aug 1 at 11:15
zzuussee
1,152419
Let $A$ be invertible. As you rightly remarked, then $mathrmadj(A)=mathrmdet(A)A^-1$. We can even prove the following stronger proposition:
Proposition: For $A$ invertible, $lambdaneq 0$ is an eigenvalue of $A$ iff $mathrmdet(A)lambda^-1$ is an eigenvalue of $mathrmadj(A)$:
Proof:
$$
Av=lambda vLeftrightarrow mathrmadj(A)(Av)=mathrmadj(A)(lambda v)Leftrightarrow mathrmdet(A)A^-1(Av)=mathrmdet(A)A^-1(lambda v)Leftrightarrow mathrmdet(A)v=mathrmdet(A)lambda A^-1(v)Leftrightarrowmathrmdet(A)lambda^-1v=mathrmdet(A)A^-1(v)Leftrightarrowmathrmdet(A)lambda^-1v=mathrmadj(A)v
$$
$Box$
You may rephrase this result in terms of roots of the characteristic polynomial, as you naturally have for $lambdaneq 0$ that $mathrmdet(A-lambda E_n)=0$ iff $lambda$ is an eigenvalue of $A$ iff $mathrmdet(A)lambda^-1$ is an eigenvalue of $mathrmadj(A)$ iff $mathrmdet(mathrmadj(A)-mathrmdet(A)lambda^-1E_n)=0$.
Obviously, the condition $lambdaneq 0$ is w.l.o.g. as $0$ is a eigenvalue of $A$ iff $det(A-0E_n)=det(A)=0$ iff $A$ is singular.
answered Aug 1 at 11:15
zzuussee
1,152419
edited Aug 1 at 21:09
answered Aug 1 at 11:15
zzuussee
1,152419
answered Aug 1 at 11:15
zzuussee
1,152419
answered Aug 1 at 11:15
zzuussee
1,152419
1,152419
add a comment |Â
add a comment |Â
up vote
0
down vote
Let $B:=det(A) A^-1$. For a scalar $mu ne 0 $ and a vector $x ne 0$ we have
$Bx= mu x iff det(A)A^-1x= mu x iff det(A)x= mu Ax iff Ax= fracdet(A)mux.$
answered Aug 1 at 11:16
Fred
37k1237
I think you mean $mathrmdet(A)$ instead of $mathrmdet(a)$.
– zzuussee
Aug 1 at 13:50
Oops, you are right !
– Fred
Aug 2 at 7:59
add a comment |Â
up vote
0
down vote
Let $B:=det(A) A^-1$. For a scalar $mu ne 0 $ and a vector $x ne 0$ we have
$Bx= mu x iff det(A)A^-1x= mu x iff det(A)x= mu Ax iff Ax= fracdet(A)mux.$
answered Aug 1 at 11:16
Fred
37k1237
I think you mean $mathrmdet(A)$ instead of $mathrmdet(a)$.
– zzuussee
Aug 1 at 13:50
Oops, you are right !
– Fred
Aug 2 at 7:59
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $B:=det(A) A^-1$. For a scalar $mu ne 0 $ and a vector $x ne 0$ we have
$Bx= mu x iff det(A)A^-1x= mu x iff det(A)x= mu Ax iff Ax= fracdet(A)mux.$
answered Aug 1 at 11:16
Fred
37k1237
Let $B:=det(A) A^-1$. For a scalar $mu ne 0 $ and a vector $x ne 0$ we have
$Bx= mu x iff det(A)A^-1x= mu x iff det(A)x= mu Ax iff Ax= fracdet(A)mux.$
answered Aug 1 at 11:16
Fred
37k1237
edited Aug 2 at 7:59
answered Aug 1 at 11:16
Fred
37k1237
answered Aug 1 at 11:16
Fred
37k1237
answered Aug 1 at 11:16
Fred
37k1237
37k1237
I think you mean $mathrmdet(A)$ instead of $mathrmdet(a)$.
– zzuussee
Aug 1 at 13:50
Oops, you are right !
– Fred
Aug 2 at 7:59
add a comment |Â
I think you mean $mathrmdet(A)$ instead of $mathrmdet(a)$.
– zzuussee
Aug 1 at 13:50
Oops, you are right !
– Fred
Aug 2 at 7:59
I think you mean $mathrmdet(A)$ instead of $mathrmdet(a)$.
– zzuussee
Aug 1 at 13:50
I think you mean $mathrmdet(A)$ instead of $mathrmdet(a)$.
– zzuussee
Aug 1 at 13:50
Oops, you are right !
– Fred
Aug 2 at 7:59
Oops, you are right !
– Fred
Aug 2 at 7:59
add a comment |Â
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$detleft(det(A)A^-1-fracdet(A)lambdaIright)=det(A)det(A^-1-frac1lambdaI)=det(A)det(A^-1)det(lambda I-A)=(-1)^ndet(A)det(A^-1)det(A-lambda I)=0$. The last equality being because $det(A-lambda I)=0$. The assumption of $A$ invertible allows to know that $lambda neq 0$ and that $A^-1$ exists.
– JessicaMcRae
Aug 1 at 11:07