If $AB-BA=A_* $, then $A_* ^2=O_n$Is adjoint of singular matrix singular? What would be its rank?If $A^2+2A+I_n=O_n$ then $A$ is invertibleProblem on matrices : $dim Eleq n^2-(n-r)^2-1$X,AX have no common eigenvalues$detbeginbmatrixdet A & det B \ det C & det Dendbmatrix=0$If $ABC = 0$ and rank$(B) = 1$, then either $AB = 0$ or $BC = 0$.Proof of $AB=BA=O_n$ when $ A^2+AB+B^2 = 2BA$Isomorphisms between $O_n$ and the direct product of $SO_n$ with another group$X^m=O_n$ implies $X^n=O_n$.Find $A$ and $B$ so that $ operatornameTr(AB) ^* =0$$(AB) ^3=O_n$ and$(BA) ^3 neq O_n$
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If $AB-BA=A_* $, then $A_* ^2=O_n$
Is adjoint of singular matrix singular? What would be its rank?If $A^2+2A+I_n=O_n$ then $A$ is invertibleProblem on matrices : $dim Eleq n^2-(n-r)^2-1$X,AX have no common eigenvalues$detbeginbmatrixdet A & det B \ det C & det Dendbmatrix=0$If $ABC = 0$ and rank$(B) = 1$, then either $AB = 0$ or $BC = 0$.Proof of $AB=BA=O_n$ when $ A^2+AB+B^2 = 2BA$Isomorphisms between $O_n$ and the direct product of $SO_n$ with another group$X^m=O_n$ implies $X^n=O_n$.Find $A$ and $B$ so that $ operatornameTr(AB) ^* =0$$(AB) ^3=O_n$ and$(BA) ^3 neq O_n$
$begingroup$
Let $A, B in M_n(mathbbC) $. If $AB-BA=A_* $, then $A_* ^2=O_n$.
So far, I have considered the cases when $rank(A) le n-2$ and $rank(A) =n-1$ and in both of them I obtained the conclusion. My problem is that I don't know what to do if $rank A=n$.
linear-algebra matrices
asked Mar 22 at 15:04
MathEnthusiastMathEnthusiast
46013
$endgroup$
add a comment |
$begingroup$
Let $A, B in M_n(mathbbC) $. If $AB-BA=A_* $, then $A_* ^2=O_n$.
So far, I have considered the cases when $rank(A) le n-2$ and $rank(A) =n-1$ and in both of them I obtained the conclusion. My problem is that I don't know what to do if $rank A=n$.
linear-algebra matrices
asked Mar 22 at 15:04
MathEnthusiastMathEnthusiast
46013
$endgroup$
1
$begingroup$
If $A_*$ is meant to be the conjugate transpose of $A$, it would be better to say so and/or to use the more standard notation $A^*$.
$endgroup$
– hardmath
Mar 22 at 15:16
$begingroup$
$A_* $ is the adjugate matrix of $A$ i.e. $Acdot A_* =det A cdot I_n$
$endgroup$
– MathEnthusiast
Mar 22 at 15:49
add a comment |
$begingroup$
Let $A, B in M_n(mathbbC) $. If $AB-BA=A_* $, then $A_* ^2=O_n$.
So far, I have considered the cases when $rank(A) le n-2$ and $rank(A) =n-1$ and in both of them I obtained the conclusion. My problem is that I don't know what to do if $rank A=n$.
linear-algebra matrices
asked Mar 22 at 15:04
MathEnthusiastMathEnthusiast
46013
$endgroup$
Let $A, B in M_n(mathbbC) $. If $AB-BA=A_* $, then $A_* ^2=O_n$.
So far, I have considered the cases when $rank(A) le n-2$ and $rank(A) =n-1$ and in both of them I obtained the conclusion. My problem is that I don't know what to do if $rank A=n$.
linear-algebra matrices
linear-algebra matrices
asked Mar 22 at 15:04
MathEnthusiastMathEnthusiast
46013
asked Mar 22 at 15:04
MathEnthusiastMathEnthusiast
46013
asked Mar 22 at 15:04
MathEnthusiastMathEnthusiast
46013
asked Mar 22 at 15:04
MathEnthusiastMathEnthusiast
46013
asked Mar 22 at 15:04
MathEnthusiastMathEnthusiast
46013
46013
1
$begingroup$
If $A_*$ is meant to be the conjugate transpose of $A$, it would be better to say so and/or to use the more standard notation $A^*$.
$endgroup$
– hardmath
Mar 22 at 15:16
$begingroup$
$A_* $ is the adjugate matrix of $A$ i.e. $Acdot A_* =det A cdot I_n$
$endgroup$
– MathEnthusiast
Mar 22 at 15:49
add a comment |
1
$begingroup$
If $A_*$ is meant to be the conjugate transpose of $A$, it would be better to say so and/or to use the more standard notation $A^*$.
$endgroup$
– hardmath
Mar 22 at 15:16
$begingroup$
$A_* $ is the adjugate matrix of $A$ i.e. $Acdot A_* =det A cdot I_n$
$endgroup$
– MathEnthusiast
Mar 22 at 15:49
1
1
$begingroup$
If $A_*$ is meant to be the conjugate transpose of $A$, it would be better to say so and/or to use the more standard notation $A^*$.
$endgroup$
– hardmath
Mar 22 at 15:16
$begingroup$
If $A_*$ is meant to be the conjugate transpose of $A$, it would be better to say so and/or to use the more standard notation $A^*$.
$endgroup$
– hardmath
Mar 22 at 15:16
$begingroup$
$A_* $ is the adjugate matrix of $A$ i.e. $Acdot A_* =det A cdot I_n$
$endgroup$
– MathEnthusiast
Mar 22 at 15:49
$begingroup$
$A_* $ is the adjugate matrix of $A$ i.e. $Acdot A_* =det A cdot I_n$
$endgroup$
– MathEnthusiast
Mar 22 at 15:49
add a comment |
1 Answer
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$begingroup$
$[A,B]=operatornameadj(A)$ commutes with $A$. Hence it is nilpotent, by Jacobson's lemma. Now the result follows, because the rank of every singular adjugate matix is at most $1$ and every nilpotent matrix of rank $le1$ must square to zero.
answered Mar 22 at 18:11
user1551user1551
74.2k566129
$endgroup$
add a comment |
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$begingroup$
$[A,B]=operatornameadj(A)$ commutes with $A$. Hence it is nilpotent, by Jacobson's lemma. Now the result follows, because the rank of every singular adjugate matix is at most $1$ and every nilpotent matrix of rank $le1$ must square to zero.
answered Mar 22 at 18:11
user1551user1551
74.2k566129
$endgroup$
add a comment |
$begingroup$
$[A,B]=operatornameadj(A)$ commutes with $A$. Hence it is nilpotent, by Jacobson's lemma. Now the result follows, because the rank of every singular adjugate matix is at most $1$ and every nilpotent matrix of rank $le1$ must square to zero.
answered Mar 22 at 18:11
user1551user1551
74.2k566129
$endgroup$
add a comment |
$begingroup$
$[A,B]=operatornameadj(A)$ commutes with $A$. Hence it is nilpotent, by Jacobson's lemma. Now the result follows, because the rank of every singular adjugate matix is at most $1$ and every nilpotent matrix of rank $le1$ must square to zero.
answered Mar 22 at 18:11
user1551user1551
74.2k566129
$endgroup$
$[A,B]=operatornameadj(A)$ commutes with $A$. Hence it is nilpotent, by Jacobson's lemma. Now the result follows, because the rank of every singular adjugate matix is at most $1$ and every nilpotent matrix of rank $le1$ must square to zero.
answered Mar 22 at 18:11
user1551user1551
74.2k566129
answered Mar 22 at 18:11
user1551user1551
74.2k566129
answered Mar 22 at 18:11
user1551user1551
74.2k566129
answered Mar 22 at 18:11
user1551user1551
74.2k566129
74.2k566129
add a comment |
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$begingroup$
If $A_*$ is meant to be the conjugate transpose of $A$, it would be better to say so and/or to use the more standard notation $A^*$.
$endgroup$
– hardmath
Mar 22 at 15:16
$begingroup$
$A_* $ is the adjugate matrix of $A$ i.e. $Acdot A_* =det A cdot I_n$
$endgroup$
– MathEnthusiast
Mar 22 at 15:49