If $B$ is nilpotent and $AB=BA$ then $det(A+B) = det(A)$ The 2019 Stack Overflow Developer Survey Results Are InSimultaneous Jordanization?Spectrum of the sum of two commuting matricesMatrix determinant lemma with adjugate matrixIf $A$ and $AB-BA$ commute, show that $AB-BA$ is nilpotentUniqueness of sum of nilpotent and diagonalizable matrices.Prove that $det(xI_m-AB)=x^m-ndet(xI_n-BA)$How to show that $D det_A (H)$ exists and equals $det( adj(A)H)$?Suppose $A$ is nilpotent, and $B = c_0I_n + c_1A + cdots + c_m-1A^m-1$, show $det(B) = 0$ iff $c_0 = 0$Show $textdet(A)$ is divisible by $a^n$Addition and multiplication of nilpotent matricesCalculate $det(p(A))$We have to show that $ntimes n$ matrices $A$ and $B$ are nilpotent.$A$ nilpotent and $A+c_iB$ is nilpotent then $B$ is nilpotent.
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If $B$ is nilpotent and $AB=BA$ then $det(A+B) = det(A)$
The 2019 Stack Overflow Developer Survey Results Are InSimultaneous Jordanization?Spectrum of the sum of two commuting matricesMatrix determinant lemma with adjugate matrixIf $A$ and $AB-BA$ commute, show that $AB-BA$ is nilpotentUniqueness of sum of nilpotent and diagonalizable matrices.Prove that $det(xI_m-AB)=x^m-ndet(xI_n-BA)$How to show that $D det_A (H)$ exists and equals $det( adj(A)H)$?Suppose $A$ is nilpotent, and $B = c_0I_n + c_1A + cdots + c_m-1A^m-1$, show $det(B) = 0$ iff $c_0 = 0$Show $textdet(A)$ is divisible by $a^n$Addition and multiplication of nilpotent matricesCalculate $det(p(A))$We have to show that $ntimes n$ matrices $A$ and $B$ are nilpotent.$A$ nilpotent and $A+c_iB$ is nilpotent then $B$ is nilpotent.
$begingroup$
The following stumps me:
Let $mathbb K$ be a field. Let $A, B in mathbb K^n times n$ where $B$ is nilpotent and commutes with $A$, i.e., $A B = B A$. Show that $$ det(A+B)=det(A) $$
I have no idea how to approach this I thought perhaps raise both sides to a power but nothing works.
Thanks for all help.
linear-algebra matrices determinant nilpotence
edited Mar 23 at 21:17
darij grinberg
11.5k33168
asked Mar 23 at 14:14
kronerkroner
1,40331027
$endgroup$
add a comment |
$begingroup$
The following stumps me:
Let $mathbb K$ be a field. Let $A, B in mathbb K^n times n$ where $B$ is nilpotent and commutes with $A$, i.e., $A B = B A$. Show that $$ det(A+B)=det(A) $$
I have no idea how to approach this I thought perhaps raise both sides to a power but nothing works.
Thanks for all help.
linear-algebra matrices determinant nilpotence
edited Mar 23 at 21:17
darij grinberg
11.5k33168
asked Mar 23 at 14:14
kronerkroner
1,40331027
$endgroup$
$begingroup$
Hint: First prove it when $A$ is invertible (in which case $BA^-1$ is nilpotent and thus $detleft(I_n + BA^-1right) = 1$). Then, argue that in the general case, you can replace $A$ by an invertible matrix $xI_n + A$ over the ring of Laurent power series in an indeterminate $x^-1$ over $mathbbK$. This new matrix $xI_n + A$ still commutes with $B$, and turns into $A$ if you substitute $x = 0$ (which you can do, because all entries are actually polynomials in $x$). I did something like that in math.stackexchange.com/questions/1514408//1847767#1847767 (albeit with $xI_n - A$).
$endgroup$
– darij grinberg
Mar 23 at 20:57
$begingroup$
This question is a special case of MSE question 19464 "Spectrum of the sum of two commuting matrices".
$endgroup$
– Somos
Mar 24 at 19:30
$begingroup$
@Somos: Not clear. We're talking about spectra with multiplicities, while the question you are citing is talking about spectra as sets. And the "WLOG all eigenvalues distinct" trick doesn't work here, since the problem space is not an affine variety (I don't think it is even an irreducible variety).
$endgroup$
– darij grinberg
Mar 24 at 22:54
add a comment |
$begingroup$
The following stumps me:
Let $mathbb K$ be a field. Let $A, B in mathbb K^n times n$ where $B$ is nilpotent and commutes with $A$, i.e., $A B = B A$. Show that $$ det(A+B)=det(A) $$
I have no idea how to approach this I thought perhaps raise both sides to a power but nothing works.
Thanks for all help.
linear-algebra matrices determinant nilpotence
edited Mar 23 at 21:17
darij grinberg
11.5k33168
asked Mar 23 at 14:14
kronerkroner
1,40331027
$endgroup$
The following stumps me:
Let $mathbb K$ be a field. Let $A, B in mathbb K^n times n$ where $B$ is nilpotent and commutes with $A$, i.e., $A B = B A$. Show that $$ det(A+B)=det(A) $$
I have no idea how to approach this I thought perhaps raise both sides to a power but nothing works.
Thanks for all help.
linear-algebra matrices determinant nilpotence
linear-algebra matrices determinant nilpotence
edited Mar 23 at 21:17
darij grinberg
11.5k33168
asked Mar 23 at 14:14
kronerkroner
1,40331027
edited Mar 23 at 21:17
darij grinberg
11.5k33168
asked Mar 23 at 14:14
kronerkroner
1,40331027
edited Mar 23 at 21:17
darij grinberg
11.5k33168
edited Mar 23 at 21:17
darij grinberg
11.5k33168
edited Mar 23 at 21:17
darij grinberg
11.5k33168
11.5k33168
asked Mar 23 at 14:14
kronerkroner
1,40331027
asked Mar 23 at 14:14
kronerkroner
1,40331027
asked Mar 23 at 14:14
kronerkroner
1,40331027
1,40331027
$begingroup$
Hint: First prove it when $A$ is invertible (in which case $BA^-1$ is nilpotent and thus $detleft(I_n + BA^-1right) = 1$). Then, argue that in the general case, you can replace $A$ by an invertible matrix $xI_n + A$ over the ring of Laurent power series in an indeterminate $x^-1$ over $mathbbK$. This new matrix $xI_n + A$ still commutes with $B$, and turns into $A$ if you substitute $x = 0$ (which you can do, because all entries are actually polynomials in $x$). I did something like that in math.stackexchange.com/questions/1514408//1847767#1847767 (albeit with $xI_n - A$).
$endgroup$
– darij grinberg
Mar 23 at 20:57
$begingroup$
This question is a special case of MSE question 19464 "Spectrum of the sum of two commuting matrices".
$endgroup$
– Somos
Mar 24 at 19:30
$begingroup$
@Somos: Not clear. We're talking about spectra with multiplicities, while the question you are citing is talking about spectra as sets. And the "WLOG all eigenvalues distinct" trick doesn't work here, since the problem space is not an affine variety (I don't think it is even an irreducible variety).
$endgroup$
– darij grinberg
Mar 24 at 22:54
add a comment |
$begingroup$
Hint: First prove it when $A$ is invertible (in which case $BA^-1$ is nilpotent and thus $detleft(I_n + BA^-1right) = 1$). Then, argue that in the general case, you can replace $A$ by an invertible matrix $xI_n + A$ over the ring of Laurent power series in an indeterminate $x^-1$ over $mathbbK$. This new matrix $xI_n + A$ still commutes with $B$, and turns into $A$ if you substitute $x = 0$ (which you can do, because all entries are actually polynomials in $x$). I did something like that in math.stackexchange.com/questions/1514408//1847767#1847767 (albeit with $xI_n - A$).
$endgroup$
– darij grinberg
Mar 23 at 20:57
$begingroup$
This question is a special case of MSE question 19464 "Spectrum of the sum of two commuting matrices".
$endgroup$
– Somos
Mar 24 at 19:30
$begingroup$
@Somos: Not clear. We're talking about spectra with multiplicities, while the question you are citing is talking about spectra as sets. And the "WLOG all eigenvalues distinct" trick doesn't work here, since the problem space is not an affine variety (I don't think it is even an irreducible variety).
$endgroup$
– darij grinberg
Mar 24 at 22:54
$begingroup$
Hint: First prove it when $A$ is invertible (in which case $BA^-1$ is nilpotent and thus $detleft(I_n + BA^-1right) = 1$). Then, argue that in the general case, you can replace $A$ by an invertible matrix $xI_n + A$ over the ring of Laurent power series in an indeterminate $x^-1$ over $mathbbK$. This new matrix $xI_n + A$ still commutes with $B$, and turns into $A$ if you substitute $x = 0$ (which you can do, because all entries are actually polynomials in $x$). I did something like that in math.stackexchange.com/questions/1514408//1847767#1847767 (albeit with $xI_n - A$).
$endgroup$
– darij grinberg
Mar 23 at 20:57
$begingroup$
Hint: First prove it when $A$ is invertible (in which case $BA^-1$ is nilpotent and thus $detleft(I_n + BA^-1right) = 1$). Then, argue that in the general case, you can replace $A$ by an invertible matrix $xI_n + A$ over the ring of Laurent power series in an indeterminate $x^-1$ over $mathbbK$. This new matrix $xI_n + A$ still commutes with $B$, and turns into $A$ if you substitute $x = 0$ (which you can do, because all entries are actually polynomials in $x$). I did something like that in math.stackexchange.com/questions/1514408//1847767#1847767 (albeit with $xI_n - A$).
$endgroup$
– darij grinberg
Mar 23 at 20:57
$begingroup$
This question is a special case of MSE question 19464 "Spectrum of the sum of two commuting matrices".
$endgroup$
– Somos
Mar 24 at 19:30
$begingroup$
This question is a special case of MSE question 19464 "Spectrum of the sum of two commuting matrices".
$endgroup$
– Somos
Mar 24 at 19:30
$begingroup$
@Somos: Not clear. We're talking about spectra with multiplicities, while the question you are citing is talking about spectra as sets. And the "WLOG all eigenvalues distinct" trick doesn't work here, since the problem space is not an affine variety (I don't think it is even an irreducible variety).
$endgroup$
– darij grinberg
Mar 24 at 22:54
$begingroup$
@Somos: Not clear. We're talking about spectra with multiplicities, while the question you are citing is talking about spectra as sets. And the "WLOG all eigenvalues distinct" trick doesn't work here, since the problem space is not an affine variety (I don't think it is even an irreducible variety).
$endgroup$
– darij grinberg
Mar 24 at 22:54
add a comment |
1 Answer
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$begingroup$
Here's a way to do it without the machinery of JNF or triangulization, although that machinery is absolutely worth learning. You really only need to know that $det(M) = 0$ if and only if there exists $xne 0$ such that $Mx = 0$ (for an $ntimes n$ matrix $M$), that $det$ is multiplicative, and that $det(M)$ is the product of the roots (in an algebraic closure) of the characteristic polynomial of $M$.
Let $m>0$ be such that $B^m = 0$. Since $A$ and $B$ commute, we may expand as follows
beginequationtag$*$(A+B)^m = sum_i=0^m binommi B^m-i A^i.endequation
Suppose that $det(A) = 0$, and let $vne 0$ be such that $Av = 0$. Observe that $(A+B)^m(v) = 0$, since each term in the expansion $(*)$ kills $v$: for $0< i le m$, we have $B^m-i(A^i(v)) = B^m-iA^i-1(Av) = 0$, and for $i=0$, we have $B^m(v) = 0$ since $B^m = 0$. Therefore $0ne vin ker (A+B)^m$, so $(det(A+B))^m = det((A+B)^m) = 0$, so $det(A+B)=0=det(A)$.
Now suppose that $det(A) ne 0$. Let $C = A^-1B$. It suffices to show that $det(I+C) = 1$. Let $lambda in bar K$ be a root of the characteristic polynomial of $I+C$, so that $det((1-lambda) I + C) = det(I + C - lambda I) = 0$. Since $A$ commutes with $B$, so does $A^-1$, and thus $-C = -A^-1B$ is nilpotent. Now since $-C$ is nilpotent and commutes with $(1-lambda)I + C$, which has determinant $0$, by the above paragraph we have
$$(1-lambda)^n = det((1-lambda)I) = det(((1-lambda)I+C)+(-C)) = det((1-lambda)I+C) = 0,$$
so $lambda = 1$. Therefore, the characteristic polynomial of $I+C$ has only $1$ as a root, and thus $det(I+C) = 1$.
answered Mar 23 at 15:15
cspruncsprun
2,804211
$endgroup$
add a comment |
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$begingroup$
Here's a way to do it without the machinery of JNF or triangulization, although that machinery is absolutely worth learning. You really only need to know that $det(M) = 0$ if and only if there exists $xne 0$ such that $Mx = 0$ (for an $ntimes n$ matrix $M$), that $det$ is multiplicative, and that $det(M)$ is the product of the roots (in an algebraic closure) of the characteristic polynomial of $M$.
Let $m>0$ be such that $B^m = 0$. Since $A$ and $B$ commute, we may expand as follows
beginequationtag$*$(A+B)^m = sum_i=0^m binommi B^m-i A^i.endequation
Suppose that $det(A) = 0$, and let $vne 0$ be such that $Av = 0$. Observe that $(A+B)^m(v) = 0$, since each term in the expansion $(*)$ kills $v$: for $0< i le m$, we have $B^m-i(A^i(v)) = B^m-iA^i-1(Av) = 0$, and for $i=0$, we have $B^m(v) = 0$ since $B^m = 0$. Therefore $0ne vin ker (A+B)^m$, so $(det(A+B))^m = det((A+B)^m) = 0$, so $det(A+B)=0=det(A)$.
Now suppose that $det(A) ne 0$. Let $C = A^-1B$. It suffices to show that $det(I+C) = 1$. Let $lambda in bar K$ be a root of the characteristic polynomial of $I+C$, so that $det((1-lambda) I + C) = det(I + C - lambda I) = 0$. Since $A$ commutes with $B$, so does $A^-1$, and thus $-C = -A^-1B$ is nilpotent. Now since $-C$ is nilpotent and commutes with $(1-lambda)I + C$, which has determinant $0$, by the above paragraph we have
$$(1-lambda)^n = det((1-lambda)I) = det(((1-lambda)I+C)+(-C)) = det((1-lambda)I+C) = 0,$$
so $lambda = 1$. Therefore, the characteristic polynomial of $I+C$ has only $1$ as a root, and thus $det(I+C) = 1$.
answered Mar 23 at 15:15
cspruncsprun
2,804211
$endgroup$
add a comment |
$begingroup$
Here's a way to do it without the machinery of JNF or triangulization, although that machinery is absolutely worth learning. You really only need to know that $det(M) = 0$ if and only if there exists $xne 0$ such that $Mx = 0$ (for an $ntimes n$ matrix $M$), that $det$ is multiplicative, and that $det(M)$ is the product of the roots (in an algebraic closure) of the characteristic polynomial of $M$.
Let $m>0$ be such that $B^m = 0$. Since $A$ and $B$ commute, we may expand as follows
beginequationtag$*$(A+B)^m = sum_i=0^m binommi B^m-i A^i.endequation
Suppose that $det(A) = 0$, and let $vne 0$ be such that $Av = 0$. Observe that $(A+B)^m(v) = 0$, since each term in the expansion $(*)$ kills $v$: for $0< i le m$, we have $B^m-i(A^i(v)) = B^m-iA^i-1(Av) = 0$, and for $i=0$, we have $B^m(v) = 0$ since $B^m = 0$. Therefore $0ne vin ker (A+B)^m$, so $(det(A+B))^m = det((A+B)^m) = 0$, so $det(A+B)=0=det(A)$.
Now suppose that $det(A) ne 0$. Let $C = A^-1B$. It suffices to show that $det(I+C) = 1$. Let $lambda in bar K$ be a root of the characteristic polynomial of $I+C$, so that $det((1-lambda) I + C) = det(I + C - lambda I) = 0$. Since $A$ commutes with $B$, so does $A^-1$, and thus $-C = -A^-1B$ is nilpotent. Now since $-C$ is nilpotent and commutes with $(1-lambda)I + C$, which has determinant $0$, by the above paragraph we have
$$(1-lambda)^n = det((1-lambda)I) = det(((1-lambda)I+C)+(-C)) = det((1-lambda)I+C) = 0,$$
so $lambda = 1$. Therefore, the characteristic polynomial of $I+C$ has only $1$ as a root, and thus $det(I+C) = 1$.
answered Mar 23 at 15:15
cspruncsprun
2,804211
$endgroup$
add a comment |
$begingroup$
Here's a way to do it without the machinery of JNF or triangulization, although that machinery is absolutely worth learning. You really only need to know that $det(M) = 0$ if and only if there exists $xne 0$ such that $Mx = 0$ (for an $ntimes n$ matrix $M$), that $det$ is multiplicative, and that $det(M)$ is the product of the roots (in an algebraic closure) of the characteristic polynomial of $M$.
Let $m>0$ be such that $B^m = 0$. Since $A$ and $B$ commute, we may expand as follows
beginequationtag$*$(A+B)^m = sum_i=0^m binommi B^m-i A^i.endequation
Suppose that $det(A) = 0$, and let $vne 0$ be such that $Av = 0$. Observe that $(A+B)^m(v) = 0$, since each term in the expansion $(*)$ kills $v$: for $0< i le m$, we have $B^m-i(A^i(v)) = B^m-iA^i-1(Av) = 0$, and for $i=0$, we have $B^m(v) = 0$ since $B^m = 0$. Therefore $0ne vin ker (A+B)^m$, so $(det(A+B))^m = det((A+B)^m) = 0$, so $det(A+B)=0=det(A)$.
Now suppose that $det(A) ne 0$. Let $C = A^-1B$. It suffices to show that $det(I+C) = 1$. Let $lambda in bar K$ be a root of the characteristic polynomial of $I+C$, so that $det((1-lambda) I + C) = det(I + C - lambda I) = 0$. Since $A$ commutes with $B$, so does $A^-1$, and thus $-C = -A^-1B$ is nilpotent. Now since $-C$ is nilpotent and commutes with $(1-lambda)I + C$, which has determinant $0$, by the above paragraph we have
$$(1-lambda)^n = det((1-lambda)I) = det(((1-lambda)I+C)+(-C)) = det((1-lambda)I+C) = 0,$$
so $lambda = 1$. Therefore, the characteristic polynomial of $I+C$ has only $1$ as a root, and thus $det(I+C) = 1$.
answered Mar 23 at 15:15
cspruncsprun
2,804211
$endgroup$
Here's a way to do it without the machinery of JNF or triangulization, although that machinery is absolutely worth learning. You really only need to know that $det(M) = 0$ if and only if there exists $xne 0$ such that $Mx = 0$ (for an $ntimes n$ matrix $M$), that $det$ is multiplicative, and that $det(M)$ is the product of the roots (in an algebraic closure) of the characteristic polynomial of $M$.
Let $m>0$ be such that $B^m = 0$. Since $A$ and $B$ commute, we may expand as follows
beginequationtag$*$(A+B)^m = sum_i=0^m binommi B^m-i A^i.endequation
Suppose that $det(A) = 0$, and let $vne 0$ be such that $Av = 0$. Observe that $(A+B)^m(v) = 0$, since each term in the expansion $(*)$ kills $v$: for $0< i le m$, we have $B^m-i(A^i(v)) = B^m-iA^i-1(Av) = 0$, and for $i=0$, we have $B^m(v) = 0$ since $B^m = 0$. Therefore $0ne vin ker (A+B)^m$, so $(det(A+B))^m = det((A+B)^m) = 0$, so $det(A+B)=0=det(A)$.
Now suppose that $det(A) ne 0$. Let $C = A^-1B$. It suffices to show that $det(I+C) = 1$. Let $lambda in bar K$ be a root of the characteristic polynomial of $I+C$, so that $det((1-lambda) I + C) = det(I + C - lambda I) = 0$. Since $A$ commutes with $B$, so does $A^-1$, and thus $-C = -A^-1B$ is nilpotent. Now since $-C$ is nilpotent and commutes with $(1-lambda)I + C$, which has determinant $0$, by the above paragraph we have
$$(1-lambda)^n = det((1-lambda)I) = det(((1-lambda)I+C)+(-C)) = det((1-lambda)I+C) = 0,$$
so $lambda = 1$. Therefore, the characteristic polynomial of $I+C$ has only $1$ as a root, and thus $det(I+C) = 1$.
answered Mar 23 at 15:15
cspruncsprun
2,804211
edited Mar 24 at 13:34
answered Mar 23 at 15:15
cspruncsprun
2,804211
answered Mar 23 at 15:15
cspruncsprun
2,804211
answered Mar 23 at 15:15
cspruncsprun
2,804211
2,804211
add a comment |
add a comment |
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Required, but never shown
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Hint: First prove it when $A$ is invertible (in which case $BA^-1$ is nilpotent and thus $detleft(I_n + BA^-1right) = 1$). Then, argue that in the general case, you can replace $A$ by an invertible matrix $xI_n + A$ over the ring of Laurent power series in an indeterminate $x^-1$ over $mathbbK$. This new matrix $xI_n + A$ still commutes with $B$, and turns into $A$ if you substitute $x = 0$ (which you can do, because all entries are actually polynomials in $x$). I did something like that in math.stackexchange.com/questions/1514408//1847767#1847767 (albeit with $xI_n - A$).
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– darij grinberg
Mar 23 at 20:57
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This question is a special case of MSE question 19464 "Spectrum of the sum of two commuting matrices".
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– Somos
Mar 24 at 19:30
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@Somos: Not clear. We're talking about spectra with multiplicities, while the question you are citing is talking about spectra as sets. And the "WLOG all eigenvalues distinct" trick doesn't work here, since the problem space is not an affine variety (I don't think it is even an irreducible variety).
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– darij grinberg
Mar 24 at 22:54