How to proceed with following determinant inequalityDeterminant of complex bordered matrixProving a determinant inequalityHow does one prove the determinant inequality $detleft(6(A^3+B^3+C^3)+I_nright)ge 5^ndet(A^2+B^2+C^2)$?How to prove the following determinant identity?Show determinant of $left[beginmatrix A & 0 \ C & Dendmatrixright] = detAcdot detD$Problem with determinantMatrix Determinant inequality.Elementary InequalityAn inequality with determinantShort inequality involving determinants
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How to proceed with following determinant inequality
Determinant of complex bordered matrixProving a determinant inequalityHow does one prove the determinant inequality $detleft(6(A^3+B^3+C^3)+I_nright)ge 5^ndet(A^2+B^2+C^2)$?How to prove the following determinant identity?Show determinant of $left[beginmatrix A & 0 \ C & Dendmatrixright] = detAcdot detD$Problem with determinantMatrix Determinant inequality.Elementary InequalityAn inequality with determinantShort inequality involving determinants
$begingroup$
Let $A,Bin M_n(mathbbR)$ be such that $B^2=I_n$ and $A^2=AB+I_n$. Prove that $$det(A)leqleft(frac1+sqrt52right)^n$$
I have been able to show that $AB=BA$, $B=A-A^-1$ and $A^4-3A^2+I=0$. Now from this how I can approach the problem.
linear-algebra inequality determinant
asked May 7 at 13:23
J.DoeJ.Doe
4618
$endgroup$
add a comment |
$begingroup$
Let $A,Bin M_n(mathbbR)$ be such that $B^2=I_n$ and $A^2=AB+I_n$. Prove that $$det(A)leqleft(frac1+sqrt52right)^n$$
I have been able to show that $AB=BA$, $B=A-A^-1$ and $A^4-3A^2+I=0$. Now from this how I can approach the problem.
linear-algebra inequality determinant
asked May 7 at 13:23
J.DoeJ.Doe
4618
$endgroup$
add a comment |
$begingroup$
Let $A,Bin M_n(mathbbR)$ be such that $B^2=I_n$ and $A^2=AB+I_n$. Prove that $$det(A)leqleft(frac1+sqrt52right)^n$$
I have been able to show that $AB=BA$, $B=A-A^-1$ and $A^4-3A^2+I=0$. Now from this how I can approach the problem.
linear-algebra inequality determinant
asked May 7 at 13:23
J.DoeJ.Doe
4618
$endgroup$
Let $A,Bin M_n(mathbbR)$ be such that $B^2=I_n$ and $A^2=AB+I_n$. Prove that $$det(A)leqleft(frac1+sqrt52right)^n$$
I have been able to show that $AB=BA$, $B=A-A^-1$ and $A^4-3A^2+I=0$. Now from this how I can approach the problem.
linear-algebra inequality determinant
linear-algebra inequality determinant
asked May 7 at 13:23
J.DoeJ.Doe
4618
asked May 7 at 13:23
J.DoeJ.Doe
4618
edited May 7 at 13:31
J.Doe
asked May 7 at 13:23
J.DoeJ.Doe
4618
asked May 7 at 13:23
J.DoeJ.Doe
4618
asked May 7 at 13:23
J.DoeJ.Doe
4618
4618
add a comment |
add a comment |
1 Answer
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$begingroup$
If $A^4-3A^2+I=0$ than any eigenvalue of $A$ should satisfy $lambda^4-3lambda^2+1=0$. This is a biquadratic equation, solving which we get 4 roots: $pmfrac1+sqrt52, pmfrac1-sqrt52$. $det(A)$ is a product of its eigenvalues, so it will be biigest when all of them will be equal to $frac1+sqrt52$.
answered May 7 at 14:06
AO1992AO1992
5038
$endgroup$
add a comment |
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$begingroup$
If $A^4-3A^2+I=0$ than any eigenvalue of $A$ should satisfy $lambda^4-3lambda^2+1=0$. This is a biquadratic equation, solving which we get 4 roots: $pmfrac1+sqrt52, pmfrac1-sqrt52$. $det(A)$ is a product of its eigenvalues, so it will be biigest when all of them will be equal to $frac1+sqrt52$.
answered May 7 at 14:06
AO1992AO1992
5038
$endgroup$
add a comment |
$begingroup$
If $A^4-3A^2+I=0$ than any eigenvalue of $A$ should satisfy $lambda^4-3lambda^2+1=0$. This is a biquadratic equation, solving which we get 4 roots: $pmfrac1+sqrt52, pmfrac1-sqrt52$. $det(A)$ is a product of its eigenvalues, so it will be biigest when all of them will be equal to $frac1+sqrt52$.
answered May 7 at 14:06
AO1992AO1992
5038
$endgroup$
add a comment |
$begingroup$
If $A^4-3A^2+I=0$ than any eigenvalue of $A$ should satisfy $lambda^4-3lambda^2+1=0$. This is a biquadratic equation, solving which we get 4 roots: $pmfrac1+sqrt52, pmfrac1-sqrt52$. $det(A)$ is a product of its eigenvalues, so it will be biigest when all of them will be equal to $frac1+sqrt52$.
answered May 7 at 14:06
AO1992AO1992
5038
$endgroup$
If $A^4-3A^2+I=0$ than any eigenvalue of $A$ should satisfy $lambda^4-3lambda^2+1=0$. This is a biquadratic equation, solving which we get 4 roots: $pmfrac1+sqrt52, pmfrac1-sqrt52$. $det(A)$ is a product of its eigenvalues, so it will be biigest when all of them will be equal to $frac1+sqrt52$.
answered May 7 at 14:06
AO1992AO1992
5038
answered May 7 at 14:06
AO1992AO1992
5038
answered May 7 at 14:06
AO1992AO1992
5038
answered May 7 at 14:06
AO1992AO1992
5038
5038
add a comment |
add a comment |
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