[Fernando Revilla]

Observe that \( \displaystyle e^A:=I+A+\frac{A^2}2 \). Thus \( \det(A^2/2+A+I)=\det(e^A)=e^{\operatorname{tr}(A)}=e^0=1 \). The other case is almost identical, just replace \( A \) by \( -A \) in the exponential.

An elegant solution. Now the question for Jacks is: What topics are covered in those olympiad problems? Can we use, for example the exponential matrix concept?

[Luis Fuentes]

Hi

 Another way. Take:

\(  f(t)=det\left(\dfrac{(At)^2}{2}+(At)+Id\right) \)

 Note that \( f(t) \) is a polynomial. Using that \( A^3=0 \) check that:

\(  f(t)^2=det(2(At)^2+2(At)+Id)=f(2t) \)

But:

\(  deg(f(t))=deg(f(2t))=deg(f(t)^2)=2deg(f(t)) \)

 so \( deg(f(t))=0 \), that is, \( f(t) \) is constant:

\( f(1)=f(-1)=f(0)=det(Id)=1 \).

Best regards.

[Fernando Revilla]

Another way. Take: \(  f(t)=det\left(\dfrac{(At)^2}{2}+(At)+Id\right) \) ....

Good, surely that is a convenient method for an olympiad problem. All the properties used are usually covered in the corresponding syllabus.

[Abdulai]

Siendo  \( A^3=0\;\;\longrightarrow\;\; \left(I+\frac{1}{2}A\right)\left(I-\frac{1}{2}A\right)^{-1}=I+A+\frac{1}{2}A^2 \)

fácil de demostrar expandiendo   \( \left(I+\frac{1}{2}A\right)  = \left(I+A+\frac{1}{2}A^2\right)\left(I-\frac{1}{2}A\right) \)

\( \therefore\quad \text{det}\left(I+A+\frac{1}{2}A^2\right)= \dfrac{\text{det}\left(I+\frac{1}{2}A\right)}{\text{det}\left(I-\frac{1}{2}A\right)} \)

Una conocida propiedad de las matrices nilpotentes es que  \( \text{det}\left(I+N\right) = 1\quad ,\quad N \) nilpotente.

Por lo tanto  \( \text{det}\left(I+A+\frac{1}{2}A^2\right)= \dfrac{1}{1} \)


Lo mismo para el otro caso  \( \text{det}\left(I-A+\frac{1}{2}A^2\right)= \dfrac{\text{det}\left(I-\frac{1}{2}A\right)}{\text{det}\left(I+\frac{1}{2}A\right)}=1 \)

[Masacroso]

Other way, more elementary in my opinion: because they are polynomials it is enough to decompose the polynomials in it factors and, due to the fact that the determinant is a multiplicative function, then we can achieve the same result, that is

\( \displaystyle \det(A^2/2+A+I)=\frac1{2^3}\det(A^2+2A+2I)=\frac1{8}\det((A-\lambda I)(A-\bar\lambda I))=\frac18\det(A-\lambda I)\det(A-\bar\lambda I)=\frac18\lambda^3\bar\lambda^3=\frac{|\lambda|^6}8=1 \)

for \( \lambda \) and \( \bar\lambda \) the two roots of the polynomial \( x^2+2x+2 \), that is \( \lambda=-1+i \) and \( \bar\lambda =-1-i \), then clearly \( |\lambda|^2=\lambda\cdot\bar\lambda=2 \) and the result follows.

The important step is to note that \( \det(A-\lambda I)=-\lambda ^3 \) because a nilpotent matrix can be transformed, under a change of basis, in an upper-triangular matrix where the diagonal is full of zeros. A change of basis doesnt change the value of the determinant so we can see \( A \) in the form that we please.

[Luis Fuentes]

Hola

The important step is notice that \( \det(A-\lambda I)=-\lambda ^3 \) because a nilpotent matrix can be transformed, under a change of basis, in an upper-triangular matrix where the diagonal is full of zeros. A change of basis doesnt change the value of the determinant so we can see \( A \) in the form that we please.

Here we are using a "Jordan matrix/upper triangular matrix" type argument again.

The result \( A \) nilpotent implies \( det(A+I)=1 \) can be proved again playing with the polynomial \( f(t)=det(At+I) \) and its powers \( f(t)^n \).   

Best regards.

[jacks]

Thanks Moderator, Admin, Masacraso got