[jacks]

if \( A \) is a \( 3\times 3 \) matrix such that \( A^3=O \).Then

 \( |\frac{A^2}{2}+A+I| \) and \( |\frac{A^2}{2}-A+I| \)

[Fernando Revilla]

if \( A \) is a \( 3\times 3 \) matrix such that \( A^3=O \).Then \( |\frac{A^2}{2}+A+I| \) and \( |\frac{A^2}{2}-A+I| \)

One way: if \( A^3=0 \) then, the possiible minimal polynomials of \( A \) are \( \mu_i(\lambda)=\lambda^i \), \( i=1,2,3 \) with respective canonical Jordan forms

          \( J_1=\begin{bmatrix}{0}&{}&{}\\{}&{0}&{}\\{}&{}&{0}\end{bmatrix},\quad J_2=\begin{bmatrix}{0}&{1}&{}\\{0}&{0}&{}\\{}&{}&{0}\end{bmatrix},\quad J_3=\begin{bmatrix}{0}&{0}&{1}\\{0}&{0}&{0}\\{0}&{0}&{0}\end{bmatrix}. \)

Then, there exists \( P \) invertible matrix such that \( A=PJ_iP^{-1} \) so,

          \( \frac{1}{2}A^2+A+I=P\left(\frac{1}{2}J_i^2+J_i+I\right)P^{-1} \)
          \( \frac{1}{2}A^2-A+I=P\left(\frac{1}{2}J_i^2-J_i+I\right)P^{-1} \)

But for all \( i=1,2,3 \) we have

          \( \frac{1}{2}J_i^2+J_i+I=\begin{bmatrix}{1}&{*}&{*}\\{0}&{1}&{*}\\{0}&{0}&{1}\end{bmatrix},\quad \frac{1}{2}J_i^2-J_i+I=\begin{bmatrix}{1}&{*}&{*}\\{0}&{1}&{*}\\{0}&{0}&{1}\end{bmatrix}. \)

As similar matrices have the same determinant,

         \( \left | \frac{1}{2}A^2+A+I \right |=\left |{\frac{1}{2}J_i^2+J_i+I}\right |=1,\quad \left | \frac{1}{2}A^2-A+I \right |=\left |{\frac{1}{2}J_i^2-J_i+I}\right |=1. \)

   

[jacks]

Thanks fernado  revilla for nice solution. Can be solve it without  without jordon form.

Please explain

[Fernando Revilla]

Thanks fernado  revilla for nice solution. Can be solve it without  without jordon form.

if \( A \) is a \( 3\times 3 \) matrix such that \( A^3=O \).Then \( |\frac{A^2}{2}+A+I| \) and \( |\frac{A^2}{2}-A+I| \)

Really, what does the problem ask? Need you find both determinants? Or only a relation between them. For instance, as \( A^4=0 \) then \( \left(\frac{1}{2}A^2+A+I\right)\left(\frac{1}{2}A^2-A+I\right)=\ldots=\frac{1}{4}A^4+I=I \) so, \( \left(\frac{1}{2}A^2+A+I\right)^{-1}=\frac{1}{2}A^2-A+I \) and \( \left|\frac{A^2}{2}+A+I\right|=\left|\frac{A^2}{2}-A+I\right|^{-1} \).
   

[jacks]

Actually Problem asked for value of both determinants.

[Fernando Revilla]

Actually Problem asked for value of both determinants.

Well, so far I can't find those values without using Jordan forms.

P.S. But possibly, tomorrow will be another day. 

[jacks]

Thanks fernedo revilla.

[Masacroso]

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.

[jacks]

To masacroso how we write \( e^x=1+x+\frac{x^2}{2} \) because it is \( e^x=\sum^{\infty}_{n=0}\frac{x^n}{n!} \).Thanks

[Fernando Revilla]

To masacroso how we write \( e^x=1+x+\frac{x^2}{2} \) because it is \( e^x=\sum^{\infty}_{n=0}\frac{x^n}{n!} \).Thanks

If \( A^3=0 \) then, \( A^4=A^5=\ldots =0 \) so, \( e^A=\sum^{\infty}_{n=0}\frac{A^n}{n!}=I+A+\frac{1}{2}A^2 \).