Autor Tema: Determinant of matrix

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18 Febrero, 2018, 03:48 am
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jacks

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if \[ A \] is a \[ 3\times 3 \] matrix such that \[ A^3=O \].Then

 \[ |\frac{A^2}{2}+A+I| \] and \[ |\frac{A^2}{2}-A+I| \]

18 Febrero, 2018, 06:25 am
Respuesta #1

Fernando Revilla

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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,

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

   

18 Febrero, 2018, 06:32 am
Respuesta #2

jacks

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

Please explain

18 Febrero, 2018, 08:08 am
Respuesta #3

Fernando Revilla

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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 \[ \left|\frac{A^2}{2}+A+I\right|=\left|\frac{A^2}{2}-A+I\right|^{-1} \].
   

18 Febrero, 2018, 09:03 am
Respuesta #4

jacks

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Actually Problem asked for value of both determinants.

18 Febrero, 2018, 02:03 pm
Respuesta #5

Fernando Revilla

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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.  :)

18 Febrero, 2018, 11:25 pm
Respuesta #6

jacks

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Thanks fernedo revilla.

19 Febrero, 2018, 01:08 am
Respuesta #7

Masacroso

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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.

19 Febrero, 2018, 03:50 am
Respuesta #8

jacks

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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

19 Febrero, 2018, 03:58 am
Respuesta #9

Fernando Revilla

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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, \[ e^A=\sum^{\infty}_{n=0}\frac{A^n}{n!}=I+A+\frac{1}{2}A^2 \].

19 Febrero, 2018, 04:50 am
Respuesta #10

Fernando Revilla

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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?

19 Febrero, 2018, 07:18 am
Respuesta #11

Luis Fuentes

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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.

19 Febrero, 2018, 07:45 am
Respuesta #12

Fernando Revilla

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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.

19 Febrero, 2018, 09:08 am
Respuesta #13

Abdulai

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



19 Febrero, 2018, 10:58 am
Respuesta #14

Masacroso

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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.

19 Febrero, 2018, 12:58 pm
Respuesta #15

Luis Fuentes

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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 \[ f(t)^n \].   ;D

Best regards.

26 Febrero, 2018, 08:59 am
Respuesta #16

jacks

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Thanks Moderator, Admin, Masacraso got