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2

$begingroup$

Find the Rational form of
$$
A=begin{pmatrix} 1&2&0&4\ :::4&1&2&0\ :::0&4&1&2\ :::2&0&4&1end{pmatrix}
$$

I don't wanna the solution, instead I would like to know a quickest way to calculate $det (lambda I-A)$.
$$
begin{vmatrix}
x-1 &-2 &0 &-4 \
-4&x-1 &-2 &0 \
0&-4 &x-1 & -2 \
-2&0 &-4 &x-1
end{vmatrix}
$$

here I could see the polynomial but the procedure is quite long.
https://www.symbolab.com/solver/matrix-eigenvectors-calculator/eigenvectors%20%5Cbegin%7Bpmatrix%7D1%262%260%264%5C%5C%20%20%20%20%204%261%262%260%5C%5C%20%20%20%20%200%264%261%262%5C%5C%20%20%20%20%202%260%264%261%5Cend%7Bpmatrix%7D

linear-algebra

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asked 8 hours ago

ipreferpiipreferpi

14511

$endgroup$

add a comment | 

2

$begingroup$

Find the Rational form of
$$
A=begin{pmatrix} 1&2&0&4\ :::4&1&2&0\ :::0&4&1&2\ :::2&0&4&1end{pmatrix}
$$

I don't wanna the solution, instead I would like to know a quickest way to calculate $det (lambda I-A)$.
$$
begin{vmatrix}
x-1 &-2 &0 &-4 \
-4&x-1 &-2 &0 \
0&-4 &x-1 & -2 \
-2&0 &-4 &x-1
end{vmatrix}
$$

here I could see the polynomial but the procedure is quite long.
https://www.symbolab.com/solver/matrix-eigenvectors-calculator/eigenvectors%20%5Cbegin%7Bpmatrix%7D1%262%260%264%5C%5C%20%20%20%20%204%261%262%260%5C%5C%20%20%20%20%200%264%261%262%5C%5C%20%20%20%20%202%260%264%261%5Cend%7Bpmatrix%7D

linear-algebra

share|cite|improve this question

asked 8 hours ago

ipreferpiipreferpi

14511

$endgroup$

add a comment | 

$begingroup$

Find the Rational form of
$$
A=begin{pmatrix} 1&2&0&4\ :::4&1&2&0\ :::0&4&1&2\ :::2&0&4&1end{pmatrix}
$$

I don't wanna the solution, instead I would like to know a quickest way to calculate $det (lambda I-A)$.
$$
begin{vmatrix}
x-1 &-2 &0 &-4 \
-4&x-1 &-2 &0 \
0&-4 &x-1 & -2 \
-2&0 &-4 &x-1
end{vmatrix}
$$

here I could see the polynomial but the procedure is quite long.
https://www.symbolab.com/solver/matrix-eigenvectors-calculator/eigenvectors%20%5Cbegin%7Bpmatrix%7D1%262%260%264%5C%5C%20%20%20%20%204%261%262%260%5C%5C%20%20%20%20%200%264%261%262%5C%5C%20%20%20%20%202%260%264%261%5Cend%7Bpmatrix%7D

linear-algebra

share|cite|improve this question

asked 8 hours ago

ipreferpiipreferpi

14511

$endgroup$

share|cite|improve this question

asked 8 hours ago

ipreferpiipreferpi

14511

share|cite|improve this question

asked 8 hours ago

ipreferpiipreferpi

14511

asked 8 hours ago

ipreferpiipreferpi

14511

asked 8 hours ago

ipreferpiipreferpi

14511

2 Answers
2

active

oldest

votes

4

$begingroup$

Add columns $,2,3,4,$ to the first one (this is a circulant matrix), and get:

$$det(t I-A)=begin{vmatrix}t-7&-2&0&-4\
t-7&t-1&-2&0\
t-7&-4&t-1&-2\
t-7&0&-4&t-1end{vmatrix}stackrel{R_i-R_1}=begin{vmatrix}t-7&-2&0&-4\
0&t+1&-2&4\
0&-2&t-1&2\
0&2&-4&t+3end{vmatrix}=$$$${}$$

$$=(t-7)begin{vmatrix}
t+1&-2&4\
-2&t-1&2\
2&-4&t+3end{vmatrix}=(t-7)left[(t^2-1)(t+3)+24+8t+8-4t-12-8t+8right]=$$

$$=(t-7)left[(t^2-1)(t+3)-4t+28right]=(t-7)left[t^3+3t^2-5t+25right]=$$

$$=(t-7)(t+5)(t^2-2t+5)=(t+t)(t+5)(t-(1-2i))(t-(1+2i))$$

share|cite|improve this answer

answered 7 hours ago

DonAntonioDonAntonio

182k1497234

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Both answers are great! However I'll choose this one, thank you.
  $endgroup$
  – ipreferpi
  3 hours ago
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

The quick way is to realise you have a circulant matrix, so you can immediately write down the determinant
begin{align*}
&detbegin{pmatrix}
x-1 &-2 &0 &-4 \
-4&x-1 &-2 &0 \
0&-4 &x-1 & -2 \
-2&0 &-4 &x-1
end{pmatrix}\
&=(x-1-2-4)(x-1-2i+4i)(x-1+2+4)(x-1+2i-4i)
end{align*}
since the eigenspaces are $langle(1,i^k,i^{2k},i^{3k})rangle=langle(1,i^k,(-1)^k,(-i)^k)rangle$, for $k=0,1,2,3$.

share|cite|improve this answer

answered 8 hours ago

user10354138user10354138

12.2k21125

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  "Since the eigenspaces are..." So you have to know first the eigenspaces?
  $endgroup$
  – DonAntonio
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Circulant matrices always have eigenspaces in this form, because these preserve the symmetry and are enough of them. No calculation needed (except to show they are independent, which follows from the vandermonde determinant).
  $endgroup$
  – user10354138
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is there any way only using determinant properties?
  $endgroup$
  – ipreferpi
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, using the basic properties of circulant matrices it is way faster and easier...+1
  $endgroup$
  – DonAntonio
  7 hours ago
  

  
  
  
  

add a comment | 

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4

$begingroup$

Add columns $,2,3,4,$ to the first one (this is a circulant matrix), and get:

$$det(t I-A)=begin{vmatrix}t-7&-2&0&-4\
t-7&t-1&-2&0\
t-7&-4&t-1&-2\
t-7&0&-4&t-1end{vmatrix}stackrel{R_i-R_1}=begin{vmatrix}t-7&-2&0&-4\
0&t+1&-2&4\
0&-2&t-1&2\
0&2&-4&t+3end{vmatrix}=$$$${}$$

$$=(t-7)begin{vmatrix}
t+1&-2&4\
-2&t-1&2\
2&-4&t+3end{vmatrix}=(t-7)left[(t^2-1)(t+3)+24+8t+8-4t-12-8t+8right]=$$

$$=(t-7)left[(t^2-1)(t+3)-4t+28right]=(t-7)left[t^3+3t^2-5t+25right]=$$

$$=(t-7)(t+5)(t^2-2t+5)=(t+t)(t+5)(t-(1-2i))(t-(1+2i))$$

share|cite|improve this answer

answered 7 hours ago

DonAntonioDonAntonio

182k1497234

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Both answers are great! However I'll choose this one, thank you.
  $endgroup$
  – ipreferpi
  3 hours ago
  
  
  

  
  
  
  

add a comment | 

4

$begingroup$

Add columns $,2,3,4,$ to the first one (this is a circulant matrix), and get:

$$det(t I-A)=begin{vmatrix}t-7&-2&0&-4\
t-7&t-1&-2&0\
t-7&-4&t-1&-2\
t-7&0&-4&t-1end{vmatrix}stackrel{R_i-R_1}=begin{vmatrix}t-7&-2&0&-4\
0&t+1&-2&4\
0&-2&t-1&2\
0&2&-4&t+3end{vmatrix}=$$$${}$$

$$=(t-7)begin{vmatrix}
t+1&-2&4\
-2&t-1&2\
2&-4&t+3end{vmatrix}=(t-7)left[(t^2-1)(t+3)+24+8t+8-4t-12-8t+8right]=$$

$$=(t-7)left[(t^2-1)(t+3)-4t+28right]=(t-7)left[t^3+3t^2-5t+25right]=$$

$$=(t-7)(t+5)(t^2-2t+5)=(t+t)(t+5)(t-(1-2i))(t-(1+2i))$$

share|cite|improve this answer

answered 7 hours ago

DonAntonioDonAntonio

182k1497234

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Both answers are great! However I'll choose this one, thank you.
  $endgroup$
  – ipreferpi
  3 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Add columns $,2,3,4,$ to the first one (this is a circulant matrix), and get:

$$det(t I-A)=begin{vmatrix}t-7&-2&0&-4\
t-7&t-1&-2&0\
t-7&-4&t-1&-2\
t-7&0&-4&t-1end{vmatrix}stackrel{R_i-R_1}=begin{vmatrix}t-7&-2&0&-4\
0&t+1&-2&4\
0&-2&t-1&2\
0&2&-4&t+3end{vmatrix}=$$$${}$$

$$=(t-7)begin{vmatrix}
t+1&-2&4\
-2&t-1&2\
2&-4&t+3end{vmatrix}=(t-7)left[(t^2-1)(t+3)+24+8t+8-4t-12-8t+8right]=$$

$$=(t-7)left[(t^2-1)(t+3)-4t+28right]=(t-7)left[t^3+3t^2-5t+25right]=$$

$$=(t-7)(t+5)(t^2-2t+5)=(t+t)(t+5)(t-(1-2i))(t-(1+2i))$$

share|cite|improve this answer

answered 7 hours ago

DonAntonioDonAntonio

182k1497234

$endgroup$

share|cite|improve this answer

answered 7 hours ago

DonAntonioDonAntonio

182k1497234

answered 7 hours ago

DonAntonioDonAntonio

182k1497234

answered 7 hours ago

DonAntonioDonAntonio

182k1497234

3

$begingroup$

The quick way is to realise you have a circulant matrix, so you can immediately write down the determinant
begin{align*}
&detbegin{pmatrix}
x-1 &-2 &0 &-4 \
-4&x-1 &-2 &0 \
0&-4 &x-1 & -2 \
-2&0 &-4 &x-1
end{pmatrix}\
&=(x-1-2-4)(x-1-2i+4i)(x-1+2+4)(x-1+2i-4i)
end{align*}
since the eigenspaces are $langle(1,i^k,i^{2k},i^{3k})rangle=langle(1,i^k,(-1)^k,(-i)^k)rangle$, for $k=0,1,2,3$.

share|cite|improve this answer

answered 8 hours ago

user10354138user10354138

12.2k21125

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  "Since the eigenspaces are..." So you have to know first the eigenspaces?
  $endgroup$
  – DonAntonio
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Circulant matrices always have eigenspaces in this form, because these preserve the symmetry and are enough of them. No calculation needed (except to show they are independent, which follows from the vandermonde determinant).
  $endgroup$
  – user10354138
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is there any way only using determinant properties?
  $endgroup$
  – ipreferpi
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, using the basic properties of circulant matrices it is way faster and easier...+1
  $endgroup$
  – DonAntonio
  7 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

The quick way is to realise you have a circulant matrix, so you can immediately write down the determinant
begin{align*}
&detbegin{pmatrix}
x-1 &-2 &0 &-4 \
-4&x-1 &-2 &0 \
0&-4 &x-1 & -2 \
-2&0 &-4 &x-1
end{pmatrix}\
&=(x-1-2-4)(x-1-2i+4i)(x-1+2+4)(x-1+2i-4i)
end{align*}
since the eigenspaces are $langle(1,i^k,i^{2k},i^{3k})rangle=langle(1,i^k,(-1)^k,(-i)^k)rangle$, for $k=0,1,2,3$.

share|cite|improve this answer

answered 8 hours ago

user10354138user10354138

12.2k21125

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  "Since the eigenspaces are..." So you have to know first the eigenspaces?
  $endgroup$
  – DonAntonio
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Circulant matrices always have eigenspaces in this form, because these preserve the symmetry and are enough of them. No calculation needed (except to show they are independent, which follows from the vandermonde determinant).
  $endgroup$
  – user10354138
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is there any way only using determinant properties?
  $endgroup$
  – ipreferpi
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, using the basic properties of circulant matrices it is way faster and easier...+1
  $endgroup$
  – DonAntonio
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

The quick way is to realise you have a circulant matrix, so you can immediately write down the determinant
begin{align*}
&detbegin{pmatrix}
x-1 &-2 &0 &-4 \
-4&x-1 &-2 &0 \
0&-4 &x-1 & -2 \
-2&0 &-4 &x-1
end{pmatrix}\
&=(x-1-2-4)(x-1-2i+4i)(x-1+2+4)(x-1+2i-4i)
end{align*}
since the eigenspaces are $langle(1,i^k,i^{2k},i^{3k})rangle=langle(1,i^k,(-1)^k,(-i)^k)rangle$, for $k=0,1,2,3$.

share|cite|improve this answer

answered 8 hours ago

user10354138user10354138

12.2k21125

$endgroup$

share|cite|improve this answer

answered 8 hours ago

user10354138user10354138

12.2k21125

answered 8 hours ago

user10354138user10354138

12.2k21125

answered 8 hours ago

user10354138user10354138

12.2k21125