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Limit of matrix existence for values of x and y

2

$begingroup$

I have been given this question to answer

The Matrix$$C = beginpmatrix
1 & 7 & 1 & 3 \
7 & 1 & 1 & 3 \
1&1&7&3 \
3 &3 &3&3
endpmatrix $$

has eigenvalues: $12 ,-6, 6, 0$

$N(x,y) = xC+yI_4$

depending on $x$ and $y$ determine when $lim_kto inftyN(x,y)^k$ exists, sketch the region for $x$ and $y$

Attempt:

From class I have the following theorem

Let A be an $n$ x $n$ matrix. Then $lim_kto inftyA^k$exists if and only if

$A$ has no eigenvalue $lambda$ with $|lambda| geq1$

If $A$ has eigenvalue $lambda$ with $lambda = 1$, then all block of Jordan Canonical for $J(A)$ of the form $J_k(1)$ must have $k=1$

Using this theorem and the eigenvalues provided my attempt would be to find the region for $x$ and $y$ such that

$$|12x+y|leq1$$
$$|-6x+y|leq1$$
$$|6x+y|leq1$$
$$|0x+y|leq1$$

However I'm finding defining the region difficult so I'm uncertain if this is the correct approach

linear-algebra matrices

share|cite|improve this question

edited Apr 8 at 20:55

Rito Lowe

asked Apr 8 at 20:34

Rito LoweRito Lowe

686

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To be changed into "$A$ has no eigenvalue λ with $|λ| geq 1$" (not $<$)
  $endgroup$
  – Jean Marie
  Apr 8 at 20:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  fixed, thank you
  $endgroup$
  – Rito Lowe
  Apr 8 at 20:56
  

  
  
  
  

add a comment | 

2

$begingroup$

I have been given this question to answer

The Matrix$$C = beginpmatrix
1 & 7 & 1 & 3 \
7 & 1 & 1 & 3 \
1&1&7&3 \
3 &3 &3&3
endpmatrix $$

has eigenvalues: $12 ,-6, 6, 0$

$N(x,y) = xC+yI_4$

depending on $x$ and $y$ determine when $lim_kto inftyN(x,y)^k$ exists, sketch the region for $x$ and $y$

Attempt:

From class I have the following theorem

Let A be an $n$ x $n$ matrix. Then $lim_kto inftyA^k$exists if and only if

$A$ has no eigenvalue $lambda$ with $|lambda| geq1$

If $A$ has eigenvalue $lambda$ with $lambda = 1$, then all block of Jordan Canonical for $J(A)$ of the form $J_k(1)$ must have $k=1$

Using this theorem and the eigenvalues provided my attempt would be to find the region for $x$ and $y$ such that

$$|12x+y|leq1$$
$$|-6x+y|leq1$$
$$|6x+y|leq1$$
$$|0x+y|leq1$$

However I'm finding defining the region difficult so I'm uncertain if this is the correct approach

linear-algebra matrices

share|cite|improve this question

edited Apr 8 at 20:55

Rito Lowe

asked Apr 8 at 20:34

Rito LoweRito Lowe

686

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To be changed into "$A$ has no eigenvalue λ with $|λ| geq 1$" (not $<$)
  $endgroup$
  – Jean Marie
  Apr 8 at 20:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  fixed, thank you
  $endgroup$
  – Rito Lowe
  Apr 8 at 20:56
  

  
  
  
  

add a comment | 

$begingroup$

I have been given this question to answer

The Matrix$$C = beginpmatrix
1 & 7 & 1 & 3 \
7 & 1 & 1 & 3 \
1&1&7&3 \
3 &3 &3&3
endpmatrix $$

has eigenvalues: $12 ,-6, 6, 0$

$N(x,y) = xC+yI_4$

depending on $x$ and $y$ determine when $lim_kto inftyN(x,y)^k$ exists, sketch the region for $x$ and $y$

Attempt:

From class I have the following theorem

Let A be an $n$ x $n$ matrix. Then $lim_kto inftyA^k$exists if and only if

$A$ has no eigenvalue $lambda$ with $|lambda| geq1$

If $A$ has eigenvalue $lambda$ with $lambda = 1$, then all block of Jordan Canonical for $J(A)$ of the form $J_k(1)$ must have $k=1$

Using this theorem and the eigenvalues provided my attempt would be to find the region for $x$ and $y$ such that

$$|12x+y|leq1$$
$$|-6x+y|leq1$$
$$|6x+y|leq1$$
$$|0x+y|leq1$$

However I'm finding defining the region difficult so I'm uncertain if this is the correct approach

linear-algebra matrices

share|cite|improve this question

edited Apr 8 at 20:55

Rito Lowe

asked Apr 8 at 20:34

Rito LoweRito Lowe

686

$endgroup$

share|cite|improve this question

edited Apr 8 at 20:55

Rito Lowe

asked Apr 8 at 20:34

Rito LoweRito Lowe

686

share|cite|improve this question

edited Apr 8 at 20:55

Rito Lowe

asked Apr 8 at 20:34

Rito LoweRito Lowe

686

asked Apr 8 at 20:34

Rito LoweRito Lowe

686

asked Apr 8 at 20:34

Rito LoweRito Lowe

686

1 Answer
1

active

oldest

votes

1

$begingroup$

Your reasoning is right, with the exception that you cannot allow the eigenvalue $-1$. So you are looking at
beginalign
-1&<12x+yleq 1\
-1&<-6x+yleq 1\
-1&<6x+yleq 1\
-1&<yleq 1\
endalign
So basically you need to plot eight lines and look at the region they enclose. Lazy people like me use some graphing calculator and then Paint:

If you want to express this analytically you will have look at the four lines that define the green region. Then you can tell that your region is given by those $x,y$ with
beginalign
-6x+yleq1 \
12x+yleq1\
12x+y>-1\
-6x+y>-1
endalign
and we can summarize this as
beginalign
-1<-6x+yleq1\
-1<12x+yleq 1
endalign

share|cite|improve this answer

answered Apr 9 at 1:46

Martin ArgeramiMartin Argerami

129k1184185

$endgroup$

add a comment | 

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1

$begingroup$

Your reasoning is right, with the exception that you cannot allow the eigenvalue $-1$. So you are looking at
beginalign
-1&<12x+yleq 1\
-1&<-6x+yleq 1\
-1&<6x+yleq 1\
-1&<yleq 1\
endalign
So basically you need to plot eight lines and look at the region they enclose. Lazy people like me use some graphing calculator and then Paint:

If you want to express this analytically you will have look at the four lines that define the green region. Then you can tell that your region is given by those $x,y$ with
beginalign
-6x+yleq1 \
12x+yleq1\
12x+y>-1\
-6x+y>-1
endalign
and we can summarize this as
beginalign
-1<-6x+yleq1\
-1<12x+yleq 1
endalign

share|cite|improve this answer

answered Apr 9 at 1:46

Martin ArgeramiMartin Argerami

129k1184185

$endgroup$

add a comment | 

1

$begingroup$

Your reasoning is right, with the exception that you cannot allow the eigenvalue $-1$. So you are looking at
beginalign
-1&<12x+yleq 1\
-1&<-6x+yleq 1\
-1&<6x+yleq 1\
-1&<yleq 1\
endalign
So basically you need to plot eight lines and look at the region they enclose. Lazy people like me use some graphing calculator and then Paint:

If you want to express this analytically you will have look at the four lines that define the green region. Then you can tell that your region is given by those $x,y$ with
beginalign
-6x+yleq1 \
12x+yleq1\
12x+y>-1\
-6x+y>-1
endalign
and we can summarize this as
beginalign
-1<-6x+yleq1\
-1<12x+yleq 1
endalign

share|cite|improve this answer

answered Apr 9 at 1:46

Martin ArgeramiMartin Argerami

129k1184185

$endgroup$

add a comment | 

$begingroup$

Your reasoning is right, with the exception that you cannot allow the eigenvalue $-1$. So you are looking at
beginalign
-1&<12x+yleq 1\
-1&<-6x+yleq 1\
-1&<6x+yleq 1\
-1&<yleq 1\
endalign
So basically you need to plot eight lines and look at the region they enclose. Lazy people like me use some graphing calculator and then Paint:

If you want to express this analytically you will have look at the four lines that define the green region. Then you can tell that your region is given by those $x,y$ with
beginalign
-6x+yleq1 \
12x+yleq1\
12x+y>-1\
-6x+y>-1
endalign
and we can summarize this as
beginalign
-1<-6x+yleq1\
-1<12x+yleq 1
endalign

share|cite|improve this answer

answered Apr 9 at 1:46

Martin ArgeramiMartin Argerami

129k1184185

$endgroup$

share|cite|improve this answer

answered Apr 9 at 1:46

Martin ArgeramiMartin Argerami

129k1184185

answered Apr 9 at 1:46

Martin ArgeramiMartin Argerami

129k1184185

answered Apr 9 at 1:46

Martin ArgeramiMartin Argerami

129k1184185