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Using eigenvectors as a basis for space

0

$begingroup$

Consider the following matrix

$A=begin{bmatrix}0&0&0&0\a&0&0&0\0&b&0&0\0&0&c&0end{bmatrix}$

Which $a,b,c$ are real numbers

What conditions are required for $a,b,c$ such that $mathbb{R}^4$ has a basis made of A eigenvectors?

linear-algebra

share|cite|improve this question

asked Mar 25 at 9:53

Alireza HosseiniAlireza Hosseini

233

$endgroup$

add a comment | 

0

$begingroup$

Consider the following matrix

$A=begin{bmatrix}0&0&0&0\a&0&0&0\0&b&0&0\0&0&c&0end{bmatrix}$

Which $a,b,c$ are real numbers

What conditions are required for $a,b,c$ such that $mathbb{R}^4$ has a basis made of A eigenvectors?

linear-algebra

share|cite|improve this question

asked Mar 25 at 9:53

Alireza HosseiniAlireza Hosseini

233

$endgroup$

add a comment | 

$begingroup$

Consider the following matrix

$A=begin{bmatrix}0&0&0&0\a&0&0&0\0&b&0&0\0&0&c&0end{bmatrix}$

Which $a,b,c$ are real numbers

What conditions are required for $a,b,c$ such that $mathbb{R}^4$ has a basis made of A eigenvectors?

linear-algebra

share|cite|improve this question

asked Mar 25 at 9:53

Alireza HosseiniAlireza Hosseini

233

$endgroup$

share|cite|improve this question

asked Mar 25 at 9:53

Alireza HosseiniAlireza Hosseini

233

share|cite|improve this question

asked Mar 25 at 9:53

Alireza HosseiniAlireza Hosseini

233

asked Mar 25 at 9:53

Alireza HosseiniAlireza Hosseini

233

asked Mar 25 at 9:53

Alireza HosseiniAlireza Hosseini

233

2 Answers
2

active

oldest

votes

0

$begingroup$

The characteristic polynomial of $A$ is $X^4$, so the only eigenvalue is $0$.

If $A$ is diagonalizable then, it means that $A=0$, so $a=b=c=0$.

Conversely, if $a=b=c=0$, all vectors are eigenvectors, so of course you can find a basis made of eigenvectors.

So the necessary and sufficient condition is that $a=b=c=0$.

share|cite|improve this answer

answered Mar 25 at 9:59

TheSilverDoeTheSilverDoe

5,608316

$endgroup$

add a comment | 

0

$begingroup$

$lambda = 0$ is the only eigenvalue (it has algebraic multiplicity 4). The only way the system $Av = 0$ has 4 degrees of freedom is if $A=0$, i.e. $a=b=c=0$.

share|cite|improve this answer

answered Mar 25 at 10:03

PierreCarrePierreCarre

2,253215

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add a comment | 

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0

$begingroup$

The characteristic polynomial of $A$ is $X^4$, so the only eigenvalue is $0$.

If $A$ is diagonalizable then, it means that $A=0$, so $a=b=c=0$.

Conversely, if $a=b=c=0$, all vectors are eigenvectors, so of course you can find a basis made of eigenvectors.

So the necessary and sufficient condition is that $a=b=c=0$.

share|cite|improve this answer

answered Mar 25 at 9:59

TheSilverDoeTheSilverDoe

5,608316

$endgroup$

add a comment | 

0

$begingroup$

The characteristic polynomial of $A$ is $X^4$, so the only eigenvalue is $0$.

If $A$ is diagonalizable then, it means that $A=0$, so $a=b=c=0$.

Conversely, if $a=b=c=0$, all vectors are eigenvectors, so of course you can find a basis made of eigenvectors.

So the necessary and sufficient condition is that $a=b=c=0$.

share|cite|improve this answer

answered Mar 25 at 9:59

TheSilverDoeTheSilverDoe

5,608316

$endgroup$

add a comment | 

$begingroup$

The characteristic polynomial of $A$ is $X^4$, so the only eigenvalue is $0$.

If $A$ is diagonalizable then, it means that $A=0$, so $a=b=c=0$.

Conversely, if $a=b=c=0$, all vectors are eigenvectors, so of course you can find a basis made of eigenvectors.

So the necessary and sufficient condition is that $a=b=c=0$.

share|cite|improve this answer

answered Mar 25 at 9:59

TheSilverDoeTheSilverDoe

5,608316

$endgroup$

share|cite|improve this answer

answered Mar 25 at 9:59

TheSilverDoeTheSilverDoe

5,608316

answered Mar 25 at 9:59

TheSilverDoeTheSilverDoe

5,608316

answered Mar 25 at 9:59

TheSilverDoeTheSilverDoe

5,608316

0

$begingroup$

$lambda = 0$ is the only eigenvalue (it has algebraic multiplicity 4). The only way the system $Av = 0$ has 4 degrees of freedom is if $A=0$, i.e. $a=b=c=0$.

share|cite|improve this answer

answered Mar 25 at 10:03

PierreCarrePierreCarre

2,253215

$endgroup$

add a comment | 

0

$begingroup$

$lambda = 0$ is the only eigenvalue (it has algebraic multiplicity 4). The only way the system $Av = 0$ has 4 degrees of freedom is if $A=0$, i.e. $a=b=c=0$.

share|cite|improve this answer

answered Mar 25 at 10:03

PierreCarrePierreCarre

2,253215

$endgroup$

add a comment | 

$begingroup$

$lambda = 0$ is the only eigenvalue (it has algebraic multiplicity 4). The only way the system $Av = 0$ has 4 degrees of freedom is if $A=0$, i.e. $a=b=c=0$.

share|cite|improve this answer

answered Mar 25 at 10:03

PierreCarrePierreCarre

2,253215

$endgroup$

share|cite|improve this answer

answered Mar 25 at 10:03

PierreCarrePierreCarre

2,253215

answered Mar 25 at 10:03

PierreCarrePierreCarre

2,253215

answered Mar 25 at 10:03

PierreCarrePierreCarre

2,253215