Finding the eigenvalues of a linear transformation without using the matrix of the linear transformation

1

$begingroup$

Define the linear transformation $L$ as

$L: mathbb{R}^3 rightarrow mathbb{R}^3: x rightarrow x - langle x, a rangle b$

Let $a, b in mathbb{R}^3$ such that $langle a, b rangle = 2$ and let $langle .,. rangle $ be the standard inner product of $mathbb{R}^3$.

How would I find the eigenvalues of the linear transformation without using the matrix of the linear transformation?

linear-algebra eigenvalues-eigenvectors linear-transformations

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asked Jan 23 at 20:04

mrMoonpenguinmrMoonpenguin

153

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you try to find $lambda$ in $f(v)=lambda v$ for some vector $vin mathbb{R}^3setminus 0$?
  $endgroup$
  – Smilia
  Jan 23 at 20:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you try to find $lambda$ in $f(v)=lambda v$ for some vector $vin mathbb{R}^3setminus 0$?
  $endgroup$
  – Smilia
  Jan 23 at 20:07
  

  
  
  
  

add a comment | 

1

$begingroup$

Define the linear transformation $L$ as

$L: mathbb{R}^3 rightarrow mathbb{R}^3: x rightarrow x - langle x, a rangle b$

Let $a, b in mathbb{R}^3$ such that $langle a, b rangle = 2$ and let $langle .,. rangle $ be the standard inner product of $mathbb{R}^3$.

How would I find the eigenvalues of the linear transformation without using the matrix of the linear transformation?

linear-algebra eigenvalues-eigenvectors linear-transformations

share|cite|improve this question

asked Jan 23 at 20:04

mrMoonpenguinmrMoonpenguin

153

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you try to find $lambda$ in $f(v)=lambda v$ for some vector $vin mathbb{R}^3setminus 0$?
  $endgroup$
  – Smilia
  Jan 23 at 20:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you try to find $lambda$ in $f(v)=lambda v$ for some vector $vin mathbb{R}^3setminus 0$?
  $endgroup$
  – Smilia
  Jan 23 at 20:07
  

  
  
  
  

add a comment | 

1

1

1

1

$begingroup$

Define the linear transformation $L$ as

$L: mathbb{R}^3 rightarrow mathbb{R}^3: x rightarrow x - langle x, a rangle b$

Let $a, b in mathbb{R}^3$ such that $langle a, b rangle = 2$ and let $langle .,. rangle $ be the standard inner product of $mathbb{R}^3$.

How would I find the eigenvalues of the linear transformation without using the matrix of the linear transformation?

linear-algebra eigenvalues-eigenvectors linear-transformations

share|cite|improve this question

asked Jan 23 at 20:04

mrMoonpenguinmrMoonpenguin

153

$endgroup$

Define the linear transformation $L$ as

$L: mathbb{R}^3 rightarrow mathbb{R}^3: x rightarrow x - langle x, a rangle b$

Let $a, b in mathbb{R}^3$ such that $langle a, b rangle = 2$ and let $langle .,. rangle $ be the standard inner product of $mathbb{R}^3$.

How would I find the eigenvalues of the linear transformation without using the matrix of the linear transformation?

linear-algebra eigenvalues-eigenvectors linear-transformations

linear-algebra eigenvalues-eigenvectors linear-transformations

share|cite|improve this question

asked Jan 23 at 20:04

mrMoonpenguinmrMoonpenguin

153

share|cite|improve this question

asked Jan 23 at 20:04

mrMoonpenguinmrMoonpenguin

153

share|cite|improve this question

share|cite|improve this question

asked Jan 23 at 20:04

mrMoonpenguinmrMoonpenguin

153

asked Jan 23 at 20:04

mrMoonpenguinmrMoonpenguin

153

asked Jan 23 at 20:04

mrMoonpenguinmrMoonpenguin

153

153

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you try to find $lambda$ in $f(v)=lambda v$ for some vector $vin mathbb{R}^3setminus 0$?
  $endgroup$
  – Smilia
  Jan 23 at 20:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you try to find $lambda$ in $f(v)=lambda v$ for some vector $vin mathbb{R}^3setminus 0$?
  $endgroup$
  – Smilia
  Jan 23 at 20:07
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you try to find $lambda$ in $f(v)=lambda v$ for some vector $vin mathbb{R}^3setminus 0$?
  $endgroup$
  – Smilia
  Jan 23 at 20:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you try to find $lambda$ in $f(v)=lambda v$ for some vector $vin mathbb{R}^3setminus 0$?
  $endgroup$
  – Smilia
  Jan 23 at 20:07
  

  
  
  
  

$begingroup$
Did you try to find $lambda$ in $f(v)=lambda v$ for some vector $vin mathbb{R}^3setminus 0$?
$endgroup$
– Smilia
Jan 23 at 20:06

$begingroup$
Did you try to find $lambda$ in $f(v)=lambda v$ for some vector $vin mathbb{R}^3setminus 0$?
$endgroup$
– Smilia
Jan 23 at 20:06

$begingroup$
Did you try to find $lambda$ in $f(v)=lambda v$ for some vector $vin mathbb{R}^3setminus 0$?
$endgroup$
– Smilia
Jan 23 at 20:07

$begingroup$
Did you try to find $lambda$ in $f(v)=lambda v$ for some vector $vin mathbb{R}^3setminus 0$?
$endgroup$
– Smilia
Jan 23 at 20:07

add a comment | 

2 Answers
2

active

oldest

votes

1

$begingroup$

It appears this all works over $Bbb R^n$:

Given that

$L(x) = x - langle x, a rangle b, tag 0$

if

$langle a, x rangle = 0, tag 1$

then

$L(x) = x, tag 2$

that is, $x$ is an eigenvector of $L$ associated with eigenvalue

$lambda = 1; tag 3$

thus the entire $n - 1$ dimensional subspace

$a^bot subset Bbb R^n tag 4$

is the $1$-eigenspace of $L$. Now if

$x = alpha b, ; alpha ne 0, tag 5$

then

$L(x) = L(alpha b) = alpha b - langle alpha b, a rangle b = alpha b - alpha langle b, a rangle b = alpha b - 2alpha b = -alpha b; tag 6$

that is, $alpha b$ is a $-1$-eigenvector of $L$; also

$langle b, a rangle = 2 ne 0 Longrightarrow b notin a^bot; tag 7$

therefore

$Bbb R^n = a^bot + {alpha b, alpha in Bbb R}; tag 8$

since $a^bot$ and $b$ generate $Bbb R^n$, there can be no other eigenvectors/eigenvalues, so we have found all the
eigenvectors/eigenvalues of $L$.

share|cite|improve this answer

answered Jan 23 at 22:02

Robert LewisRobert Lewis

48k23067

$endgroup$

add a comment | 

1

$begingroup$

Hint: Assuming that $L(x)=x-langle x,arangle b=lambda x$ one obtains the equation $(1-lambda)x=langle x,arangle b$. It is easy to find the eigenvalues from this equation.

share|cite|improve this answer

answered Jan 23 at 20:24

Oleg SmirnovOleg Smirnov

658

$endgroup$

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

It appears this all works over $Bbb R^n$:

Given that

$L(x) = x - langle x, a rangle b, tag 0$

if

$langle a, x rangle = 0, tag 1$

then

$L(x) = x, tag 2$

that is, $x$ is an eigenvector of $L$ associated with eigenvalue

$lambda = 1; tag 3$

thus the entire $n - 1$ dimensional subspace

$a^bot subset Bbb R^n tag 4$

is the $1$-eigenspace of $L$. Now if

$x = alpha b, ; alpha ne 0, tag 5$

then

$L(x) = L(alpha b) = alpha b - langle alpha b, a rangle b = alpha b - alpha langle b, a rangle b = alpha b - 2alpha b = -alpha b; tag 6$

that is, $alpha b$ is a $-1$-eigenvector of $L$; also

$langle b, a rangle = 2 ne 0 Longrightarrow b notin a^bot; tag 7$

therefore

$Bbb R^n = a^bot + {alpha b, alpha in Bbb R}; tag 8$

since $a^bot$ and $b$ generate $Bbb R^n$, there can be no other eigenvectors/eigenvalues, so we have found all the
eigenvectors/eigenvalues of $L$.

share|cite|improve this answer

answered Jan 23 at 22:02

Robert LewisRobert Lewis

48k23067

$endgroup$

add a comment | 

1

$begingroup$

It appears this all works over $Bbb R^n$:

Given that

$L(x) = x - langle x, a rangle b, tag 0$

if

$langle a, x rangle = 0, tag 1$

then

$L(x) = x, tag 2$

that is, $x$ is an eigenvector of $L$ associated with eigenvalue

$lambda = 1; tag 3$

thus the entire $n - 1$ dimensional subspace

$a^bot subset Bbb R^n tag 4$

is the $1$-eigenspace of $L$. Now if

$x = alpha b, ; alpha ne 0, tag 5$

then

$L(x) = L(alpha b) = alpha b - langle alpha b, a rangle b = alpha b - alpha langle b, a rangle b = alpha b - 2alpha b = -alpha b; tag 6$

that is, $alpha b$ is a $-1$-eigenvector of $L$; also

$langle b, a rangle = 2 ne 0 Longrightarrow b notin a^bot; tag 7$

therefore

$Bbb R^n = a^bot + {alpha b, alpha in Bbb R}; tag 8$

since $a^bot$ and $b$ generate $Bbb R^n$, there can be no other eigenvectors/eigenvalues, so we have found all the
eigenvectors/eigenvalues of $L$.

share|cite|improve this answer

answered Jan 23 at 22:02

Robert LewisRobert Lewis

48k23067

$endgroup$

add a comment | 

1

1

1

$begingroup$

It appears this all works over $Bbb R^n$:

Given that

$L(x) = x - langle x, a rangle b, tag 0$

if

$langle a, x rangle = 0, tag 1$

then

$L(x) = x, tag 2$

that is, $x$ is an eigenvector of $L$ associated with eigenvalue

$lambda = 1; tag 3$

thus the entire $n - 1$ dimensional subspace

$a^bot subset Bbb R^n tag 4$

is the $1$-eigenspace of $L$. Now if

$x = alpha b, ; alpha ne 0, tag 5$

then

$L(x) = L(alpha b) = alpha b - langle alpha b, a rangle b = alpha b - alpha langle b, a rangle b = alpha b - 2alpha b = -alpha b; tag 6$

that is, $alpha b$ is a $-1$-eigenvector of $L$; also

$langle b, a rangle = 2 ne 0 Longrightarrow b notin a^bot; tag 7$

therefore

$Bbb R^n = a^bot + {alpha b, alpha in Bbb R}; tag 8$

since $a^bot$ and $b$ generate $Bbb R^n$, there can be no other eigenvectors/eigenvalues, so we have found all the
eigenvectors/eigenvalues of $L$.

share|cite|improve this answer

answered Jan 23 at 22:02

Robert LewisRobert Lewis

48k23067

$endgroup$

It appears this all works over $Bbb R^n$:

Given that

$L(x) = x - langle x, a rangle b, tag 0$

if

$langle a, x rangle = 0, tag 1$

then

$L(x) = x, tag 2$

that is, $x$ is an eigenvector of $L$ associated with eigenvalue

$lambda = 1; tag 3$

thus the entire $n - 1$ dimensional subspace

$a^bot subset Bbb R^n tag 4$

is the $1$-eigenspace of $L$. Now if

$x = alpha b, ; alpha ne 0, tag 5$

then

$L(x) = L(alpha b) = alpha b - langle alpha b, a rangle b = alpha b - alpha langle b, a rangle b = alpha b - 2alpha b = -alpha b; tag 6$

that is, $alpha b$ is a $-1$-eigenvector of $L$; also

$langle b, a rangle = 2 ne 0 Longrightarrow b notin a^bot; tag 7$

therefore

$Bbb R^n = a^bot + {alpha b, alpha in Bbb R}; tag 8$

since $a^bot$ and $b$ generate $Bbb R^n$, there can be no other eigenvectors/eigenvalues, so we have found all the
eigenvectors/eigenvalues of $L$.

share|cite|improve this answer

answered Jan 23 at 22:02

Robert LewisRobert Lewis

48k23067

share|cite|improve this answer

share|cite|improve this answer

answered Jan 23 at 22:02

Robert LewisRobert Lewis

48k23067

answered Jan 23 at 22:02

Robert LewisRobert Lewis

48k23067

answered Jan 23 at 22:02

Robert LewisRobert Lewis

48k23067

48k23067

add a comment | 

add a comment | 

1

$begingroup$

Hint: Assuming that $L(x)=x-langle x,arangle b=lambda x$ one obtains the equation $(1-lambda)x=langle x,arangle b$. It is easy to find the eigenvalues from this equation.

share|cite|improve this answer

answered Jan 23 at 20:24

Oleg SmirnovOleg Smirnov

658

$endgroup$

add a comment | 

1

$begingroup$

Hint: Assuming that $L(x)=x-langle x,arangle b=lambda x$ one obtains the equation $(1-lambda)x=langle x,arangle b$. It is easy to find the eigenvalues from this equation.

share|cite|improve this answer

answered Jan 23 at 20:24

Oleg SmirnovOleg Smirnov

658

$endgroup$

add a comment | 

1

1

1

$begingroup$

Hint: Assuming that $L(x)=x-langle x,arangle b=lambda x$ one obtains the equation $(1-lambda)x=langle x,arangle b$. It is easy to find the eigenvalues from this equation.

share|cite|improve this answer

answered Jan 23 at 20:24

Oleg SmirnovOleg Smirnov

658

$endgroup$

Hint: Assuming that $L(x)=x-langle x,arangle b=lambda x$ one obtains the equation $(1-lambda)x=langle x,arangle b$. It is easy to find the eigenvalues from this equation.

share|cite|improve this answer

answered Jan 23 at 20:24

Oleg SmirnovOleg Smirnov

658

share|cite|improve this answer

share|cite|improve this answer

answered Jan 23 at 20:24

Oleg SmirnovOleg Smirnov

658

answered Jan 23 at 20:24

Oleg SmirnovOleg Smirnov

658

answered Jan 23 at 20:24

Oleg SmirnovOleg Smirnov

658

658

add a comment | 

add a comment | 

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Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

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Use MathJax to format equations. MathJax reference.

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