Intuition behind subtraction of diagonal matrices build from eigenvalues

0

$begingroup$

Suppose we have a matrix A:
begin{array}{ccc}
3 & 1 & 1 \
3 & 2 & 3 \
3 & 2 & 1 \
end{array}

It's eigenvalues are:
{6, -1, 1}

If I make a diagonal matrix out of any of them for example:
begin{array}{ccc}
6 & 0 & 0 \
0 & 6 & 0 \
0 & 0 & 6 \
end{array}

And subtract it from A the rank of the resulting matrix is one less than the rank of matrix A. Why is that? What role does the eigenvalue play here? I'd be very thankful for any intuition.

linear-algebra abstract-algebra eigenvalues-eigenvectors

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edited Jan 14 at 12:10

wklm

asked Jan 14 at 12:04

wklmwklm

1105

$endgroup$

add a comment | 

0

$begingroup$

Suppose we have a matrix A:
begin{array}{ccc}
3 & 1 & 1 \
3 & 2 & 3 \
3 & 2 & 1 \
end{array}

It's eigenvalues are:
{6, -1, 1}

If I make a diagonal matrix out of any of them for example:
begin{array}{ccc}
6 & 0 & 0 \
0 & 6 & 0 \
0 & 0 & 6 \
end{array}

And subtract it from A the rank of the resulting matrix is one less than the rank of matrix A. Why is that? What role does the eigenvalue play here? I'd be very thankful for any intuition.

linear-algebra abstract-algebra eigenvalues-eigenvectors

share|cite|improve this question

edited Jan 14 at 12:10

wklm

asked Jan 14 at 12:04

wklmwklm

1105

$endgroup$

add a comment | 

0

0

0

$begingroup$

Suppose we have a matrix A:
begin{array}{ccc}
3 & 1 & 1 \
3 & 2 & 3 \
3 & 2 & 1 \
end{array}

It's eigenvalues are:
{6, -1, 1}

If I make a diagonal matrix out of any of them for example:
begin{array}{ccc}
6 & 0 & 0 \
0 & 6 & 0 \
0 & 0 & 6 \
end{array}

And subtract it from A the rank of the resulting matrix is one less than the rank of matrix A. Why is that? What role does the eigenvalue play here? I'd be very thankful for any intuition.

linear-algebra abstract-algebra eigenvalues-eigenvectors

share|cite|improve this question

edited Jan 14 at 12:10

wklm

asked Jan 14 at 12:04

wklmwklm

1105

$endgroup$

Suppose we have a matrix A:
begin{array}{ccc}
3 & 1 & 1 \
3 & 2 & 3 \
3 & 2 & 1 \
end{array}

It's eigenvalues are:
{6, -1, 1}

If I make a diagonal matrix out of any of them for example:
begin{array}{ccc}
6 & 0 & 0 \
0 & 6 & 0 \
0 & 0 & 6 \
end{array}

And subtract it from A the rank of the resulting matrix is one less than the rank of matrix A. Why is that? What role does the eigenvalue play here? I'd be very thankful for any intuition.

linear-algebra abstract-algebra eigenvalues-eigenvectors

linear-algebra abstract-algebra eigenvalues-eigenvectors

share|cite|improve this question

edited Jan 14 at 12:10

wklm

asked Jan 14 at 12:04

wklmwklm

1105

share|cite|improve this question

edited Jan 14 at 12:10

wklm

asked Jan 14 at 12:04

wklmwklm

1105

share|cite|improve this question

share|cite|improve this question

edited Jan 14 at 12:10

wklm

edited Jan 14 at 12:10

wklm

edited Jan 14 at 12:10

wklm

asked Jan 14 at 12:04

wklmwklm

1105

asked Jan 14 at 12:04

wklmwklm

1105

asked Jan 14 at 12:04

wklmwklm

1105

1105

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

2

$begingroup$

Hint: Because $det(A-lambda I) = 0$ by the definition of the eigenvalue. Also, as there three distinct eigenvalues for $A$ there two distinct eigenvalues for $B = A-6I$ too. ${-5, -7}$ are eigenvalues of $B$ as $B+5I = A-I$ and $B+7I = A+I$.

share|cite|improve this answer

edited Jan 14 at 12:16

answered Jan 14 at 12:08

OmGOmG

2,502822

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks a lot! Could you also please enlighten me why do A and B have the same eigenvectors?
  $endgroup$
  – wklm
  Jan 14 at 12:22
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @wklm my pleasure. as I mentioned $B+5I = A-I$ and $B+7I = A+I$, and $1, -1$ are eigenvalues of $A$. Hence, $A$ and $B$ have two eigenvectors in common.
  $endgroup$
  – OmG
  Jan 14 at 12:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @wklm You can check that for yourself: If $Av=lambda v$, then what does $Bv$ equal?
  $endgroup$
  – amd
  Jan 14 at 20:06
  

  
  
  
  

add a comment | 

1

$begingroup$

Remember: $Av= lambda v$, then you can rewrite this as: $(A - lambda I)v = 0$, which means that $det(A-lambda I) = 0$ (assuming $v$ is not 0, $v$ is the eigenvector corresponding to $lambda$).

share|cite|improve this answer

answered Jan 14 at 12:11

lightxbulblightxbulb

900211

$endgroup$

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

2

$begingroup$

Hint: Because $det(A-lambda I) = 0$ by the definition of the eigenvalue. Also, as there three distinct eigenvalues for $A$ there two distinct eigenvalues for $B = A-6I$ too. ${-5, -7}$ are eigenvalues of $B$ as $B+5I = A-I$ and $B+7I = A+I$.

share|cite|improve this answer

edited Jan 14 at 12:16

answered Jan 14 at 12:08

OmGOmG

2,502822

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks a lot! Could you also please enlighten me why do A and B have the same eigenvectors?
  $endgroup$
  – wklm
  Jan 14 at 12:22
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @wklm my pleasure. as I mentioned $B+5I = A-I$ and $B+7I = A+I$, and $1, -1$ are eigenvalues of $A$. Hence, $A$ and $B$ have two eigenvectors in common.
  $endgroup$
  – OmG
  Jan 14 at 12:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @wklm You can check that for yourself: If $Av=lambda v$, then what does $Bv$ equal?
  $endgroup$
  – amd
  Jan 14 at 20:06
  

  
  
  
  

add a comment | 

2

$begingroup$

Hint: Because $det(A-lambda I) = 0$ by the definition of the eigenvalue. Also, as there three distinct eigenvalues for $A$ there two distinct eigenvalues for $B = A-6I$ too. ${-5, -7}$ are eigenvalues of $B$ as $B+5I = A-I$ and $B+7I = A+I$.

share|cite|improve this answer

edited Jan 14 at 12:16

answered Jan 14 at 12:08

OmGOmG

2,502822

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks a lot! Could you also please enlighten me why do A and B have the same eigenvectors?
  $endgroup$
  – wklm
  Jan 14 at 12:22
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @wklm my pleasure. as I mentioned $B+5I = A-I$ and $B+7I = A+I$, and $1, -1$ are eigenvalues of $A$. Hence, $A$ and $B$ have two eigenvectors in common.
  $endgroup$
  – OmG
  Jan 14 at 12:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @wklm You can check that for yourself: If $Av=lambda v$, then what does $Bv$ equal?
  $endgroup$
  – amd
  Jan 14 at 20:06
  

  
  
  
  

add a comment | 

2

2

2

$begingroup$

Hint: Because $det(A-lambda I) = 0$ by the definition of the eigenvalue. Also, as there three distinct eigenvalues for $A$ there two distinct eigenvalues for $B = A-6I$ too. ${-5, -7}$ are eigenvalues of $B$ as $B+5I = A-I$ and $B+7I = A+I$.

share|cite|improve this answer

edited Jan 14 at 12:16

answered Jan 14 at 12:08

OmGOmG

2,502822

$endgroup$

Hint: Because $det(A-lambda I) = 0$ by the definition of the eigenvalue. Also, as there three distinct eigenvalues for $A$ there two distinct eigenvalues for $B = A-6I$ too. ${-5, -7}$ are eigenvalues of $B$ as $B+5I = A-I$ and $B+7I = A+I$.

share|cite|improve this answer

edited Jan 14 at 12:16

answered Jan 14 at 12:08

OmGOmG

2,502822

share|cite|improve this answer

share|cite|improve this answer

edited Jan 14 at 12:16

edited Jan 14 at 12:16

edited Jan 14 at 12:16

answered Jan 14 at 12:08

OmGOmG

2,502822

answered Jan 14 at 12:08

OmGOmG

2,502822

answered Jan 14 at 12:08

OmGOmG

2,502822

2,502822

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks a lot! Could you also please enlighten me why do A and B have the same eigenvectors?
  $endgroup$
  – wklm
  Jan 14 at 12:22
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @wklm my pleasure. as I mentioned $B+5I = A-I$ and $B+7I = A+I$, and $1, -1$ are eigenvalues of $A$. Hence, $A$ and $B$ have two eigenvectors in common.
  $endgroup$
  – OmG
  Jan 14 at 12:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @wklm You can check that for yourself: If $Av=lambda v$, then what does $Bv$ equal?
  $endgroup$
  – amd
  Jan 14 at 20:06
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks a lot! Could you also please enlighten me why do A and B have the same eigenvectors?
  $endgroup$
  – wklm
  Jan 14 at 12:22
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @wklm my pleasure. as I mentioned $B+5I = A-I$ and $B+7I = A+I$, and $1, -1$ are eigenvalues of $A$. Hence, $A$ and $B$ have two eigenvectors in common.
  $endgroup$
  – OmG
  Jan 14 at 12:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @wklm You can check that for yourself: If $Av=lambda v$, then what does $Bv$ equal?
  $endgroup$
  – amd
  Jan 14 at 20:06
  

  
  
  
  

$begingroup$
thanks a lot! Could you also please enlighten me why do A and B have the same eigenvectors?
$endgroup$
– wklm
Jan 14 at 12:22

$begingroup$
thanks a lot! Could you also please enlighten me why do A and B have the same eigenvectors?
$endgroup$
– wklm
Jan 14 at 12:22

1

1

$begingroup$
@wklm my pleasure. as I mentioned $B+5I = A-I$ and $B+7I = A+I$, and $1, -1$ are eigenvalues of $A$. Hence, $A$ and $B$ have two eigenvectors in common.
$endgroup$
– OmG
Jan 14 at 12:25

$begingroup$
@wklm my pleasure. as I mentioned $B+5I = A-I$ and $B+7I = A+I$, and $1, -1$ are eigenvalues of $A$. Hence, $A$ and $B$ have two eigenvectors in common.
$endgroup$
– OmG
Jan 14 at 12:25

1

1

$begingroup$
@wklm You can check that for yourself: If $Av=lambda v$, then what does $Bv$ equal?
$endgroup$
– amd
Jan 14 at 20:06

$begingroup$
@wklm You can check that for yourself: If $Av=lambda v$, then what does $Bv$ equal?
$endgroup$
– amd
Jan 14 at 20:06

add a comment | 

1

$begingroup$

Remember: $Av= lambda v$, then you can rewrite this as: $(A - lambda I)v = 0$, which means that $det(A-lambda I) = 0$ (assuming $v$ is not 0, $v$ is the eigenvector corresponding to $lambda$).

share|cite|improve this answer

answered Jan 14 at 12:11

lightxbulblightxbulb

900211

$endgroup$

add a comment | 

1

$begingroup$

Remember: $Av= lambda v$, then you can rewrite this as: $(A - lambda I)v = 0$, which means that $det(A-lambda I) = 0$ (assuming $v$ is not 0, $v$ is the eigenvector corresponding to $lambda$).

share|cite|improve this answer

answered Jan 14 at 12:11

lightxbulblightxbulb

900211

$endgroup$

add a comment | 

1

1

1

$begingroup$

Remember: $Av= lambda v$, then you can rewrite this as: $(A - lambda I)v = 0$, which means that $det(A-lambda I) = 0$ (assuming $v$ is not 0, $v$ is the eigenvector corresponding to $lambda$).

share|cite|improve this answer

answered Jan 14 at 12:11

lightxbulblightxbulb

900211

$endgroup$

Remember: $Av= lambda v$, then you can rewrite this as: $(A - lambda I)v = 0$, which means that $det(A-lambda I) = 0$ (assuming $v$ is not 0, $v$ is the eigenvector corresponding to $lambda$).

share|cite|improve this answer

answered Jan 14 at 12:11

lightxbulblightxbulb

900211

share|cite|improve this answer

share|cite|improve this answer

answered Jan 14 at 12:11

lightxbulblightxbulb

900211

answered Jan 14 at 12:11

lightxbulblightxbulb

900211

answered Jan 14 at 12:11

lightxbulblightxbulb

900211

900211

add a comment | 

add a comment | 

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