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Need help finding eigenvectors of matrix

2

$begingroup$

I have the following Matrix:

$$B= beginbmatrix
8 & -6 & -6 \
30 & -22 & -30 \
-30 & 30 & 38
endbmatrix $$

I find the characteristic polynomial as:

$$det(B-lambda E)=-(lambda-8)^3$$

So we have the root 8 (with multiplicty 3). Therefore $B$ has the eigenvalue 8

Now I solve for

$$(B-lambda E)x= beginbmatrix
8-8 & -6 & -6 \
30 & -22-8 & -30 \
-30 & 30 & 38-8
endbmatrix
beginbmatrix
x \
y \
z
endbmatrix=
beginbmatrix
0 \
0 \
0
endbmatrix $$

$$0x-6y-6z=0$$
$$30x-30y-30z=0$$
$$-30x+30y+30z=0$$

I find the solutions to the equations as:

$$x=0$$
$$y=-1$$
$$z=1$$

Am I going in the right direction? What does the results tell me about which eigenvectors exist for the matrix B?

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited Mar 24 at 8:06

Rodrigo de Azevedo

13.2k41962

asked Mar 24 at 7:38

Boris GrunwaldBoris Grunwald

2797

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, this tells you it has exactly one eigenvector, of eigenvalue 8. This is correct. Unless the matrix is hermitian, the number of eigenvectors (ie. dimension of the eigenspace) for a specified eigenvalue can be less than its multiplicity. You should look up on generalized eigenvectors (eigenvectors for $(B−λE)^k$ ). These are indeed equal in number to the multiplicity of the eigenvalue. You should look up generalized eigenvectors and the Jordan normal form: en.wikipedia.org/wiki/Generalized_eigenvector
  $endgroup$
  – user3257842
  Mar 24 at 7:59
  

  
  
  
  

add a comment | 

2

$begingroup$

I have the following Matrix:

$$B= beginbmatrix
8 & -6 & -6 \
30 & -22 & -30 \
-30 & 30 & 38
endbmatrix $$

I find the characteristic polynomial as:

$$det(B-lambda E)=-(lambda-8)^3$$

So we have the root 8 (with multiplicty 3). Therefore $B$ has the eigenvalue 8

Now I solve for

$$(B-lambda E)x= beginbmatrix
8-8 & -6 & -6 \
30 & -22-8 & -30 \
-30 & 30 & 38-8
endbmatrix
beginbmatrix
x \
y \
z
endbmatrix=
beginbmatrix
0 \
0 \
0
endbmatrix $$

$$0x-6y-6z=0$$
$$30x-30y-30z=0$$
$$-30x+30y+30z=0$$

I find the solutions to the equations as:

$$x=0$$
$$y=-1$$
$$z=1$$

Am I going in the right direction? What does the results tell me about which eigenvectors exist for the matrix B?

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited Mar 24 at 8:06

Rodrigo de Azevedo

13.2k41962

asked Mar 24 at 7:38

Boris GrunwaldBoris Grunwald

2797

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, this tells you it has exactly one eigenvector, of eigenvalue 8. This is correct. Unless the matrix is hermitian, the number of eigenvectors (ie. dimension of the eigenspace) for a specified eigenvalue can be less than its multiplicity. You should look up on generalized eigenvectors (eigenvectors for $(B−λE)^k$ ). These are indeed equal in number to the multiplicity of the eigenvalue. You should look up generalized eigenvectors and the Jordan normal form: en.wikipedia.org/wiki/Generalized_eigenvector
  $endgroup$
  – user3257842
  Mar 24 at 7:59
  

  
  
  
  

add a comment | 

$begingroup$

I have the following Matrix:

$$B= beginbmatrix
8 & -6 & -6 \
30 & -22 & -30 \
-30 & 30 & 38
endbmatrix $$

I find the characteristic polynomial as:

$$det(B-lambda E)=-(lambda-8)^3$$

So we have the root 8 (with multiplicty 3). Therefore $B$ has the eigenvalue 8

Now I solve for

$$(B-lambda E)x= beginbmatrix
8-8 & -6 & -6 \
30 & -22-8 & -30 \
-30 & 30 & 38-8
endbmatrix
beginbmatrix
x \
y \
z
endbmatrix=
beginbmatrix
0 \
0 \
0
endbmatrix $$

$$0x-6y-6z=0$$
$$30x-30y-30z=0$$
$$-30x+30y+30z=0$$

I find the solutions to the equations as:

$$x=0$$
$$y=-1$$
$$z=1$$

Am I going in the right direction? What does the results tell me about which eigenvectors exist for the matrix B?

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited Mar 24 at 8:06

Rodrigo de Azevedo

13.2k41962

asked Mar 24 at 7:38

Boris GrunwaldBoris Grunwald

2797

$endgroup$

share|cite|improve this question

edited Mar 24 at 8:06

Rodrigo de Azevedo

13.2k41962

asked Mar 24 at 7:38

Boris GrunwaldBoris Grunwald

2797

share|cite|improve this question

edited Mar 24 at 8:06

Rodrigo de Azevedo

13.2k41962

asked Mar 24 at 7:38

Boris GrunwaldBoris Grunwald

2797

edited Mar 24 at 8:06

Rodrigo de Azevedo

13.2k41962

edited Mar 24 at 8:06

Rodrigo de Azevedo

13.2k41962

asked Mar 24 at 7:38

Boris GrunwaldBoris Grunwald

2797

asked Mar 24 at 7:38

Boris GrunwaldBoris Grunwald

2797

1 Answer
1

active

oldest

votes

1

$begingroup$

What you've done so far is find an eigenvector corresponding to the eigenvalue $lambda=8$. This doesn't tell you anything about other possible (linearly independent) eigenvectors. To determine their existence, you'll need to determine $mathrmdimN(lambda I - B)$, where $N(cdot)$ denotes the null space. Looking at the linear system, we can see that we must have $x=0$. Then, we can easily see that, though there are infinitely many solutions, they are all of the form
$$v=beginpmatrix 0\ v_2\ -v_2 endpmatrix.$$
Therefore, there is only the one linearly independent eigenvector (i.e., $mathrmdimN(lambda I - B) = 1$).

share|cite|improve this answer

edited Mar 24 at 8:05

answered Mar 24 at 7:58

Gary MoonGary Moon

921127

$endgroup$

add a comment | 

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1

$begingroup$

What you've done so far is find an eigenvector corresponding to the eigenvalue $lambda=8$. This doesn't tell you anything about other possible (linearly independent) eigenvectors. To determine their existence, you'll need to determine $mathrmdimN(lambda I - B)$, where $N(cdot)$ denotes the null space. Looking at the linear system, we can see that we must have $x=0$. Then, we can easily see that, though there are infinitely many solutions, they are all of the form
$$v=beginpmatrix 0\ v_2\ -v_2 endpmatrix.$$
Therefore, there is only the one linearly independent eigenvector (i.e., $mathrmdimN(lambda I - B) = 1$).

share|cite|improve this answer

edited Mar 24 at 8:05

answered Mar 24 at 7:58

Gary MoonGary Moon

921127

$endgroup$

add a comment | 

1

$begingroup$

What you've done so far is find an eigenvector corresponding to the eigenvalue $lambda=8$. This doesn't tell you anything about other possible (linearly independent) eigenvectors. To determine their existence, you'll need to determine $mathrmdimN(lambda I - B)$, where $N(cdot)$ denotes the null space. Looking at the linear system, we can see that we must have $x=0$. Then, we can easily see that, though there are infinitely many solutions, they are all of the form
$$v=beginpmatrix 0\ v_2\ -v_2 endpmatrix.$$
Therefore, there is only the one linearly independent eigenvector (i.e., $mathrmdimN(lambda I - B) = 1$).

share|cite|improve this answer

edited Mar 24 at 8:05

answered Mar 24 at 7:58

Gary MoonGary Moon

921127

$endgroup$

add a comment | 

$begingroup$

What you've done so far is find an eigenvector corresponding to the eigenvalue $lambda=8$. This doesn't tell you anything about other possible (linearly independent) eigenvectors. To determine their existence, you'll need to determine $mathrmdimN(lambda I - B)$, where $N(cdot)$ denotes the null space. Looking at the linear system, we can see that we must have $x=0$. Then, we can easily see that, though there are infinitely many solutions, they are all of the form
$$v=beginpmatrix 0\ v_2\ -v_2 endpmatrix.$$
Therefore, there is only the one linearly independent eigenvector (i.e., $mathrmdimN(lambda I - B) = 1$).

share|cite|improve this answer

edited Mar 24 at 8:05

answered Mar 24 at 7:58

Gary MoonGary Moon

921127

$endgroup$

share|cite|improve this answer

edited Mar 24 at 8:05

answered Mar 24 at 7:58

Gary MoonGary Moon

921127

answered Mar 24 at 7:58

Gary MoonGary Moon

921127

answered Mar 24 at 7:58

Gary MoonGary Moon

921127