How to find dimension of eigenspace?
$begingroup$
Given $lambda$ = 2 and matrix A:
A =
$begin{bmatrix}
2 & 1 & 0 \
0 & 2 & 0 \
0 & 0 & 2 \
end{bmatrix}$
My textbook says that to find the eigenvectors of $lambda = 2$, solve the homogeneous linear system represented by $(2I - A)x = 0$.
$2I - A$ =
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$
"This implies that $x_2$ = 0".
How is the matrix
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$
computed and how does this imply that $x_2$ = 0? My textbook does not give an explanation.
linear-algebra
asked Dec 4 '18 at 20:18
Evan KimEvan Kim
998
$endgroup$
|
show 3 more comments
$begingroup$
Given $lambda$ = 2 and matrix A:
A =
$begin{bmatrix}
2 & 1 & 0 \
0 & 2 & 0 \
0 & 0 & 2 \
end{bmatrix}$
My textbook says that to find the eigenvectors of $lambda = 2$, solve the homogeneous linear system represented by $(2I - A)x = 0$.
$2I - A$ =
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$
"This implies that $x_2$ = 0".
How is the matrix
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$
computed and how does this imply that $x_2$ = 0? My textbook does not give an explanation.
linear-algebra
asked Dec 4 '18 at 20:18
Evan KimEvan Kim
998
$endgroup$
$begingroup$
The matrix that you're asking about is really $2I-A$, just the operations directly on the matrices, there's not really anything to the computation...
$endgroup$
– Michael Burr
Dec 4 '18 at 20:22
$begingroup$
The rank of $2I-A$ is the codimension of the eigenspace $E_2$.
$endgroup$
– Bernard
Dec 4 '18 at 20:27
$begingroup$
@Evan Kim This is the second question regarding this matrix and this exercise in general. Maybe you should read again about eigenvalues, characteristic polynomial, kernel, rank, dimension... It seems that the definitions are not completely clear. I feel it would be better for you if you first go back to study the concepts
$endgroup$
– Federico
Dec 4 '18 at 20:27
$begingroup$
$dim ker(2I-A) = 3-mathrm{rank}(2I-A)=3-1=2$
$endgroup$
– Federico
Dec 4 '18 at 20:29
$begingroup$
@Federico The issue is that I am having a difficult time grasping the definitions in the study material assigned to me in class. I do agree that these are trivial questions that should be self-explanatory though yet I have still struggled the entire semester. An example is the book explains rank and dimension and I understand that, but it fails to provide a connection between concepts in different chapters and perhaps expect us to see these connections by ourselves.
$endgroup$
– Evan Kim
Dec 4 '18 at 21:25
|
show 3 more comments
$begingroup$
Given $lambda$ = 2 and matrix A:
A =
$begin{bmatrix}
2 & 1 & 0 \
0 & 2 & 0 \
0 & 0 & 2 \
end{bmatrix}$
My textbook says that to find the eigenvectors of $lambda = 2$, solve the homogeneous linear system represented by $(2I - A)x = 0$.
$2I - A$ =
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$
"This implies that $x_2$ = 0".
How is the matrix
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$
computed and how does this imply that $x_2$ = 0? My textbook does not give an explanation.
linear-algebra
asked Dec 4 '18 at 20:18
Evan KimEvan Kim
998
$endgroup$
Given $lambda$ = 2 and matrix A:
A =
$begin{bmatrix}
2 & 1 & 0 \
0 & 2 & 0 \
0 & 0 & 2 \
end{bmatrix}$
My textbook says that to find the eigenvectors of $lambda = 2$, solve the homogeneous linear system represented by $(2I - A)x = 0$.
$2I - A$ =
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$
"This implies that $x_2$ = 0".
How is the matrix
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$
computed and how does this imply that $x_2$ = 0? My textbook does not give an explanation.
linear-algebra
linear-algebra
asked Dec 4 '18 at 20:18
Evan KimEvan Kim
998
asked Dec 4 '18 at 20:18
Evan KimEvan Kim
998
asked Dec 4 '18 at 20:18
Evan KimEvan Kim
998
asked Dec 4 '18 at 20:18
Evan KimEvan Kim
998
asked Dec 4 '18 at 20:18
Evan KimEvan Kim
998
998
$begingroup$
The matrix that you're asking about is really $2I-A$, just the operations directly on the matrices, there's not really anything to the computation...
$endgroup$
– Michael Burr
Dec 4 '18 at 20:22
$begingroup$
The rank of $2I-A$ is the codimension of the eigenspace $E_2$.
$endgroup$
– Bernard
Dec 4 '18 at 20:27
$begingroup$
@Evan Kim This is the second question regarding this matrix and this exercise in general. Maybe you should read again about eigenvalues, characteristic polynomial, kernel, rank, dimension... It seems that the definitions are not completely clear. I feel it would be better for you if you first go back to study the concepts
$endgroup$
– Federico
Dec 4 '18 at 20:27
$begingroup$
$dim ker(2I-A) = 3-mathrm{rank}(2I-A)=3-1=2$
$endgroup$
– Federico
Dec 4 '18 at 20:29
$begingroup$
@Federico The issue is that I am having a difficult time grasping the definitions in the study material assigned to me in class. I do agree that these are trivial questions that should be self-explanatory though yet I have still struggled the entire semester. An example is the book explains rank and dimension and I understand that, but it fails to provide a connection between concepts in different chapters and perhaps expect us to see these connections by ourselves.
$endgroup$
– Evan Kim
Dec 4 '18 at 21:25
|
show 3 more comments
$begingroup$
The matrix that you're asking about is really $2I-A$, just the operations directly on the matrices, there's not really anything to the computation...
$endgroup$
– Michael Burr
Dec 4 '18 at 20:22
$begingroup$
The rank of $2I-A$ is the codimension of the eigenspace $E_2$.
$endgroup$
– Bernard
Dec 4 '18 at 20:27
$begingroup$
@Evan Kim This is the second question regarding this matrix and this exercise in general. Maybe you should read again about eigenvalues, characteristic polynomial, kernel, rank, dimension... It seems that the definitions are not completely clear. I feel it would be better for you if you first go back to study the concepts
$endgroup$
– Federico
Dec 4 '18 at 20:27
$begingroup$
$dim ker(2I-A) = 3-mathrm{rank}(2I-A)=3-1=2$
$endgroup$
– Federico
Dec 4 '18 at 20:29
$begingroup$
@Federico The issue is that I am having a difficult time grasping the definitions in the study material assigned to me in class. I do agree that these are trivial questions that should be self-explanatory though yet I have still struggled the entire semester. An example is the book explains rank and dimension and I understand that, but it fails to provide a connection between concepts in different chapters and perhaps expect us to see these connections by ourselves.
$endgroup$
– Evan Kim
Dec 4 '18 at 21:25
$begingroup$
The matrix that you're asking about is really $2I-A$, just the operations directly on the matrices, there's not really anything to the computation...
$endgroup$
– Michael Burr
Dec 4 '18 at 20:22
$begingroup$
The matrix that you're asking about is really $2I-A$, just the operations directly on the matrices, there's not really anything to the computation...
$endgroup$
– Michael Burr
Dec 4 '18 at 20:22
$begingroup$
The rank of $2I-A$ is the codimension of the eigenspace $E_2$.
$endgroup$
– Bernard
Dec 4 '18 at 20:27
$begingroup$
The rank of $2I-A$ is the codimension of the eigenspace $E_2$.
$endgroup$
– Bernard
Dec 4 '18 at 20:27
$begingroup$
@Evan Kim This is the second question regarding this matrix and this exercise in general. Maybe you should read again about eigenvalues, characteristic polynomial, kernel, rank, dimension... It seems that the definitions are not completely clear. I feel it would be better for you if you first go back to study the concepts
$endgroup$
– Federico
Dec 4 '18 at 20:27
$begingroup$
@Evan Kim This is the second question regarding this matrix and this exercise in general. Maybe you should read again about eigenvalues, characteristic polynomial, kernel, rank, dimension... It seems that the definitions are not completely clear. I feel it would be better for you if you first go back to study the concepts
$endgroup$
– Federico
Dec 4 '18 at 20:27
$begingroup$
$dim ker(2I-A) = 3-mathrm{rank}(2I-A)=3-1=2$
$endgroup$
– Federico
Dec 4 '18 at 20:29
$begingroup$
$dim ker(2I-A) = 3-mathrm{rank}(2I-A)=3-1=2$
$endgroup$
– Federico
Dec 4 '18 at 20:29
$begingroup$
@Federico The issue is that I am having a difficult time grasping the definitions in the study material assigned to me in class. I do agree that these are trivial questions that should be self-explanatory though yet I have still struggled the entire semester. An example is the book explains rank and dimension and I understand that, but it fails to provide a connection between concepts in different chapters and perhaps expect us to see these connections by ourselves.
$endgroup$
– Evan Kim
Dec 4 '18 at 21:25
$begingroup$
@Federico The issue is that I am having a difficult time grasping the definitions in the study material assigned to me in class. I do agree that these are trivial questions that should be self-explanatory though yet I have still struggled the entire semester. An example is the book explains rank and dimension and I understand that, but it fails to provide a connection between concepts in different chapters and perhaps expect us to see these connections by ourselves.
$endgroup$
– Evan Kim
Dec 4 '18 at 21:25
|
show 3 more comments
3 Answers
3
active
oldest
votes
$begingroup$
The matrix $2I-A$ is simply computed as $$begin{bmatrix}2 & 0 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}-begin{bmatrix}2 & 1 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}=begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$$
If a vector $x$ with coordinates $x_1,x_2,x_3$ satisfies $(2I-A)x=0$, then it must be that $x_2=0$. This is because $$begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}begin{bmatrix}x_1\x_2\x_3end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}implies begin{bmatrix}-x_2\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$ so we must have $x_2=0$. Notice that $x_1$ and $x_3$ can be any number, so what does this tell you about the dimension of the eigenspace (i.e. the dimension of ${xinmathbb C^3:(2A-I)x=0}$)?
answered Dec 4 '18 at 20:31
DaveDave
8,76711033
$endgroup$
add a comment |
$begingroup$
We have that
$$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}vec x=vec 0 implies vec x=(x_1,0,x_3)$$
therefore what is the dimension of the eigenspace?
answered Dec 4 '18 at 20:20
gimusigimusi
92.9k94494
$endgroup$
add a comment |
$begingroup$
You must find all a basis for the space of vectors such that:
$Amathbf v = lambda mathbf v$
and with a little algebra
$(A- lambda I)mathbf v = mathbf 0$
If $x_2 ne 0$
$(A- 2I)mathbf v = begin{bmatrix} x_2\0\0 end{bmatrix} ne mathbf 0$
answered Dec 4 '18 at 20:39
Doug MDoug M
44.5k31854
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
The matrix $2I-A$ is simply computed as $$begin{bmatrix}2 & 0 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}-begin{bmatrix}2 & 1 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}=begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$$
If a vector $x$ with coordinates $x_1,x_2,x_3$ satisfies $(2I-A)x=0$, then it must be that $x_2=0$. This is because $$begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}begin{bmatrix}x_1\x_2\x_3end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}implies begin{bmatrix}-x_2\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$ so we must have $x_2=0$. Notice that $x_1$ and $x_3$ can be any number, so what does this tell you about the dimension of the eigenspace (i.e. the dimension of ${xinmathbb C^3:(2A-I)x=0}$)?
answered Dec 4 '18 at 20:31
DaveDave
8,76711033
$endgroup$
add a comment |
$begingroup$
The matrix $2I-A$ is simply computed as $$begin{bmatrix}2 & 0 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}-begin{bmatrix}2 & 1 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}=begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$$
If a vector $x$ with coordinates $x_1,x_2,x_3$ satisfies $(2I-A)x=0$, then it must be that $x_2=0$. This is because $$begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}begin{bmatrix}x_1\x_2\x_3end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}implies begin{bmatrix}-x_2\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$ so we must have $x_2=0$. Notice that $x_1$ and $x_3$ can be any number, so what does this tell you about the dimension of the eigenspace (i.e. the dimension of ${xinmathbb C^3:(2A-I)x=0}$)?
answered Dec 4 '18 at 20:31
DaveDave
8,76711033
$endgroup$
add a comment |
$begingroup$
The matrix $2I-A$ is simply computed as $$begin{bmatrix}2 & 0 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}-begin{bmatrix}2 & 1 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}=begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$$
If a vector $x$ with coordinates $x_1,x_2,x_3$ satisfies $(2I-A)x=0$, then it must be that $x_2=0$. This is because $$begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}begin{bmatrix}x_1\x_2\x_3end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}implies begin{bmatrix}-x_2\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$ so we must have $x_2=0$. Notice that $x_1$ and $x_3$ can be any number, so what does this tell you about the dimension of the eigenspace (i.e. the dimension of ${xinmathbb C^3:(2A-I)x=0}$)?
answered Dec 4 '18 at 20:31
DaveDave
8,76711033
$endgroup$
The matrix $2I-A$ is simply computed as $$begin{bmatrix}2 & 0 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}-begin{bmatrix}2 & 1 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}=begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$$
If a vector $x$ with coordinates $x_1,x_2,x_3$ satisfies $(2I-A)x=0$, then it must be that $x_2=0$. This is because $$begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}begin{bmatrix}x_1\x_2\x_3end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}implies begin{bmatrix}-x_2\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$ so we must have $x_2=0$. Notice that $x_1$ and $x_3$ can be any number, so what does this tell you about the dimension of the eigenspace (i.e. the dimension of ${xinmathbb C^3:(2A-I)x=0}$)?
answered Dec 4 '18 at 20:31
DaveDave
8,76711033
answered Dec 4 '18 at 20:31
DaveDave
8,76711033
answered Dec 4 '18 at 20:31
DaveDave
8,76711033
answered Dec 4 '18 at 20:31
DaveDave
8,76711033
8,76711033
add a comment |
add a comment |
$begingroup$
We have that
$$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}vec x=vec 0 implies vec x=(x_1,0,x_3)$$
therefore what is the dimension of the eigenspace?
answered Dec 4 '18 at 20:20
gimusigimusi
92.9k94494
$endgroup$
add a comment |
$begingroup$
We have that
$$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}vec x=vec 0 implies vec x=(x_1,0,x_3)$$
therefore what is the dimension of the eigenspace?
answered Dec 4 '18 at 20:20
gimusigimusi
92.9k94494
$endgroup$
add a comment |
$begingroup$
We have that
$$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}vec x=vec 0 implies vec x=(x_1,0,x_3)$$
therefore what is the dimension of the eigenspace?
answered Dec 4 '18 at 20:20
gimusigimusi
92.9k94494
$endgroup$
We have that
$$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}vec x=vec 0 implies vec x=(x_1,0,x_3)$$
therefore what is the dimension of the eigenspace?
answered Dec 4 '18 at 20:20
gimusigimusi
92.9k94494
answered Dec 4 '18 at 20:20
gimusigimusi
92.9k94494
answered Dec 4 '18 at 20:20
gimusigimusi
92.9k94494
answered Dec 4 '18 at 20:20
gimusigimusi
92.9k94494
92.9k94494
add a comment |
add a comment |
$begingroup$
You must find all a basis for the space of vectors such that:
$Amathbf v = lambda mathbf v$
and with a little algebra
$(A- lambda I)mathbf v = mathbf 0$
If $x_2 ne 0$
$(A- 2I)mathbf v = begin{bmatrix} x_2\0\0 end{bmatrix} ne mathbf 0$
answered Dec 4 '18 at 20:39
Doug MDoug M
44.5k31854
$endgroup$
add a comment |
$begingroup$
You must find all a basis for the space of vectors such that:
$Amathbf v = lambda mathbf v$
and with a little algebra
$(A- lambda I)mathbf v = mathbf 0$
If $x_2 ne 0$
$(A- 2I)mathbf v = begin{bmatrix} x_2\0\0 end{bmatrix} ne mathbf 0$
answered Dec 4 '18 at 20:39
Doug MDoug M
44.5k31854
$endgroup$
add a comment |
$begingroup$
You must find all a basis for the space of vectors such that:
$Amathbf v = lambda mathbf v$
and with a little algebra
$(A- lambda I)mathbf v = mathbf 0$
If $x_2 ne 0$
$(A- 2I)mathbf v = begin{bmatrix} x_2\0\0 end{bmatrix} ne mathbf 0$
answered Dec 4 '18 at 20:39
Doug MDoug M
44.5k31854
$endgroup$
You must find all a basis for the space of vectors such that:
$Amathbf v = lambda mathbf v$
and with a little algebra
$(A- lambda I)mathbf v = mathbf 0$
If $x_2 ne 0$
$(A- 2I)mathbf v = begin{bmatrix} x_2\0\0 end{bmatrix} ne mathbf 0$
answered Dec 4 '18 at 20:39
Doug MDoug M
44.5k31854
answered Dec 4 '18 at 20:39
Doug MDoug M
44.5k31854
answered Dec 4 '18 at 20:39
Doug MDoug M
44.5k31854
answered Dec 4 '18 at 20:39
Doug MDoug M
44.5k31854
44.5k31854
add a comment |
add a comment |
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$begingroup$
The matrix that you're asking about is really $2I-A$, just the operations directly on the matrices, there's not really anything to the computation...
$endgroup$
– Michael Burr
Dec 4 '18 at 20:22
$begingroup$
The rank of $2I-A$ is the codimension of the eigenspace $E_2$.
$endgroup$
– Bernard
Dec 4 '18 at 20:27
$begingroup$
@Evan Kim This is the second question regarding this matrix and this exercise in general. Maybe you should read again about eigenvalues, characteristic polynomial, kernel, rank, dimension... It seems that the definitions are not completely clear. I feel it would be better for you if you first go back to study the concepts
$endgroup$
– Federico
Dec 4 '18 at 20:27
$begingroup$
$dim ker(2I-A) = 3-mathrm{rank}(2I-A)=3-1=2$
$endgroup$
– Federico
Dec 4 '18 at 20:29
$begingroup$
@Federico The issue is that I am having a difficult time grasping the definitions in the study material assigned to me in class. I do agree that these are trivial questions that should be self-explanatory though yet I have still struggled the entire semester. An example is the book explains rank and dimension and I understand that, but it fails to provide a connection between concepts in different chapters and perhaps expect us to see these connections by ourselves.
$endgroup$
– Evan Kim
Dec 4 '18 at 21:25