How to find a basis of eigenvectors??
$begingroup$
DISCLAIMER ! THIS IS NOT A DUPLICATE OF MY OTHER QUESTIONS. THE QUESTION I AM ASKING NOW ONLY TAKES INTO CONSIDERATION THE SAME MATRIX I HAVE USED IN OTHER QUESTIONS. THAT'S IT.
So i have this matrix
A= $$
begin{pmatrix}
-2 & 0 & 0 \
3 & 1 & -6 \
0 & 0 & -2 \
end{pmatrix}
$$
We compute the characteristic polynomial
p(λ) = det(A − λId) = (λ + 2)²· (1 − λ)
which gives me two roots namely λ1 = −2 of multiplicity 2, and λ2 = 1 of multiplicity 1.
First of all , I solve the homogeneous system associated with A + 2Id which gives us
Nul(A+2Id)=Span $$
begin{pmatrix}
2 & -1 \
0 & 1 \
1 & 0 \
end{pmatrix}
$$
I know how to solve the system associated with A+2Id since the matrix i get by doing (A+2Id) is pretty simple (there are two 0 rows) and the homogeneous equation associated with it is x+y-2z=0
However my confusion arises when , in order to find a basis of eigenvectors, I need to compute E(λ2), so we need to solve the homogeneous system associated with(A − Id).
I do not now how to go on from here.
Computing A-Id (please correct me if im wrong, since A-1(Id)=A-Id), gives me the matrix
A-Id= $$
begin{pmatrix}
-3 & 0 & 0 \
3 & 0 & -6 \
0 & 0 & -3 \
end{pmatrix}
$$
How do i proceed from here? On the answers sheet it says that Nul(A − Id) = Span (column vector)= (0,1,0).
Should i compute the characteristic polynomial of the (A-Id) matrix? How do i do that ? Is that the right way to do that?
Please if anybody can show me some simple and clear steps on how to get
Nul(A − Id) = Span (column vector)= (0,1,0) and tell me if what I'm doing is right! Thanks!
linear-algebra matrices matrix-equations
asked Dec 2 '18 at 12:35
BM97BM97
758
$endgroup$
|
show 8 more comments
$begingroup$
DISCLAIMER ! THIS IS NOT A DUPLICATE OF MY OTHER QUESTIONS. THE QUESTION I AM ASKING NOW ONLY TAKES INTO CONSIDERATION THE SAME MATRIX I HAVE USED IN OTHER QUESTIONS. THAT'S IT.
So i have this matrix
A= $$
begin{pmatrix}
-2 & 0 & 0 \
3 & 1 & -6 \
0 & 0 & -2 \
end{pmatrix}
$$
We compute the characteristic polynomial
p(λ) = det(A − λId) = (λ + 2)²· (1 − λ)
which gives me two roots namely λ1 = −2 of multiplicity 2, and λ2 = 1 of multiplicity 1.
First of all , I solve the homogeneous system associated with A + 2Id which gives us
Nul(A+2Id)=Span $$
begin{pmatrix}
2 & -1 \
0 & 1 \
1 & 0 \
end{pmatrix}
$$
I know how to solve the system associated with A+2Id since the matrix i get by doing (A+2Id) is pretty simple (there are two 0 rows) and the homogeneous equation associated with it is x+y-2z=0
However my confusion arises when , in order to find a basis of eigenvectors, I need to compute E(λ2), so we need to solve the homogeneous system associated with(A − Id).
I do not now how to go on from here.
Computing A-Id (please correct me if im wrong, since A-1(Id)=A-Id), gives me the matrix
A-Id= $$
begin{pmatrix}
-3 & 0 & 0 \
3 & 0 & -6 \
0 & 0 & -3 \
end{pmatrix}
$$
How do i proceed from here? On the answers sheet it says that Nul(A − Id) = Span (column vector)= (0,1,0).
Should i compute the characteristic polynomial of the (A-Id) matrix? How do i do that ? Is that the right way to do that?
Please if anybody can show me some simple and clear steps on how to get
Nul(A − Id) = Span (column vector)= (0,1,0) and tell me if what I'm doing is right! Thanks!
linear-algebra matrices matrix-equations
asked Dec 2 '18 at 12:35
BM97BM97
758
$endgroup$
$begingroup$
You calculated $A-I$ incorrectly.
$endgroup$
– David C. Ullrich
Dec 2 '18 at 13:02
$begingroup$
How come ? where did i do it wrong
$endgroup$
– BM97
Dec 2 '18 at 13:02
$begingroup$
Come on now! I mean really - figuring out $A-I$ is not that hard. Do the subtraction again, carefully; what you get is not what you say $A-I$ is above.
$endgroup$
– David C. Ullrich
Dec 2 '18 at 13:19
$begingroup$
I did it again . On the second row i still got -6, but the user below got 6?
$endgroup$
– BM97
Dec 2 '18 at 13:20
1
$begingroup$
It seems you found the error, as you just edited it away.
$endgroup$
– Ingix
Dec 2 '18 at 13:29
|
show 8 more comments
$begingroup$
DISCLAIMER ! THIS IS NOT A DUPLICATE OF MY OTHER QUESTIONS. THE QUESTION I AM ASKING NOW ONLY TAKES INTO CONSIDERATION THE SAME MATRIX I HAVE USED IN OTHER QUESTIONS. THAT'S IT.
So i have this matrix
A= $$
begin{pmatrix}
-2 & 0 & 0 \
3 & 1 & -6 \
0 & 0 & -2 \
end{pmatrix}
$$
We compute the characteristic polynomial
p(λ) = det(A − λId) = (λ + 2)²· (1 − λ)
which gives me two roots namely λ1 = −2 of multiplicity 2, and λ2 = 1 of multiplicity 1.
First of all , I solve the homogeneous system associated with A + 2Id which gives us
Nul(A+2Id)=Span $$
begin{pmatrix}
2 & -1 \
0 & 1 \
1 & 0 \
end{pmatrix}
$$
I know how to solve the system associated with A+2Id since the matrix i get by doing (A+2Id) is pretty simple (there are two 0 rows) and the homogeneous equation associated with it is x+y-2z=0
However my confusion arises when , in order to find a basis of eigenvectors, I need to compute E(λ2), so we need to solve the homogeneous system associated with(A − Id).
I do not now how to go on from here.
Computing A-Id (please correct me if im wrong, since A-1(Id)=A-Id), gives me the matrix
A-Id= $$
begin{pmatrix}
-3 & 0 & 0 \
3 & 0 & -6 \
0 & 0 & -3 \
end{pmatrix}
$$
How do i proceed from here? On the answers sheet it says that Nul(A − Id) = Span (column vector)= (0,1,0).
Should i compute the characteristic polynomial of the (A-Id) matrix? How do i do that ? Is that the right way to do that?
Please if anybody can show me some simple and clear steps on how to get
Nul(A − Id) = Span (column vector)= (0,1,0) and tell me if what I'm doing is right! Thanks!
linear-algebra matrices matrix-equations
asked Dec 2 '18 at 12:35
BM97BM97
758
$endgroup$
DISCLAIMER ! THIS IS NOT A DUPLICATE OF MY OTHER QUESTIONS. THE QUESTION I AM ASKING NOW ONLY TAKES INTO CONSIDERATION THE SAME MATRIX I HAVE USED IN OTHER QUESTIONS. THAT'S IT.
So i have this matrix
A= $$
begin{pmatrix}
-2 & 0 & 0 \
3 & 1 & -6 \
0 & 0 & -2 \
end{pmatrix}
$$
We compute the characteristic polynomial
p(λ) = det(A − λId) = (λ + 2)²· (1 − λ)
which gives me two roots namely λ1 = −2 of multiplicity 2, and λ2 = 1 of multiplicity 1.
First of all , I solve the homogeneous system associated with A + 2Id which gives us
Nul(A+2Id)=Span $$
begin{pmatrix}
2 & -1 \
0 & 1 \
1 & 0 \
end{pmatrix}
$$
I know how to solve the system associated with A+2Id since the matrix i get by doing (A+2Id) is pretty simple (there are two 0 rows) and the homogeneous equation associated with it is x+y-2z=0
However my confusion arises when , in order to find a basis of eigenvectors, I need to compute E(λ2), so we need to solve the homogeneous system associated with(A − Id).
I do not now how to go on from here.
Computing A-Id (please correct me if im wrong, since A-1(Id)=A-Id), gives me the matrix
A-Id= $$
begin{pmatrix}
-3 & 0 & 0 \
3 & 0 & -6 \
0 & 0 & -3 \
end{pmatrix}
$$
How do i proceed from here? On the answers sheet it says that Nul(A − Id) = Span (column vector)= (0,1,0).
Should i compute the characteristic polynomial of the (A-Id) matrix? How do i do that ? Is that the right way to do that?
Please if anybody can show me some simple and clear steps on how to get
Nul(A − Id) = Span (column vector)= (0,1,0) and tell me if what I'm doing is right! Thanks!
linear-algebra matrices matrix-equations
linear-algebra matrices matrix-equations
asked Dec 2 '18 at 12:35
BM97BM97
758
asked Dec 2 '18 at 12:35
BM97BM97
758
edited Dec 2 '18 at 13:28
BM97
asked Dec 2 '18 at 12:35
BM97BM97
758
asked Dec 2 '18 at 12:35
BM97BM97
758
asked Dec 2 '18 at 12:35
BM97BM97
758
758
$begingroup$
You calculated $A-I$ incorrectly.
$endgroup$
– David C. Ullrich
Dec 2 '18 at 13:02
$begingroup$
How come ? where did i do it wrong
$endgroup$
– BM97
Dec 2 '18 at 13:02
$begingroup$
Come on now! I mean really - figuring out $A-I$ is not that hard. Do the subtraction again, carefully; what you get is not what you say $A-I$ is above.
$endgroup$
– David C. Ullrich
Dec 2 '18 at 13:19
$begingroup$
I did it again . On the second row i still got -6, but the user below got 6?
$endgroup$
– BM97
Dec 2 '18 at 13:20
1
$begingroup$
It seems you found the error, as you just edited it away.
$endgroup$
– Ingix
Dec 2 '18 at 13:29
|
show 8 more comments
$begingroup$
You calculated $A-I$ incorrectly.
$endgroup$
– David C. Ullrich
Dec 2 '18 at 13:02
$begingroup$
How come ? where did i do it wrong
$endgroup$
– BM97
Dec 2 '18 at 13:02
$begingroup$
Come on now! I mean really - figuring out $A-I$ is not that hard. Do the subtraction again, carefully; what you get is not what you say $A-I$ is above.
$endgroup$
– David C. Ullrich
Dec 2 '18 at 13:19
$begingroup$
I did it again . On the second row i still got -6, but the user below got 6?
$endgroup$
– BM97
Dec 2 '18 at 13:20
1
$begingroup$
It seems you found the error, as you just edited it away.
$endgroup$
– Ingix
Dec 2 '18 at 13:29
$begingroup$
You calculated $A-I$ incorrectly.
$endgroup$
– David C. Ullrich
Dec 2 '18 at 13:02
$begingroup$
You calculated $A-I$ incorrectly.
$endgroup$
– David C. Ullrich
Dec 2 '18 at 13:02
$begingroup$
How come ? where did i do it wrong
$endgroup$
– BM97
Dec 2 '18 at 13:02
$begingroup$
How come ? where did i do it wrong
$endgroup$
– BM97
Dec 2 '18 at 13:02
$begingroup$
Come on now! I mean really - figuring out $A-I$ is not that hard. Do the subtraction again, carefully; what you get is not what you say $A-I$ is above.
$endgroup$
– David C. Ullrich
Dec 2 '18 at 13:19
$begingroup$
Come on now! I mean really - figuring out $A-I$ is not that hard. Do the subtraction again, carefully; what you get is not what you say $A-I$ is above.
$endgroup$
– David C. Ullrich
Dec 2 '18 at 13:19
$begingroup$
I did it again . On the second row i still got -6, but the user below got 6?
$endgroup$
– BM97
Dec 2 '18 at 13:20
$begingroup$
I did it again . On the second row i still got -6, but the user below got 6?
$endgroup$
– BM97
Dec 2 '18 at 13:20
1
1
$begingroup$
It seems you found the error, as you just edited it away.
$endgroup$
– Ingix
Dec 2 '18 at 13:29
$begingroup$
It seems you found the error, as you just edited it away.
$endgroup$
– Ingix
Dec 2 '18 at 13:29
|
show 8 more comments
1 Answer
1
active
oldest
votes
$begingroup$
You put $A-I$ in reduced row echelon form:
$$begin{pmatrix}-3&0&0\3&0&-6\0&0&-3end{pmatrix}rightsquigarrowbegin{pmatrix}1&0&0\1&0&-2\0&0&1end{pmatrix}rightsquigarrowbegin{pmatrix}1&0&0\0&0&-2\0&0&1end{pmatrix}rightsquigarrowbegin{pmatrix}1&0&0\0&0&1\0&0&0end{pmatrix}$$
so we have the equations
$$x=0,quad z=0.$$
answered Dec 2 '18 at 12:45
BernardBernard
119k740113
$endgroup$
$begingroup$
Thanks ! What if i try to to it like this. From the matrix (A-Id) i can directly see that x=0 and z=0. This leaves me with the second row being 3(0)+y-6(0)=0. How do i solve for y which gives me as a y basis 1?
$endgroup$
– BM97
Dec 2 '18 at 13:02
$begingroup$
Shouldn't it be -6 on the second row? (-6-0)=-6
$endgroup$
– BM97
Dec 2 '18 at 13:15
$begingroup$
You're right – a mistyping. I'll fix that.
$endgroup$
– Bernard
Dec 2 '18 at 13:36
$begingroup$
Since y on the second equation is mutliplied by 0, can i say that it is a free variable as so equate it to 1?
$endgroup$
– BM97
Dec 2 '18 at 13:38
$begingroup$
You're absolutely right.
$endgroup$
– Bernard
Dec 2 '18 at 13:42
|
show 1 more comment
Your Answer
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$begingroup$
You put $A-I$ in reduced row echelon form:
$$begin{pmatrix}-3&0&0\3&0&-6\0&0&-3end{pmatrix}rightsquigarrowbegin{pmatrix}1&0&0\1&0&-2\0&0&1end{pmatrix}rightsquigarrowbegin{pmatrix}1&0&0\0&0&-2\0&0&1end{pmatrix}rightsquigarrowbegin{pmatrix}1&0&0\0&0&1\0&0&0end{pmatrix}$$
so we have the equations
$$x=0,quad z=0.$$
answered Dec 2 '18 at 12:45
BernardBernard
119k740113
$endgroup$
$begingroup$
Thanks ! What if i try to to it like this. From the matrix (A-Id) i can directly see that x=0 and z=0. This leaves me with the second row being 3(0)+y-6(0)=0. How do i solve for y which gives me as a y basis 1?
$endgroup$
– BM97
Dec 2 '18 at 13:02
$begingroup$
Shouldn't it be -6 on the second row? (-6-0)=-6
$endgroup$
– BM97
Dec 2 '18 at 13:15
$begingroup$
You're right – a mistyping. I'll fix that.
$endgroup$
– Bernard
Dec 2 '18 at 13:36
$begingroup$
Since y on the second equation is mutliplied by 0, can i say that it is a free variable as so equate it to 1?
$endgroup$
– BM97
Dec 2 '18 at 13:38
$begingroup$
You're absolutely right.
$endgroup$
– Bernard
Dec 2 '18 at 13:42
|
show 1 more comment
$begingroup$
You put $A-I$ in reduced row echelon form:
$$begin{pmatrix}-3&0&0\3&0&-6\0&0&-3end{pmatrix}rightsquigarrowbegin{pmatrix}1&0&0\1&0&-2\0&0&1end{pmatrix}rightsquigarrowbegin{pmatrix}1&0&0\0&0&-2\0&0&1end{pmatrix}rightsquigarrowbegin{pmatrix}1&0&0\0&0&1\0&0&0end{pmatrix}$$
so we have the equations
$$x=0,quad z=0.$$
answered Dec 2 '18 at 12:45
BernardBernard
119k740113
$endgroup$
$begingroup$
Thanks ! What if i try to to it like this. From the matrix (A-Id) i can directly see that x=0 and z=0. This leaves me with the second row being 3(0)+y-6(0)=0. How do i solve for y which gives me as a y basis 1?
$endgroup$
– BM97
Dec 2 '18 at 13:02
$begingroup$
Shouldn't it be -6 on the second row? (-6-0)=-6
$endgroup$
– BM97
Dec 2 '18 at 13:15
$begingroup$
You're right – a mistyping. I'll fix that.
$endgroup$
– Bernard
Dec 2 '18 at 13:36
$begingroup$
Since y on the second equation is mutliplied by 0, can i say that it is a free variable as so equate it to 1?
$endgroup$
– BM97
Dec 2 '18 at 13:38
$begingroup$
You're absolutely right.
$endgroup$
– Bernard
Dec 2 '18 at 13:42
|
show 1 more comment
$begingroup$
You put $A-I$ in reduced row echelon form:
$$begin{pmatrix}-3&0&0\3&0&-6\0&0&-3end{pmatrix}rightsquigarrowbegin{pmatrix}1&0&0\1&0&-2\0&0&1end{pmatrix}rightsquigarrowbegin{pmatrix}1&0&0\0&0&-2\0&0&1end{pmatrix}rightsquigarrowbegin{pmatrix}1&0&0\0&0&1\0&0&0end{pmatrix}$$
so we have the equations
$$x=0,quad z=0.$$
answered Dec 2 '18 at 12:45
BernardBernard
119k740113
$endgroup$
You put $A-I$ in reduced row echelon form:
$$begin{pmatrix}-3&0&0\3&0&-6\0&0&-3end{pmatrix}rightsquigarrowbegin{pmatrix}1&0&0\1&0&-2\0&0&1end{pmatrix}rightsquigarrowbegin{pmatrix}1&0&0\0&0&-2\0&0&1end{pmatrix}rightsquigarrowbegin{pmatrix}1&0&0\0&0&1\0&0&0end{pmatrix}$$
so we have the equations
$$x=0,quad z=0.$$
answered Dec 2 '18 at 12:45
BernardBernard
119k740113
edited Dec 2 '18 at 13:37
answered Dec 2 '18 at 12:45
BernardBernard
119k740113
answered Dec 2 '18 at 12:45
BernardBernard
119k740113
answered Dec 2 '18 at 12:45
BernardBernard
119k740113
119k740113
$begingroup$
Thanks ! What if i try to to it like this. From the matrix (A-Id) i can directly see that x=0 and z=0. This leaves me with the second row being 3(0)+y-6(0)=0. How do i solve for y which gives me as a y basis 1?
$endgroup$
– BM97
Dec 2 '18 at 13:02
$begingroup$
Shouldn't it be -6 on the second row? (-6-0)=-6
$endgroup$
– BM97
Dec 2 '18 at 13:15
$begingroup$
You're right – a mistyping. I'll fix that.
$endgroup$
– Bernard
Dec 2 '18 at 13:36
$begingroup$
Since y on the second equation is mutliplied by 0, can i say that it is a free variable as so equate it to 1?
$endgroup$
– BM97
Dec 2 '18 at 13:38
$begingroup$
You're absolutely right.
$endgroup$
– Bernard
Dec 2 '18 at 13:42
|
show 1 more comment
$begingroup$
Thanks ! What if i try to to it like this. From the matrix (A-Id) i can directly see that x=0 and z=0. This leaves me with the second row being 3(0)+y-6(0)=0. How do i solve for y which gives me as a y basis 1?
$endgroup$
– BM97
Dec 2 '18 at 13:02
$begingroup$
Shouldn't it be -6 on the second row? (-6-0)=-6
$endgroup$
– BM97
Dec 2 '18 at 13:15
$begingroup$
You're right – a mistyping. I'll fix that.
$endgroup$
– Bernard
Dec 2 '18 at 13:36
$begingroup$
Since y on the second equation is mutliplied by 0, can i say that it is a free variable as so equate it to 1?
$endgroup$
– BM97
Dec 2 '18 at 13:38
$begingroup$
You're absolutely right.
$endgroup$
– Bernard
Dec 2 '18 at 13:42
$begingroup$
Thanks ! What if i try to to it like this. From the matrix (A-Id) i can directly see that x=0 and z=0. This leaves me with the second row being 3(0)+y-6(0)=0. How do i solve for y which gives me as a y basis 1?
$endgroup$
– BM97
Dec 2 '18 at 13:02
$begingroup$
Thanks ! What if i try to to it like this. From the matrix (A-Id) i can directly see that x=0 and z=0. This leaves me with the second row being 3(0)+y-6(0)=0. How do i solve for y which gives me as a y basis 1?
$endgroup$
– BM97
Dec 2 '18 at 13:02
$begingroup$
Shouldn't it be -6 on the second row? (-6-0)=-6
$endgroup$
– BM97
Dec 2 '18 at 13:15
$begingroup$
Shouldn't it be -6 on the second row? (-6-0)=-6
$endgroup$
– BM97
Dec 2 '18 at 13:15
$begingroup$
You're right – a mistyping. I'll fix that.
$endgroup$
– Bernard
Dec 2 '18 at 13:36
$begingroup$
You're right – a mistyping. I'll fix that.
$endgroup$
– Bernard
Dec 2 '18 at 13:36
$begingroup$
Since y on the second equation is mutliplied by 0, can i say that it is a free variable as so equate it to 1?
$endgroup$
– BM97
Dec 2 '18 at 13:38
$begingroup$
Since y on the second equation is mutliplied by 0, can i say that it is a free variable as so equate it to 1?
$endgroup$
– BM97
Dec 2 '18 at 13:38
$begingroup$
You're absolutely right.
$endgroup$
– Bernard
Dec 2 '18 at 13:42
$begingroup$
You're absolutely right.
$endgroup$
– Bernard
Dec 2 '18 at 13:42
|
show 1 more comment
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You calculated $A-I$ incorrectly.
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– David C. Ullrich
Dec 2 '18 at 13:02
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How come ? where did i do it wrong
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– BM97
Dec 2 '18 at 13:02
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Come on now! I mean really - figuring out $A-I$ is not that hard. Do the subtraction again, carefully; what you get is not what you say $A-I$ is above.
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– David C. Ullrich
Dec 2 '18 at 13:19
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I did it again . On the second row i still got -6, but the user below got 6?
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– BM97
Dec 2 '18 at 13:20
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It seems you found the error, as you just edited it away.
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– Ingix
Dec 2 '18 at 13:29