How to check if a matrix is diagonalizable? [duplicate]
This question already has an answer here:
How to check if a matrix is diagonizable?
2 answers
So I have this matrix
$$
begin{pmatrix}
-2 & 0 & 0 \
3 & 1 & -6 \
0 & 0 & -2 \
end{pmatrix}
$$
Finding Nul(A+2Id) gives me
$$
begin{pmatrix}
0 & 0 & 0 \
3 & 3 & -6 \
0 & 0 & 0 \
end{pmatrix}
$$
On the answers sheet it says to solve the homogeneous system associated with A + 2Id which gives us
Nul(A+2Id)=Span $$
begin{pmatrix}
1 & -1 \
1 & 1 \
1 & 0 \
end{pmatrix}
$$
I am completely lost on how to find the Span Nul(A+2Id)=Span $$
begin{pmatrix}
1 & -1 \
1 & 1 \
1 & 0 \
end{pmatrix}
$$
Can somebody please help me with the steps in order to get to this solution ?
Moreover, on the answers sheet is states
Since the dimension of the eigenspace E(λ1) is equal to 2 we observe that the ma- trix A is diagonalizable.
How do i know from this solution that the eigenspace E(λ1) is equal to 2?
Thanks!
linear-algebra matrices diagonalization
edited Dec 1 '18 at 15:55
Martin Sleziak
44.7k7115270
asked Dec 1 '18 at 12:53
BM97
758
marked as duplicate by user10354138, amWhy linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
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Dec 1 '18 at 19:11
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
How to check if a matrix is diagonizable?
2 answers
So I have this matrix
$$
begin{pmatrix}
-2 & 0 & 0 \
3 & 1 & -6 \
0 & 0 & -2 \
end{pmatrix}
$$
Finding Nul(A+2Id) gives me
$$
begin{pmatrix}
0 & 0 & 0 \
3 & 3 & -6 \
0 & 0 & 0 \
end{pmatrix}
$$
On the answers sheet it says to solve the homogeneous system associated with A + 2Id which gives us
Nul(A+2Id)=Span $$
begin{pmatrix}
1 & -1 \
1 & 1 \
1 & 0 \
end{pmatrix}
$$
I am completely lost on how to find the Span Nul(A+2Id)=Span $$
begin{pmatrix}
1 & -1 \
1 & 1 \
1 & 0 \
end{pmatrix}
$$
Can somebody please help me with the steps in order to get to this solution ?
Moreover, on the answers sheet is states
Since the dimension of the eigenspace E(λ1) is equal to 2 we observe that the ma- trix A is diagonalizable.
How do i know from this solution that the eigenspace E(λ1) is equal to 2?
Thanks!
linear-algebra matrices diagonalization
edited Dec 1 '18 at 15:55
Martin Sleziak
44.7k7115270
asked Dec 1 '18 at 12:53
BM97
758
marked as duplicate by user10354138, amWhy linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
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Dec 1 '18 at 19:11
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
You've already posted this question here
– user10354138
Dec 1 '18 at 12:56
Nope I haven't :) Please don't comment if you don't read the question properperly. This is a continuation and my question is different from the one before
– BM97
Dec 1 '18 at 13:19
add a comment |
This question already has an answer here:
How to check if a matrix is diagonizable?
2 answers
So I have this matrix
$$
begin{pmatrix}
-2 & 0 & 0 \
3 & 1 & -6 \
0 & 0 & -2 \
end{pmatrix}
$$
Finding Nul(A+2Id) gives me
$$
begin{pmatrix}
0 & 0 & 0 \
3 & 3 & -6 \
0 & 0 & 0 \
end{pmatrix}
$$
On the answers sheet it says to solve the homogeneous system associated with A + 2Id which gives us
Nul(A+2Id)=Span $$
begin{pmatrix}
1 & -1 \
1 & 1 \
1 & 0 \
end{pmatrix}
$$
I am completely lost on how to find the Span Nul(A+2Id)=Span $$
begin{pmatrix}
1 & -1 \
1 & 1 \
1 & 0 \
end{pmatrix}
$$
Can somebody please help me with the steps in order to get to this solution ?
Moreover, on the answers sheet is states
Since the dimension of the eigenspace E(λ1) is equal to 2 we observe that the ma- trix A is diagonalizable.
How do i know from this solution that the eigenspace E(λ1) is equal to 2?
Thanks!
linear-algebra matrices diagonalization
edited Dec 1 '18 at 15:55
Martin Sleziak
44.7k7115270
asked Dec 1 '18 at 12:53
BM97
758
This question already has an answer here:
How to check if a matrix is diagonizable?
2 answers
So I have this matrix
$$
begin{pmatrix}
-2 & 0 & 0 \
3 & 1 & -6 \
0 & 0 & -2 \
end{pmatrix}
$$
Finding Nul(A+2Id) gives me
$$
begin{pmatrix}
0 & 0 & 0 \
3 & 3 & -6 \
0 & 0 & 0 \
end{pmatrix}
$$
On the answers sheet it says to solve the homogeneous system associated with A + 2Id which gives us
Nul(A+2Id)=Span $$
begin{pmatrix}
1 & -1 \
1 & 1 \
1 & 0 \
end{pmatrix}
$$
I am completely lost on how to find the Span Nul(A+2Id)=Span $$
begin{pmatrix}
1 & -1 \
1 & 1 \
1 & 0 \
end{pmatrix}
$$
Can somebody please help me with the steps in order to get to this solution ?
Moreover, on the answers sheet is states
Since the dimension of the eigenspace E(λ1) is equal to 2 we observe that the ma- trix A is diagonalizable.
How do i know from this solution that the eigenspace E(λ1) is equal to 2?
Thanks!
This question already has an answer here:
How to check if a matrix is diagonizable?
2 answers
linear-algebra matrices diagonalization
linear-algebra matrices diagonalization
edited Dec 1 '18 at 15:55
Martin Sleziak
44.7k7115270
asked Dec 1 '18 at 12:53
BM97
758
edited Dec 1 '18 at 15:55
Martin Sleziak
44.7k7115270
asked Dec 1 '18 at 12:53
BM97
758
edited Dec 1 '18 at 15:55
Martin Sleziak
44.7k7115270
edited Dec 1 '18 at 15:55
Martin Sleziak
44.7k7115270
edited Dec 1 '18 at 15:55
Martin Sleziak
44.7k7115270
44.7k7115270
asked Dec 1 '18 at 12:53
BM97
758
asked Dec 1 '18 at 12:53
BM97
758
asked Dec 1 '18 at 12:53
BM97
758
758
marked as duplicate by user10354138, amWhy linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
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Dec 1 '18 at 19:11
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by user10354138, amWhy linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
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Dec 1 '18 at 19:11
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
You've already posted this question here
– user10354138
Dec 1 '18 at 12:56
Nope I haven't :) Please don't comment if you don't read the question properperly. This is a continuation and my question is different from the one before
– BM97
Dec 1 '18 at 13:19
add a comment |
1
You've already posted this question here
– user10354138
Dec 1 '18 at 12:56
Nope I haven't :) Please don't comment if you don't read the question properperly. This is a continuation and my question is different from the one before
– BM97
Dec 1 '18 at 13:19
1
1
You've already posted this question here
– user10354138
Dec 1 '18 at 12:56
You've already posted this question here
– user10354138
Dec 1 '18 at 12:56
Nope I haven't :) Please don't comment if you don't read the question properperly. This is a continuation and my question is different from the one before
– BM97
Dec 1 '18 at 13:19
Nope I haven't :) Please don't comment if you don't read the question properperly. This is a continuation and my question is different from the one before
– BM97
Dec 1 '18 at 13:19
add a comment |
1 Answer
1
active
oldest
votes
You know that $dim E_{-2}=2$ because you computed $operatorname{Nul}(A+2operatorname{Id})$ and it turned out that it has dimension $2$: it is spanned by two linearly independent vectors.
answered Dec 1 '18 at 12:56
José Carlos Santos
150k22121221
How do i know that it turned out to have dimension 2 and that it is spanned by two linearly independent vectors? Where are the two vectors (1,1,1) and (-1,1,0) coming from ? Thanks for the help!
– BM97
Dec 1 '18 at 13:22
What operations do i need to do in order to find these two vectors?
– BM97
Dec 1 '18 at 13:23
You solve the equation$$begin{bmatrix}0&0&0\3&3&-6\0&0&0end{bmatrix}.begin{bmatrix}x\y\zend{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$and you see that its solutions are precisely the linear combinations of $(1,1,1)$ and $(1,-1,0)$.
– José Carlos Santos
Dec 1 '18 at 13:25
Thank you so much !!!! You are great ...Last question , if (1,1,1) and (1,-1,0) were not linearly independent, what would happen?
– BM97
Dec 1 '18 at 13:29
Then $dim E_{-2}=1$.
– José Carlos Santos
Dec 1 '18 at 13:33
|
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You know that $dim E_{-2}=2$ because you computed $operatorname{Nul}(A+2operatorname{Id})$ and it turned out that it has dimension $2$: it is spanned by two linearly independent vectors.
answered Dec 1 '18 at 12:56
José Carlos Santos
150k22121221
How do i know that it turned out to have dimension 2 and that it is spanned by two linearly independent vectors? Where are the two vectors (1,1,1) and (-1,1,0) coming from ? Thanks for the help!
– BM97
Dec 1 '18 at 13:22
What operations do i need to do in order to find these two vectors?
– BM97
Dec 1 '18 at 13:23
You solve the equation$$begin{bmatrix}0&0&0\3&3&-6\0&0&0end{bmatrix}.begin{bmatrix}x\y\zend{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$and you see that its solutions are precisely the linear combinations of $(1,1,1)$ and $(1,-1,0)$.
– José Carlos Santos
Dec 1 '18 at 13:25
Thank you so much !!!! You are great ...Last question , if (1,1,1) and (1,-1,0) were not linearly independent, what would happen?
– BM97
Dec 1 '18 at 13:29
Then $dim E_{-2}=1$.
– José Carlos Santos
Dec 1 '18 at 13:33
|
show 2 more comments
You know that $dim E_{-2}=2$ because you computed $operatorname{Nul}(A+2operatorname{Id})$ and it turned out that it has dimension $2$: it is spanned by two linearly independent vectors.
answered Dec 1 '18 at 12:56
José Carlos Santos
150k22121221
How do i know that it turned out to have dimension 2 and that it is spanned by two linearly independent vectors? Where are the two vectors (1,1,1) and (-1,1,0) coming from ? Thanks for the help!
– BM97
Dec 1 '18 at 13:22
What operations do i need to do in order to find these two vectors?
– BM97
Dec 1 '18 at 13:23
You solve the equation$$begin{bmatrix}0&0&0\3&3&-6\0&0&0end{bmatrix}.begin{bmatrix}x\y\zend{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$and you see that its solutions are precisely the linear combinations of $(1,1,1)$ and $(1,-1,0)$.
– José Carlos Santos
Dec 1 '18 at 13:25
Thank you so much !!!! You are great ...Last question , if (1,1,1) and (1,-1,0) were not linearly independent, what would happen?
– BM97
Dec 1 '18 at 13:29
Then $dim E_{-2}=1$.
– José Carlos Santos
Dec 1 '18 at 13:33
|
show 2 more comments
You know that $dim E_{-2}=2$ because you computed $operatorname{Nul}(A+2operatorname{Id})$ and it turned out that it has dimension $2$: it is spanned by two linearly independent vectors.
answered Dec 1 '18 at 12:56
José Carlos Santos
150k22121221
You know that $dim E_{-2}=2$ because you computed $operatorname{Nul}(A+2operatorname{Id})$ and it turned out that it has dimension $2$: it is spanned by two linearly independent vectors.
answered Dec 1 '18 at 12:56
José Carlos Santos
150k22121221
answered Dec 1 '18 at 12:56
José Carlos Santos
150k22121221
answered Dec 1 '18 at 12:56
José Carlos Santos
150k22121221
answered Dec 1 '18 at 12:56
José Carlos Santos
150k22121221
150k22121221
How do i know that it turned out to have dimension 2 and that it is spanned by two linearly independent vectors? Where are the two vectors (1,1,1) and (-1,1,0) coming from ? Thanks for the help!
– BM97
Dec 1 '18 at 13:22
What operations do i need to do in order to find these two vectors?
– BM97
Dec 1 '18 at 13:23
You solve the equation$$begin{bmatrix}0&0&0\3&3&-6\0&0&0end{bmatrix}.begin{bmatrix}x\y\zend{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$and you see that its solutions are precisely the linear combinations of $(1,1,1)$ and $(1,-1,0)$.
– José Carlos Santos
Dec 1 '18 at 13:25
Thank you so much !!!! You are great ...Last question , if (1,1,1) and (1,-1,0) were not linearly independent, what would happen?
– BM97
Dec 1 '18 at 13:29
Then $dim E_{-2}=1$.
– José Carlos Santos
Dec 1 '18 at 13:33
|
show 2 more comments
How do i know that it turned out to have dimension 2 and that it is spanned by two linearly independent vectors? Where are the two vectors (1,1,1) and (-1,1,0) coming from ? Thanks for the help!
– BM97
Dec 1 '18 at 13:22
What operations do i need to do in order to find these two vectors?
– BM97
Dec 1 '18 at 13:23
You solve the equation$$begin{bmatrix}0&0&0\3&3&-6\0&0&0end{bmatrix}.begin{bmatrix}x\y\zend{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$and you see that its solutions are precisely the linear combinations of $(1,1,1)$ and $(1,-1,0)$.
– José Carlos Santos
Dec 1 '18 at 13:25
Thank you so much !!!! You are great ...Last question , if (1,1,1) and (1,-1,0) were not linearly independent, what would happen?
– BM97
Dec 1 '18 at 13:29
Then $dim E_{-2}=1$.
– José Carlos Santos
Dec 1 '18 at 13:33
How do i know that it turned out to have dimension 2 and that it is spanned by two linearly independent vectors? Where are the two vectors (1,1,1) and (-1,1,0) coming from ? Thanks for the help!
– BM97
Dec 1 '18 at 13:22
How do i know that it turned out to have dimension 2 and that it is spanned by two linearly independent vectors? Where are the two vectors (1,1,1) and (-1,1,0) coming from ? Thanks for the help!
– BM97
Dec 1 '18 at 13:22
What operations do i need to do in order to find these two vectors?
– BM97
Dec 1 '18 at 13:23
What operations do i need to do in order to find these two vectors?
– BM97
Dec 1 '18 at 13:23
You solve the equation$$begin{bmatrix}0&0&0\3&3&-6\0&0&0end{bmatrix}.begin{bmatrix}x\y\zend{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$and you see that its solutions are precisely the linear combinations of $(1,1,1)$ and $(1,-1,0)$.
– José Carlos Santos
Dec 1 '18 at 13:25
You solve the equation$$begin{bmatrix}0&0&0\3&3&-6\0&0&0end{bmatrix}.begin{bmatrix}x\y\zend{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$and you see that its solutions are precisely the linear combinations of $(1,1,1)$ and $(1,-1,0)$.
– José Carlos Santos
Dec 1 '18 at 13:25
Thank you so much !!!! You are great ...Last question , if (1,1,1) and (1,-1,0) were not linearly independent, what would happen?
– BM97
Dec 1 '18 at 13:29
Thank you so much !!!! You are great ...Last question , if (1,1,1) and (1,-1,0) were not linearly independent, what would happen?
– BM97
Dec 1 '18 at 13:29
Then $dim E_{-2}=1$.
– José Carlos Santos
Dec 1 '18 at 13:33
Then $dim E_{-2}=1$.
– José Carlos Santos
Dec 1 '18 at 13:33
|
show 2 more comments
1
You've already posted this question here
– user10354138
Dec 1 '18 at 12:56
Nope I haven't :) Please don't comment if you don't read the question properperly. This is a continuation and my question is different from the one before
– BM97
Dec 1 '18 at 13:19