Mixing
How to construct a 3 by 3 matrix with a given eigenspace in the standard basis
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I am having a great difficulty wrapping my head around this question, I need help and hopefully this will help others.
The question is as follows:
Assume that the matrix $F: Bbb R^3 rightarrow Bbb R^3$ is a linear transformation with an eigenvalue of 1 with a respective eigenspace of [(1,0,0) (0,1,0)] and an eigenvalue of 3 with an eigenspace of [(1,0,1)]. Define $F$'s matrix in the standard basis
So I have attempted multiple things. I have watched 3blue1brown on youtube and he explains the topic on eigenvalues, eigenvectors, eigenspaces and change of basis quite well. I thought I could apply this knowledge unto this question.
This is how I read the question. So we have a 3 by 3 matrix that's, let's say it belongs to Bob, Bob's basis vectors (his (1,0,0) (0,1,0) (0,0,1)) is ( (1,0,0) (0,1,0) (3,0,3) ). This is now our change of basis matrix.
and to translate between the basis' the formula for that is $F vec v = vec u$ this will 'translate' the input vector $vec v$ into what Bob really meant in our language. So how I see it $F vec [1,0,0]$ (should be) $=vec [1, 0, 0]$ and $Fvec[0,1,0] = vec[0,1,0]$ and finally $Fvec[0,0,1] = vec[3,0,3]$
but this is wrong and according to the answer (and their solution which is non-intuitive ) it should be $Fvec[1,0,0] = vec[1,0,0]$, $Fvec[0,1,0] = vec[0,1,0]$ and $Fvec[0,0,1] = Fvec[1,0,1] - Fvec[1,0,0]$ I understand the calculation, but it just isn't intuitive to me.
Why is my solution incorrect? I mean $F$ is our change of basis matrix, why shouldn't I just be able to plug in the vectors and get the vector written in our language?
Also, please explain the textbook answer intuitively to help me and others.
Thank you very much.
((on a side note I also played with the inverse. THe inverse, as I understand it, is playing the transformation in reverse i.e this should give us the basis vector pre-transformation, idk it didn't get me anywhere))
linear-algebra matrices eigenvalues-eigenvectors change-of-basis
asked Jul 27 at 11:58
Hoaz
375
add a comment |Â
up vote
0
down vote
favorite
I am having a great difficulty wrapping my head around this question, I need help and hopefully this will help others.
The question is as follows:
Assume that the matrix $F: Bbb R^3 rightarrow Bbb R^3$ is a linear transformation with an eigenvalue of 1 with a respective eigenspace of [(1,0,0) (0,1,0)] and an eigenvalue of 3 with an eigenspace of [(1,0,1)]. Define $F$'s matrix in the standard basis
So I have attempted multiple things. I have watched 3blue1brown on youtube and he explains the topic on eigenvalues, eigenvectors, eigenspaces and change of basis quite well. I thought I could apply this knowledge unto this question.
This is how I read the question. So we have a 3 by 3 matrix that's, let's say it belongs to Bob, Bob's basis vectors (his (1,0,0) (0,1,0) (0,0,1)) is ( (1,0,0) (0,1,0) (3,0,3) ). This is now our change of basis matrix.
and to translate between the basis' the formula for that is $F vec v = vec u$ this will 'translate' the input vector $vec v$ into what Bob really meant in our language. So how I see it $F vec [1,0,0]$ (should be) $=vec [1, 0, 0]$ and $Fvec[0,1,0] = vec[0,1,0]$ and finally $Fvec[0,0,1] = vec[3,0,3]$
but this is wrong and according to the answer (and their solution which is non-intuitive ) it should be $Fvec[1,0,0] = vec[1,0,0]$, $Fvec[0,1,0] = vec[0,1,0]$ and $Fvec[0,0,1] = Fvec[1,0,1] - Fvec[1,0,0]$ I understand the calculation, but it just isn't intuitive to me.
Why is my solution incorrect? I mean $F$ is our change of basis matrix, why shouldn't I just be able to plug in the vectors and get the vector written in our language?
Also, please explain the textbook answer intuitively to help me and others.
Thank you very much.
((on a side note I also played with the inverse. THe inverse, as I understand it, is playing the transformation in reverse i.e this should give us the basis vector pre-transformation, idk it didn't get me anywhere))
linear-algebra matrices eigenvalues-eigenvectors change-of-basis
asked Jul 27 at 11:58
Hoaz
375
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am having a great difficulty wrapping my head around this question, I need help and hopefully this will help others.
The question is as follows:
Assume that the matrix $F: Bbb R^3 rightarrow Bbb R^3$ is a linear transformation with an eigenvalue of 1 with a respective eigenspace of [(1,0,0) (0,1,0)] and an eigenvalue of 3 with an eigenspace of [(1,0,1)]. Define $F$'s matrix in the standard basis
So I have attempted multiple things. I have watched 3blue1brown on youtube and he explains the topic on eigenvalues, eigenvectors, eigenspaces and change of basis quite well. I thought I could apply this knowledge unto this question.
This is how I read the question. So we have a 3 by 3 matrix that's, let's say it belongs to Bob, Bob's basis vectors (his (1,0,0) (0,1,0) (0,0,1)) is ( (1,0,0) (0,1,0) (3,0,3) ). This is now our change of basis matrix.
and to translate between the basis' the formula for that is $F vec v = vec u$ this will 'translate' the input vector $vec v$ into what Bob really meant in our language. So how I see it $F vec [1,0,0]$ (should be) $=vec [1, 0, 0]$ and $Fvec[0,1,0] = vec[0,1,0]$ and finally $Fvec[0,0,1] = vec[3,0,3]$
but this is wrong and according to the answer (and their solution which is non-intuitive ) it should be $Fvec[1,0,0] = vec[1,0,0]$, $Fvec[0,1,0] = vec[0,1,0]$ and $Fvec[0,0,1] = Fvec[1,0,1] - Fvec[1,0,0]$ I understand the calculation, but it just isn't intuitive to me.
Why is my solution incorrect? I mean $F$ is our change of basis matrix, why shouldn't I just be able to plug in the vectors and get the vector written in our language?
Also, please explain the textbook answer intuitively to help me and others.
Thank you very much.
((on a side note I also played with the inverse. THe inverse, as I understand it, is playing the transformation in reverse i.e this should give us the basis vector pre-transformation, idk it didn't get me anywhere))
linear-algebra matrices eigenvalues-eigenvectors change-of-basis
asked Jul 27 at 11:58
Hoaz
375
I am having a great difficulty wrapping my head around this question, I need help and hopefully this will help others.
The question is as follows:
Assume that the matrix $F: Bbb R^3 rightarrow Bbb R^3$ is a linear transformation with an eigenvalue of 1 with a respective eigenspace of [(1,0,0) (0,1,0)] and an eigenvalue of 3 with an eigenspace of [(1,0,1)]. Define $F$'s matrix in the standard basis
So I have attempted multiple things. I have watched 3blue1brown on youtube and he explains the topic on eigenvalues, eigenvectors, eigenspaces and change of basis quite well. I thought I could apply this knowledge unto this question.
This is how I read the question. So we have a 3 by 3 matrix that's, let's say it belongs to Bob, Bob's basis vectors (his (1,0,0) (0,1,0) (0,0,1)) is ( (1,0,0) (0,1,0) (3,0,3) ). This is now our change of basis matrix.
and to translate between the basis' the formula for that is $F vec v = vec u$ this will 'translate' the input vector $vec v$ into what Bob really meant in our language. So how I see it $F vec [1,0,0]$ (should be) $=vec [1, 0, 0]$ and $Fvec[0,1,0] = vec[0,1,0]$ and finally $Fvec[0,0,1] = vec[3,0,3]$
but this is wrong and according to the answer (and their solution which is non-intuitive ) it should be $Fvec[1,0,0] = vec[1,0,0]$, $Fvec[0,1,0] = vec[0,1,0]$ and $Fvec[0,0,1] = Fvec[1,0,1] - Fvec[1,0,0]$ I understand the calculation, but it just isn't intuitive to me.
Why is my solution incorrect? I mean $F$ is our change of basis matrix, why shouldn't I just be able to plug in the vectors and get the vector written in our language?
Also, please explain the textbook answer intuitively to help me and others.
Thank you very much.
((on a side note I also played with the inverse. THe inverse, as I understand it, is playing the transformation in reverse i.e this should give us the basis vector pre-transformation, idk it didn't get me anywhere))
linear-algebra matrices eigenvalues-eigenvectors change-of-basis
asked Jul 27 at 11:58
Hoaz
375
asked Jul 27 at 11:58
Hoaz
375
asked Jul 27 at 11:58
Hoaz
375
asked Jul 27 at 11:58
Hoaz
375
375
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
The definition of eigenvectors let's you know how your map $F$ acts on some specific vectors, the eigenvectors. In this manner you know that if you take from $mathbbR^3$ the vectors $(1,0,0),(0,1,0)$ and $(1,0,1)$ they will be mapped to the same vectors (times some scalar) onto $mathbbR^3$. Eigenvectors, indeed, are very good vectors, because they form a base for $mathbbR^3$, not the standard base but a base nonetheless. With this notions we can now tackle your problem.
The definition of a $lambda$-eigenvector is as follows:
$mathbfvinmathbbR^3$ is a $lambda$-eigenvector for the linear map $$F:mathbbR^3rightarrowmathbbR^3$$ iif $$F(mathbfv)=lambdamathbfv$$
If we take as a base for either domain and image of this linear map the eigenvectors, the associated matrix for $F$ in this base is the following $$D=left(beginmatrix1&0&0\0&1&0\0&0&3endmatrixright)$$ as you can see is a diagonal matrix with elements the eigenvectors taken with their respective geometric multiplicity. You clearly have found a base in that the matrix associated with $F$ is diagonal. This is called diagonalization. From this matrix you can go back to the matrix in standard base applying a matrix of basis transformation which in this case is the matrix that has as columns the eigenvectors associated with their respective eigenvalue. So, by definition, $$A=PDP^-1$$ where $A$ is the matrix of $F$ in standard base and $$P=left(beginmatrix1&0&1\0&1&0\0&0&1endmatrixright)$$ the matrix of the eigenvectors.
So, then $$A=left(beginmatrix1&0&2\0&1&0\0&0&3endmatrixright)$$
I give you some links where you can find useful explanations on this topic: Diagonalizable matrix and the spectral theorem
Edit
Obviously there is a lot to talk about this subject, I've tried to be more concise as possible auto not go on a tangent. If you have any other questions you can read any linear algebra. The topic of diagonalization, change of basis and the spectral theorem and of central importance in lots of fields, even applied fields as physics, and it's very important to have a firm idea of what's going on! If I should give an advice on what book to read I will say "Linear Algebra" by Serge Lang. But there are so many valuable books out there, just do a simple google search
answered Jul 27 at 12:16
Davide Morgante
1,751220
Ah yes. Makes sense. It's like we convert any vector into bob's language, apply our transformation and then convert it back into our language ($A = PDP^-1$) cool thanks! :)
– Hoaz
Jul 27 at 16:22
That's it! I too loved 3Blue1Brown series on linear algebra, I watched it when I was about too follow my fist course in algebra an linear algebra, it was very insightful
– Davide Morgante
Jul 27 at 16:24
yes. I thought I understood linear algebra until I watched his videos, then I really started to understand. really intuitive. It's really a must watch if you're doing linear algebra. And thanks again :)
– Hoaz
Jul 27 at 16:26
You're welcome!
– Davide Morgante
Jul 27 at 16:26
add a comment |Â
up vote
0
down vote
HINT
Recall that
$$D=P^-1F P implies F=PFP^-1$$
with
$$P=beginbmatrix1&0&1\0&1&0\0&0&1endbmatrixquad D=beginbmatrix1&0&0\0&1&0\0&0&3endbmatrix$$
answered Jul 27 at 12:13
gimusi
64.9k73583
add a comment |Â
up vote
0
down vote
Here is a very simple solution:
Denoting $e_1, e_2, e_3$ the canonical basis, you need to express $F(e_1), F(e_2)$ and $F(e_3)$ in the canonical basis.
For $F(e_1)$ and $F(e_2)$, its in the hypotheses, so the first two columns are
$$left[beginmatrix1&0\0&1\0&0endmatrixright.$$
Now, setting $u=e_1+e_3$, we know that $F(u)=F(e_1)+F(e_3)=3e_1+3e_3$, so $F(e_3)=2e_1+3e_3$, and we have the third column of the matrix, which is
$$ beginbmatrix1&0&2\0&1&0\0&0&3endbmatrix$$
answered Jul 27 at 12:22
Bernard
110k635102
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The definition of eigenvectors let's you know how your map $F$ acts on some specific vectors, the eigenvectors. In this manner you know that if you take from $mathbbR^3$ the vectors $(1,0,0),(0,1,0)$ and $(1,0,1)$ they will be mapped to the same vectors (times some scalar) onto $mathbbR^3$. Eigenvectors, indeed, are very good vectors, because they form a base for $mathbbR^3$, not the standard base but a base nonetheless. With this notions we can now tackle your problem.
The definition of a $lambda$-eigenvector is as follows:
$mathbfvinmathbbR^3$ is a $lambda$-eigenvector for the linear map $$F:mathbbR^3rightarrowmathbbR^3$$ iif $$F(mathbfv)=lambdamathbfv$$
If we take as a base for either domain and image of this linear map the eigenvectors, the associated matrix for $F$ in this base is the following $$D=left(beginmatrix1&0&0\0&1&0\0&0&3endmatrixright)$$ as you can see is a diagonal matrix with elements the eigenvectors taken with their respective geometric multiplicity. You clearly have found a base in that the matrix associated with $F$ is diagonal. This is called diagonalization. From this matrix you can go back to the matrix in standard base applying a matrix of basis transformation which in this case is the matrix that has as columns the eigenvectors associated with their respective eigenvalue. So, by definition, $$A=PDP^-1$$ where $A$ is the matrix of $F$ in standard base and $$P=left(beginmatrix1&0&1\0&1&0\0&0&1endmatrixright)$$ the matrix of the eigenvectors.
So, then $$A=left(beginmatrix1&0&2\0&1&0\0&0&3endmatrixright)$$
I give you some links where you can find useful explanations on this topic: Diagonalizable matrix and the spectral theorem
Edit
Obviously there is a lot to talk about this subject, I've tried to be more concise as possible auto not go on a tangent. If you have any other questions you can read any linear algebra. The topic of diagonalization, change of basis and the spectral theorem and of central importance in lots of fields, even applied fields as physics, and it's very important to have a firm idea of what's going on! If I should give an advice on what book to read I will say "Linear Algebra" by Serge Lang. But there are so many valuable books out there, just do a simple google search
answered Jul 27 at 12:16
Davide Morgante
1,751220
Ah yes. Makes sense. It's like we convert any vector into bob's language, apply our transformation and then convert it back into our language ($A = PDP^-1$) cool thanks! :)
– Hoaz
Jul 27 at 16:22
That's it! I too loved 3Blue1Brown series on linear algebra, I watched it when I was about too follow my fist course in algebra an linear algebra, it was very insightful
– Davide Morgante
Jul 27 at 16:24
yes. I thought I understood linear algebra until I watched his videos, then I really started to understand. really intuitive. It's really a must watch if you're doing linear algebra. And thanks again :)
– Hoaz
Jul 27 at 16:26
You're welcome!
– Davide Morgante
Jul 27 at 16:26
add a comment |Â
up vote
2
down vote
accepted
The definition of eigenvectors let's you know how your map $F$ acts on some specific vectors, the eigenvectors. In this manner you know that if you take from $mathbbR^3$ the vectors $(1,0,0),(0,1,0)$ and $(1,0,1)$ they will be mapped to the same vectors (times some scalar) onto $mathbbR^3$. Eigenvectors, indeed, are very good vectors, because they form a base for $mathbbR^3$, not the standard base but a base nonetheless. With this notions we can now tackle your problem.
The definition of a $lambda$-eigenvector is as follows:
$mathbfvinmathbbR^3$ is a $lambda$-eigenvector for the linear map $$F:mathbbR^3rightarrowmathbbR^3$$ iif $$F(mathbfv)=lambdamathbfv$$
If we take as a base for either domain and image of this linear map the eigenvectors, the associated matrix for $F$ in this base is the following $$D=left(beginmatrix1&0&0\0&1&0\0&0&3endmatrixright)$$ as you can see is a diagonal matrix with elements the eigenvectors taken with their respective geometric multiplicity. You clearly have found a base in that the matrix associated with $F$ is diagonal. This is called diagonalization. From this matrix you can go back to the matrix in standard base applying a matrix of basis transformation which in this case is the matrix that has as columns the eigenvectors associated with their respective eigenvalue. So, by definition, $$A=PDP^-1$$ where $A$ is the matrix of $F$ in standard base and $$P=left(beginmatrix1&0&1\0&1&0\0&0&1endmatrixright)$$ the matrix of the eigenvectors.
So, then $$A=left(beginmatrix1&0&2\0&1&0\0&0&3endmatrixright)$$
I give you some links where you can find useful explanations on this topic: Diagonalizable matrix and the spectral theorem
Edit
Obviously there is a lot to talk about this subject, I've tried to be more concise as possible auto not go on a tangent. If you have any other questions you can read any linear algebra. The topic of diagonalization, change of basis and the spectral theorem and of central importance in lots of fields, even applied fields as physics, and it's very important to have a firm idea of what's going on! If I should give an advice on what book to read I will say "Linear Algebra" by Serge Lang. But there are so many valuable books out there, just do a simple google search
answered Jul 27 at 12:16
Davide Morgante
1,751220
Ah yes. Makes sense. It's like we convert any vector into bob's language, apply our transformation and then convert it back into our language ($A = PDP^-1$) cool thanks! :)
– Hoaz
Jul 27 at 16:22
That's it! I too loved 3Blue1Brown series on linear algebra, I watched it when I was about too follow my fist course in algebra an linear algebra, it was very insightful
– Davide Morgante
Jul 27 at 16:24
yes. I thought I understood linear algebra until I watched his videos, then I really started to understand. really intuitive. It's really a must watch if you're doing linear algebra. And thanks again :)
– Hoaz
Jul 27 at 16:26
You're welcome!
– Davide Morgante
Jul 27 at 16:26
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The definition of eigenvectors let's you know how your map $F$ acts on some specific vectors, the eigenvectors. In this manner you know that if you take from $mathbbR^3$ the vectors $(1,0,0),(0,1,0)$ and $(1,0,1)$ they will be mapped to the same vectors (times some scalar) onto $mathbbR^3$. Eigenvectors, indeed, are very good vectors, because they form a base for $mathbbR^3$, not the standard base but a base nonetheless. With this notions we can now tackle your problem.
The definition of a $lambda$-eigenvector is as follows:
$mathbfvinmathbbR^3$ is a $lambda$-eigenvector for the linear map $$F:mathbbR^3rightarrowmathbbR^3$$ iif $$F(mathbfv)=lambdamathbfv$$
If we take as a base for either domain and image of this linear map the eigenvectors, the associated matrix for $F$ in this base is the following $$D=left(beginmatrix1&0&0\0&1&0\0&0&3endmatrixright)$$ as you can see is a diagonal matrix with elements the eigenvectors taken with their respective geometric multiplicity. You clearly have found a base in that the matrix associated with $F$ is diagonal. This is called diagonalization. From this matrix you can go back to the matrix in standard base applying a matrix of basis transformation which in this case is the matrix that has as columns the eigenvectors associated with their respective eigenvalue. So, by definition, $$A=PDP^-1$$ where $A$ is the matrix of $F$ in standard base and $$P=left(beginmatrix1&0&1\0&1&0\0&0&1endmatrixright)$$ the matrix of the eigenvectors.
So, then $$A=left(beginmatrix1&0&2\0&1&0\0&0&3endmatrixright)$$
I give you some links where you can find useful explanations on this topic: Diagonalizable matrix and the spectral theorem
Edit
Obviously there is a lot to talk about this subject, I've tried to be more concise as possible auto not go on a tangent. If you have any other questions you can read any linear algebra. The topic of diagonalization, change of basis and the spectral theorem and of central importance in lots of fields, even applied fields as physics, and it's very important to have a firm idea of what's going on! If I should give an advice on what book to read I will say "Linear Algebra" by Serge Lang. But there are so many valuable books out there, just do a simple google search
answered Jul 27 at 12:16
Davide Morgante
1,751220
The definition of eigenvectors let's you know how your map $F$ acts on some specific vectors, the eigenvectors. In this manner you know that if you take from $mathbbR^3$ the vectors $(1,0,0),(0,1,0)$ and $(1,0,1)$ they will be mapped to the same vectors (times some scalar) onto $mathbbR^3$. Eigenvectors, indeed, are very good vectors, because they form a base for $mathbbR^3$, not the standard base but a base nonetheless. With this notions we can now tackle your problem.
The definition of a $lambda$-eigenvector is as follows:
$mathbfvinmathbbR^3$ is a $lambda$-eigenvector for the linear map $$F:mathbbR^3rightarrowmathbbR^3$$ iif $$F(mathbfv)=lambdamathbfv$$
If we take as a base for either domain and image of this linear map the eigenvectors, the associated matrix for $F$ in this base is the following $$D=left(beginmatrix1&0&0\0&1&0\0&0&3endmatrixright)$$ as you can see is a diagonal matrix with elements the eigenvectors taken with their respective geometric multiplicity. You clearly have found a base in that the matrix associated with $F$ is diagonal. This is called diagonalization. From this matrix you can go back to the matrix in standard base applying a matrix of basis transformation which in this case is the matrix that has as columns the eigenvectors associated with their respective eigenvalue. So, by definition, $$A=PDP^-1$$ where $A$ is the matrix of $F$ in standard base and $$P=left(beginmatrix1&0&1\0&1&0\0&0&1endmatrixright)$$ the matrix of the eigenvectors.
So, then $$A=left(beginmatrix1&0&2\0&1&0\0&0&3endmatrixright)$$
I give you some links where you can find useful explanations on this topic: Diagonalizable matrix and the spectral theorem
Edit
Obviously there is a lot to talk about this subject, I've tried to be more concise as possible auto not go on a tangent. If you have any other questions you can read any linear algebra. The topic of diagonalization, change of basis and the spectral theorem and of central importance in lots of fields, even applied fields as physics, and it's very important to have a firm idea of what's going on! If I should give an advice on what book to read I will say "Linear Algebra" by Serge Lang. But there are so many valuable books out there, just do a simple google search
answered Jul 27 at 12:16
Davide Morgante
1,751220
edited Jul 27 at 12:49
answered Jul 27 at 12:16
Davide Morgante
1,751220
answered Jul 27 at 12:16
Davide Morgante
1,751220
answered Jul 27 at 12:16
Davide Morgante
1,751220
1,751220
Ah yes. Makes sense. It's like we convert any vector into bob's language, apply our transformation and then convert it back into our language ($A = PDP^-1$) cool thanks! :)
– Hoaz
Jul 27 at 16:22
That's it! I too loved 3Blue1Brown series on linear algebra, I watched it when I was about too follow my fist course in algebra an linear algebra, it was very insightful
– Davide Morgante
Jul 27 at 16:24
yes. I thought I understood linear algebra until I watched his videos, then I really started to understand. really intuitive. It's really a must watch if you're doing linear algebra. And thanks again :)
– Hoaz
Jul 27 at 16:26
You're welcome!
– Davide Morgante
Jul 27 at 16:26
add a comment |Â
Ah yes. Makes sense. It's like we convert any vector into bob's language, apply our transformation and then convert it back into our language ($A = PDP^-1$) cool thanks! :)
– Hoaz
Jul 27 at 16:22
That's it! I too loved 3Blue1Brown series on linear algebra, I watched it when I was about too follow my fist course in algebra an linear algebra, it was very insightful
– Davide Morgante
Jul 27 at 16:24
yes. I thought I understood linear algebra until I watched his videos, then I really started to understand. really intuitive. It's really a must watch if you're doing linear algebra. And thanks again :)
– Hoaz
Jul 27 at 16:26
You're welcome!
– Davide Morgante
Jul 27 at 16:26
Ah yes. Makes sense. It's like we convert any vector into bob's language, apply our transformation and then convert it back into our language ($A = PDP^-1$) cool thanks! :)
– Hoaz
Jul 27 at 16:22
Ah yes. Makes sense. It's like we convert any vector into bob's language, apply our transformation and then convert it back into our language ($A = PDP^-1$) cool thanks! :)
– Hoaz
Jul 27 at 16:22
That's it! I too loved 3Blue1Brown series on linear algebra, I watched it when I was about too follow my fist course in algebra an linear algebra, it was very insightful
– Davide Morgante
Jul 27 at 16:24
That's it! I too loved 3Blue1Brown series on linear algebra, I watched it when I was about too follow my fist course in algebra an linear algebra, it was very insightful
– Davide Morgante
Jul 27 at 16:24
yes. I thought I understood linear algebra until I watched his videos, then I really started to understand. really intuitive. It's really a must watch if you're doing linear algebra. And thanks again :)
– Hoaz
Jul 27 at 16:26
yes. I thought I understood linear algebra until I watched his videos, then I really started to understand. really intuitive. It's really a must watch if you're doing linear algebra. And thanks again :)
– Hoaz
Jul 27 at 16:26
You're welcome!
– Davide Morgante
Jul 27 at 16:26
You're welcome!
– Davide Morgante
Jul 27 at 16:26
add a comment |Â
up vote
0
down vote
HINT
Recall that
$$D=P^-1F P implies F=PFP^-1$$
with
$$P=beginbmatrix1&0&1\0&1&0\0&0&1endbmatrixquad D=beginbmatrix1&0&0\0&1&0\0&0&3endbmatrix$$
answered Jul 27 at 12:13
gimusi
64.9k73583
add a comment |Â
up vote
0
down vote
HINT
Recall that
$$D=P^-1F P implies F=PFP^-1$$
with
$$P=beginbmatrix1&0&1\0&1&0\0&0&1endbmatrixquad D=beginbmatrix1&0&0\0&1&0\0&0&3endbmatrix$$
answered Jul 27 at 12:13
gimusi
64.9k73583
add a comment |Â
up vote
0
down vote
up vote
0
down vote
HINT
Recall that
$$D=P^-1F P implies F=PFP^-1$$
with
$$P=beginbmatrix1&0&1\0&1&0\0&0&1endbmatrixquad D=beginbmatrix1&0&0\0&1&0\0&0&3endbmatrix$$
answered Jul 27 at 12:13
gimusi
64.9k73583
HINT
Recall that
$$D=P^-1F P implies F=PFP^-1$$
with
$$P=beginbmatrix1&0&1\0&1&0\0&0&1endbmatrixquad D=beginbmatrix1&0&0\0&1&0\0&0&3endbmatrix$$
answered Jul 27 at 12:13
gimusi
64.9k73583
answered Jul 27 at 12:13
gimusi
64.9k73583
answered Jul 27 at 12:13
gimusi
64.9k73583
answered Jul 27 at 12:13
gimusi
64.9k73583
64.9k73583
add a comment |Â
add a comment |Â
up vote
0
down vote
Here is a very simple solution:
Denoting $e_1, e_2, e_3$ the canonical basis, you need to express $F(e_1), F(e_2)$ and $F(e_3)$ in the canonical basis.
For $F(e_1)$ and $F(e_2)$, its in the hypotheses, so the first two columns are
$$left[beginmatrix1&0\0&1\0&0endmatrixright.$$
Now, setting $u=e_1+e_3$, we know that $F(u)=F(e_1)+F(e_3)=3e_1+3e_3$, so $F(e_3)=2e_1+3e_3$, and we have the third column of the matrix, which is
$$ beginbmatrix1&0&2\0&1&0\0&0&3endbmatrix$$
answered Jul 27 at 12:22
Bernard
110k635102
add a comment |Â
up vote
0
down vote
Here is a very simple solution:
Denoting $e_1, e_2, e_3$ the canonical basis, you need to express $F(e_1), F(e_2)$ and $F(e_3)$ in the canonical basis.
For $F(e_1)$ and $F(e_2)$, its in the hypotheses, so the first two columns are
$$left[beginmatrix1&0\0&1\0&0endmatrixright.$$
Now, setting $u=e_1+e_3$, we know that $F(u)=F(e_1)+F(e_3)=3e_1+3e_3$, so $F(e_3)=2e_1+3e_3$, and we have the third column of the matrix, which is
$$ beginbmatrix1&0&2\0&1&0\0&0&3endbmatrix$$
answered Jul 27 at 12:22
Bernard
110k635102
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Here is a very simple solution:
Denoting $e_1, e_2, e_3$ the canonical basis, you need to express $F(e_1), F(e_2)$ and $F(e_3)$ in the canonical basis.
For $F(e_1)$ and $F(e_2)$, its in the hypotheses, so the first two columns are
$$left[beginmatrix1&0\0&1\0&0endmatrixright.$$
Now, setting $u=e_1+e_3$, we know that $F(u)=F(e_1)+F(e_3)=3e_1+3e_3$, so $F(e_3)=2e_1+3e_3$, and we have the third column of the matrix, which is
$$ beginbmatrix1&0&2\0&1&0\0&0&3endbmatrix$$
answered Jul 27 at 12:22
Bernard
110k635102
Here is a very simple solution:
Denoting $e_1, e_2, e_3$ the canonical basis, you need to express $F(e_1), F(e_2)$ and $F(e_3)$ in the canonical basis.
For $F(e_1)$ and $F(e_2)$, its in the hypotheses, so the first two columns are
$$left[beginmatrix1&0\0&1\0&0endmatrixright.$$
Now, setting $u=e_1+e_3$, we know that $F(u)=F(e_1)+F(e_3)=3e_1+3e_3$, so $F(e_3)=2e_1+3e_3$, and we have the third column of the matrix, which is
$$ beginbmatrix1&0&2\0&1&0\0&0&3endbmatrix$$
answered Jul 27 at 12:22
Bernard
110k635102
answered Jul 27 at 12:22
Bernard
110k635102
answered Jul 27 at 12:22
Bernard
110k635102
answered Jul 27 at 12:22
Bernard
110k635102
110k635102
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2864344%2fhow-to-construct-a-3-by-3-matrix-with-a-given-eigenspace-in-the-standard-basis%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password