what is Matrix of a linear transformation?
$begingroup$
I am a student and I'm studying linear algebra. in the Sheldon Axler book in the part "The Spectral Theorem" and in this video he mentions the operator $T$ as
$$T=begin{pmatrix}2&-3\3&2end{pmatrix}$$
then after finding it's eigenvectors $2 + 3i$ and $2 - 3i$ (in the video I've linked), he says: "with respect to this basis, the matrix of $T$ is the diagonal matrix":
$$
begin{pmatrix}
2 + 3i&0 \
0&2 - 3i
end{pmatrix}
$$
I am confused. $T$ already mentions as $T=begin{pmatrix}2&-3\3&2end{pmatrix}$ so the matrix of $T$ must be $begin{pmatrix}2&-3\3&2end{pmatrix}$ so I think that I don't know the meaning of the matrix of $T$.
could you help me figure it out?
linear-algebra matrices operator-theory spectral-theory
asked Jan 5 at 22:31
Peyman mohseni kiasariPeyman mohseni kiasari
13711
$endgroup$
add a comment |
$begingroup$
I am a student and I'm studying linear algebra. in the Sheldon Axler book in the part "The Spectral Theorem" and in this video he mentions the operator $T$ as
$$T=begin{pmatrix}2&-3\3&2end{pmatrix}$$
then after finding it's eigenvectors $2 + 3i$ and $2 - 3i$ (in the video I've linked), he says: "with respect to this basis, the matrix of $T$ is the diagonal matrix":
$$
begin{pmatrix}
2 + 3i&0 \
0&2 - 3i
end{pmatrix}
$$
I am confused. $T$ already mentions as $T=begin{pmatrix}2&-3\3&2end{pmatrix}$ so the matrix of $T$ must be $begin{pmatrix}2&-3\3&2end{pmatrix}$ so I think that I don't know the meaning of the matrix of $T$.
could you help me figure it out?
linear-algebra matrices operator-theory spectral-theory
asked Jan 5 at 22:31
Peyman mohseni kiasariPeyman mohseni kiasari
13711
$endgroup$
add a comment |
$begingroup$
I am a student and I'm studying linear algebra. in the Sheldon Axler book in the part "The Spectral Theorem" and in this video he mentions the operator $T$ as
$$T=begin{pmatrix}2&-3\3&2end{pmatrix}$$
then after finding it's eigenvectors $2 + 3i$ and $2 - 3i$ (in the video I've linked), he says: "with respect to this basis, the matrix of $T$ is the diagonal matrix":
$$
begin{pmatrix}
2 + 3i&0 \
0&2 - 3i
end{pmatrix}
$$
I am confused. $T$ already mentions as $T=begin{pmatrix}2&-3\3&2end{pmatrix}$ so the matrix of $T$ must be $begin{pmatrix}2&-3\3&2end{pmatrix}$ so I think that I don't know the meaning of the matrix of $T$.
could you help me figure it out?
linear-algebra matrices operator-theory spectral-theory
asked Jan 5 at 22:31
Peyman mohseni kiasariPeyman mohseni kiasari
13711
$endgroup$
I am a student and I'm studying linear algebra. in the Sheldon Axler book in the part "The Spectral Theorem" and in this video he mentions the operator $T$ as
$$T=begin{pmatrix}2&-3\3&2end{pmatrix}$$
then after finding it's eigenvectors $2 + 3i$ and $2 - 3i$ (in the video I've linked), he says: "with respect to this basis, the matrix of $T$ is the diagonal matrix":
$$
begin{pmatrix}
2 + 3i&0 \
0&2 - 3i
end{pmatrix}
$$
I am confused. $T$ already mentions as $T=begin{pmatrix}2&-3\3&2end{pmatrix}$ so the matrix of $T$ must be $begin{pmatrix}2&-3\3&2end{pmatrix}$ so I think that I don't know the meaning of the matrix of $T$.
could you help me figure it out?
linear-algebra matrices operator-theory spectral-theory
linear-algebra matrices operator-theory spectral-theory
asked Jan 5 at 22:31
Peyman mohseni kiasariPeyman mohseni kiasari
13711
asked Jan 5 at 22:31
Peyman mohseni kiasariPeyman mohseni kiasari
13711
edited Jan 5 at 22:52
Peyman mohseni kiasari
asked Jan 5 at 22:31
Peyman mohseni kiasariPeyman mohseni kiasari
13711
asked Jan 5 at 22:31
Peyman mohseni kiasariPeyman mohseni kiasari
13711
asked Jan 5 at 22:31
Peyman mohseni kiasariPeyman mohseni kiasari
13711
13711
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Matrices only define linear transformations relative to some basis. They don't describe a linear transformation on their own. Thus implicit in $$T=begin{pmatrix} 2&-3 \ 3 & 2 end{pmatrix}$$
is the statement that the vector space has some basis already and that this is the matrix for $T$ with respect to that basis. E.g., perhaps it's $k^2$, so it has the standard basis vectors $begin{pmatrix} 1 \ 0end{pmatrix}$ and $begin{pmatrix} 0 \ 1end{pmatrix}$ already, and $T$ has that matrix with respect to that basis.
If $V$ is a vector space, then a linear transformation $T:Vto V$ is not a matrix, but rather a function with nice properties that respect the vector space structure. We can then describe it using bases and a matrix, but that's only a description, and the description depends on the basis used to compute the matrix.
As for how to compute the matrix of a linear transformation with respect to some basis, look here for some random notes I found online, or in any decent linear algebra textbook, like presumably in Axler's book somewhere (It appears to be section 3C). For an example of how to do this in a particular case, you can look at this question.
answered Jan 5 at 22:42
jgonjgon
15.7k32143
$endgroup$
$begingroup$
great answer, thanks.
$endgroup$
– Peyman mohseni kiasari
Jan 5 at 22:46
$begingroup$
@Peyman Glad it was helpful, also the matrix for a linear transformation appears to be defined in section 3C of Axler's book based on a table of contents I found online.
$endgroup$
– jgon
Jan 5 at 22:48
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Matrices only define linear transformations relative to some basis. They don't describe a linear transformation on their own. Thus implicit in $$T=begin{pmatrix} 2&-3 \ 3 & 2 end{pmatrix}$$
is the statement that the vector space has some basis already and that this is the matrix for $T$ with respect to that basis. E.g., perhaps it's $k^2$, so it has the standard basis vectors $begin{pmatrix} 1 \ 0end{pmatrix}$ and $begin{pmatrix} 0 \ 1end{pmatrix}$ already, and $T$ has that matrix with respect to that basis.
If $V$ is a vector space, then a linear transformation $T:Vto V$ is not a matrix, but rather a function with nice properties that respect the vector space structure. We can then describe it using bases and a matrix, but that's only a description, and the description depends on the basis used to compute the matrix.
As for how to compute the matrix of a linear transformation with respect to some basis, look here for some random notes I found online, or in any decent linear algebra textbook, like presumably in Axler's book somewhere (It appears to be section 3C). For an example of how to do this in a particular case, you can look at this question.
answered Jan 5 at 22:42
jgonjgon
15.7k32143
$endgroup$
$begingroup$
great answer, thanks.
$endgroup$
– Peyman mohseni kiasari
Jan 5 at 22:46
$begingroup$
@Peyman Glad it was helpful, also the matrix for a linear transformation appears to be defined in section 3C of Axler's book based on a table of contents I found online.
$endgroup$
– jgon
Jan 5 at 22:48
add a comment |
$begingroup$
Matrices only define linear transformations relative to some basis. They don't describe a linear transformation on their own. Thus implicit in $$T=begin{pmatrix} 2&-3 \ 3 & 2 end{pmatrix}$$
is the statement that the vector space has some basis already and that this is the matrix for $T$ with respect to that basis. E.g., perhaps it's $k^2$, so it has the standard basis vectors $begin{pmatrix} 1 \ 0end{pmatrix}$ and $begin{pmatrix} 0 \ 1end{pmatrix}$ already, and $T$ has that matrix with respect to that basis.
If $V$ is a vector space, then a linear transformation $T:Vto V$ is not a matrix, but rather a function with nice properties that respect the vector space structure. We can then describe it using bases and a matrix, but that's only a description, and the description depends on the basis used to compute the matrix.
As for how to compute the matrix of a linear transformation with respect to some basis, look here for some random notes I found online, or in any decent linear algebra textbook, like presumably in Axler's book somewhere (It appears to be section 3C). For an example of how to do this in a particular case, you can look at this question.
answered Jan 5 at 22:42
jgonjgon
15.7k32143
$endgroup$
$begingroup$
great answer, thanks.
$endgroup$
– Peyman mohseni kiasari
Jan 5 at 22:46
$begingroup$
@Peyman Glad it was helpful, also the matrix for a linear transformation appears to be defined in section 3C of Axler's book based on a table of contents I found online.
$endgroup$
– jgon
Jan 5 at 22:48
add a comment |
$begingroup$
Matrices only define linear transformations relative to some basis. They don't describe a linear transformation on their own. Thus implicit in $$T=begin{pmatrix} 2&-3 \ 3 & 2 end{pmatrix}$$
is the statement that the vector space has some basis already and that this is the matrix for $T$ with respect to that basis. E.g., perhaps it's $k^2$, so it has the standard basis vectors $begin{pmatrix} 1 \ 0end{pmatrix}$ and $begin{pmatrix} 0 \ 1end{pmatrix}$ already, and $T$ has that matrix with respect to that basis.
If $V$ is a vector space, then a linear transformation $T:Vto V$ is not a matrix, but rather a function with nice properties that respect the vector space structure. We can then describe it using bases and a matrix, but that's only a description, and the description depends on the basis used to compute the matrix.
As for how to compute the matrix of a linear transformation with respect to some basis, look here for some random notes I found online, or in any decent linear algebra textbook, like presumably in Axler's book somewhere (It appears to be section 3C). For an example of how to do this in a particular case, you can look at this question.
answered Jan 5 at 22:42
jgonjgon
15.7k32143
$endgroup$
Matrices only define linear transformations relative to some basis. They don't describe a linear transformation on their own. Thus implicit in $$T=begin{pmatrix} 2&-3 \ 3 & 2 end{pmatrix}$$
is the statement that the vector space has some basis already and that this is the matrix for $T$ with respect to that basis. E.g., perhaps it's $k^2$, so it has the standard basis vectors $begin{pmatrix} 1 \ 0end{pmatrix}$ and $begin{pmatrix} 0 \ 1end{pmatrix}$ already, and $T$ has that matrix with respect to that basis.
If $V$ is a vector space, then a linear transformation $T:Vto V$ is not a matrix, but rather a function with nice properties that respect the vector space structure. We can then describe it using bases and a matrix, but that's only a description, and the description depends on the basis used to compute the matrix.
As for how to compute the matrix of a linear transformation with respect to some basis, look here for some random notes I found online, or in any decent linear algebra textbook, like presumably in Axler's book somewhere (It appears to be section 3C). For an example of how to do this in a particular case, you can look at this question.
answered Jan 5 at 22:42
jgonjgon
15.7k32143
edited Jan 5 at 22:48
answered Jan 5 at 22:42
jgonjgon
15.7k32143
answered Jan 5 at 22:42
jgonjgon
15.7k32143
answered Jan 5 at 22:42
jgonjgon
15.7k32143
15.7k32143
$begingroup$
great answer, thanks.
$endgroup$
– Peyman mohseni kiasari
Jan 5 at 22:46
$begingroup$
@Peyman Glad it was helpful, also the matrix for a linear transformation appears to be defined in section 3C of Axler's book based on a table of contents I found online.
$endgroup$
– jgon
Jan 5 at 22:48
add a comment |
$begingroup$
great answer, thanks.
$endgroup$
– Peyman mohseni kiasari
Jan 5 at 22:46
$begingroup$
@Peyman Glad it was helpful, also the matrix for a linear transformation appears to be defined in section 3C of Axler's book based on a table of contents I found online.
$endgroup$
– jgon
Jan 5 at 22:48
$begingroup$
great answer, thanks.
$endgroup$
– Peyman mohseni kiasari
Jan 5 at 22:46
$begingroup$
great answer, thanks.
$endgroup$
– Peyman mohseni kiasari
Jan 5 at 22:46
$begingroup$
@Peyman Glad it was helpful, also the matrix for a linear transformation appears to be defined in section 3C of Axler's book based on a table of contents I found online.
$endgroup$
– jgon
Jan 5 at 22:48
$begingroup$
@Peyman Glad it was helpful, also the matrix for a linear transformation appears to be defined in section 3C of Axler's book based on a table of contents I found online.
$endgroup$
– jgon
Jan 5 at 22:48
add a comment |
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