Mixing
basis of space of all $3times3$ matrices and basis of space of all $3times3$ matrices with rank $0, 1$ or $2$
I have a set of all $3times3$ matrices $B$ that have rank $0, 1$ or $2$ - what is the basis of a subspace that this set generates?
I came to the conclusion that this subspace consists of lines, planes and the null vector and it generates the whole space of $Bbb R^3$, but I don't know what is the basis of this subspace - is it just the set of all the $B$'s?
EDIT: my original thinking was very wrong, the questions I ended up with are what is the basis of the space of all $3times3$ matrices and what is the basis of the subspace - all $3times3$ matrices with rank $0, 1$ or $2$
linear-algebra matrices matrix-rank
asked Nov 19 '18 at 12:04
agromekagromek
345
add a comment |
I have a set of all $3times3$ matrices $B$ that have rank $0, 1$ or $2$ - what is the basis of a subspace that this set generates?
I came to the conclusion that this subspace consists of lines, planes and the null vector and it generates the whole space of $Bbb R^3$, but I don't know what is the basis of this subspace - is it just the set of all the $B$'s?
EDIT: my original thinking was very wrong, the questions I ended up with are what is the basis of the space of all $3times3$ matrices and what is the basis of the subspace - all $3times3$ matrices with rank $0, 1$ or $2$
linear-algebra matrices matrix-rank
asked Nov 19 '18 at 12:04
agromekagromek
345
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 19 '18 at 12:06
add a comment |
I have a set of all $3times3$ matrices $B$ that have rank $0, 1$ or $2$ - what is the basis of a subspace that this set generates?
I came to the conclusion that this subspace consists of lines, planes and the null vector and it generates the whole space of $Bbb R^3$, but I don't know what is the basis of this subspace - is it just the set of all the $B$'s?
EDIT: my original thinking was very wrong, the questions I ended up with are what is the basis of the space of all $3times3$ matrices and what is the basis of the subspace - all $3times3$ matrices with rank $0, 1$ or $2$
linear-algebra matrices matrix-rank
asked Nov 19 '18 at 12:04
agromekagromek
345
I have a set of all $3times3$ matrices $B$ that have rank $0, 1$ or $2$ - what is the basis of a subspace that this set generates?
I came to the conclusion that this subspace consists of lines, planes and the null vector and it generates the whole space of $Bbb R^3$, but I don't know what is the basis of this subspace - is it just the set of all the $B$'s?
EDIT: my original thinking was very wrong, the questions I ended up with are what is the basis of the space of all $3times3$ matrices and what is the basis of the subspace - all $3times3$ matrices with rank $0, 1$ or $2$
linear-algebra matrices matrix-rank
linear-algebra matrices matrix-rank
asked Nov 19 '18 at 12:04
agromekagromek
345
asked Nov 19 '18 at 12:04
agromekagromek
345
edited Nov 21 '18 at 22:59
agromek
asked Nov 19 '18 at 12:04
agromekagromek
345
asked Nov 19 '18 at 12:04
agromekagromek
345
asked Nov 19 '18 at 12:04
agromekagromek
345
345
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 19 '18 at 12:06
add a comment |
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 19 '18 at 12:06
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 19 '18 at 12:06
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 19 '18 at 12:06
add a comment |
2 Answers
2
active
oldest
votes
The vector space of all $3 x 3$ matrices is not $R^3$. You can verify that the space has dimension $9$ because you will need $9$ vectors for a basis. Probably the most likely ones would be $begin{bmatrix}1 & 0 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$, $begin{bmatrix}0 & 1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$ ... $begin{bmatrix}0 & 0 & 1\0 & 0 & 0\0 & 0 & 0end{bmatrix}$,$begin{bmatrix}0 & 0 & 0\1 & 0 & 0\0 & 0 & 0end{bmatrix}$, $begin{bmatrix}0 & 0 & 0\0 & 1 & 0\0 & 0 & 0end{bmatrix}$, $begin{bmatrix}0 & 0 & 0\0 & 0 & 1\0 & 0 & 0end{bmatrix}$,$begin{bmatrix}0 & 0 & 0\0 & 0 & 0\1 & 0 & 0end{bmatrix}$, $begin{bmatrix}0 & 0 & 0\0 & 0 & 0\0 & 1 & 0end{bmatrix}$, $begin{bmatrix}0 & 0 & 0\0 & 0 & 0\0 & 0 & 1end{bmatrix}$
Now considering that your subspace only contains matrices with rank $leq2$ the basis for that space will contain only $6$ of these $9$.
answered Nov 19 '18 at 13:56
ChaiChai
1255
SInce all those matrices have rank $le2$, they all belong to the given set.
– egreg
Nov 19 '18 at 14:23
Is that true? Taking linear combinations of all $9$ will yield all possible matrices, including rank $3$ ones. If we leave out the ones that have 1's in the last column ofr instance, all combinations will have rank $leq2$ , or am I missing something?
– Chai
Nov 19 '18 at 14:28
Well, every matrix belongs to the span of the rank $1$ matrices.
– egreg
Nov 19 '18 at 14:33
I'm not quite sure what you mean by that. As I understand it, OP is looking for a basis of the subspace of $3x3$ matrices that do not include the ones with rank$3$. I believe you could find a basis in this manner. But if I am wrong I'd like to learn as well !
– Chai
Nov 19 '18 at 14:38
I don't really understand why the vector space of all 3x3 matrices has dimension 9? can someone please explain?
– agromek
Nov 19 '18 at 17:45
|
show 3 more comments
It's unclear which space you're talking about---subspace of what? If you are viewing $3times3$ matrices as elements of $V=mathbb R^{3,3}$ i.e. $3times3$ matrices under addition, then the answer is all of $V$.
answered Nov 19 '18 at 12:21
Richard MartinRichard Martin
1,61118
the set of B matrices is the subspace of R^3
– agromek
Nov 19 '18 at 12:31
No, I don't think so.
– Richard Martin
Nov 19 '18 at 12:33
the whole task was that I have two 3x3 matrices: A and C - A has rank 2 and C can be any 3x3 matrix and the question is: what is the basis of a subspace of the space of all 3x3 matrices that is generated by the set of 3x3 matrices B that satisfy the equation A*C=B (I'm trying to translate it into english as exact as I can)
– agromek
Nov 19 '18 at 12:34
maybe I'm interpreting the phrase "the space of all 3x3 matrices" wrongly, I thought it is just R^3 but maybe it's something else?
– agromek
Nov 19 '18 at 12:37
The space of all 3x3 matrices acts on $mathbb R^3$ but it is not of itself $mathbb R^3$.
– Richard Martin
Nov 19 '18 at 12:38
|
show 2 more comments
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2 Answers
2
active
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votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The vector space of all $3 x 3$ matrices is not $R^3$. You can verify that the space has dimension $9$ because you will need $9$ vectors for a basis. Probably the most likely ones would be $begin{bmatrix}1 & 0 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$, $begin{bmatrix}0 & 1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$ ... $begin{bmatrix}0 & 0 & 1\0 & 0 & 0\0 & 0 & 0end{bmatrix}$,$begin{bmatrix}0 & 0 & 0\1 & 0 & 0\0 & 0 & 0end{bmatrix}$, $begin{bmatrix}0 & 0 & 0\0 & 1 & 0\0 & 0 & 0end{bmatrix}$, $begin{bmatrix}0 & 0 & 0\0 & 0 & 1\0 & 0 & 0end{bmatrix}$,$begin{bmatrix}0 & 0 & 0\0 & 0 & 0\1 & 0 & 0end{bmatrix}$, $begin{bmatrix}0 & 0 & 0\0 & 0 & 0\0 & 1 & 0end{bmatrix}$, $begin{bmatrix}0 & 0 & 0\0 & 0 & 0\0 & 0 & 1end{bmatrix}$
Now considering that your subspace only contains matrices with rank $leq2$ the basis for that space will contain only $6$ of these $9$.
answered Nov 19 '18 at 13:56
ChaiChai
1255
SInce all those matrices have rank $le2$, they all belong to the given set.
– egreg
Nov 19 '18 at 14:23
Is that true? Taking linear combinations of all $9$ will yield all possible matrices, including rank $3$ ones. If we leave out the ones that have 1's in the last column ofr instance, all combinations will have rank $leq2$ , or am I missing something?
– Chai
Nov 19 '18 at 14:28
Well, every matrix belongs to the span of the rank $1$ matrices.
– egreg
Nov 19 '18 at 14:33
I'm not quite sure what you mean by that. As I understand it, OP is looking for a basis of the subspace of $3x3$ matrices that do not include the ones with rank$3$. I believe you could find a basis in this manner. But if I am wrong I'd like to learn as well !
– Chai
Nov 19 '18 at 14:38
I don't really understand why the vector space of all 3x3 matrices has dimension 9? can someone please explain?
– agromek
Nov 19 '18 at 17:45
|
show 3 more comments
The vector space of all $3 x 3$ matrices is not $R^3$. You can verify that the space has dimension $9$ because you will need $9$ vectors for a basis. Probably the most likely ones would be $begin{bmatrix}1 & 0 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$, $begin{bmatrix}0 & 1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$ ... $begin{bmatrix}0 & 0 & 1\0 & 0 & 0\0 & 0 & 0end{bmatrix}$,$begin{bmatrix}0 & 0 & 0\1 & 0 & 0\0 & 0 & 0end{bmatrix}$, $begin{bmatrix}0 & 0 & 0\0 & 1 & 0\0 & 0 & 0end{bmatrix}$, $begin{bmatrix}0 & 0 & 0\0 & 0 & 1\0 & 0 & 0end{bmatrix}$,$begin{bmatrix}0 & 0 & 0\0 & 0 & 0\1 & 0 & 0end{bmatrix}$, $begin{bmatrix}0 & 0 & 0\0 & 0 & 0\0 & 1 & 0end{bmatrix}$, $begin{bmatrix}0 & 0 & 0\0 & 0 & 0\0 & 0 & 1end{bmatrix}$
Now considering that your subspace only contains matrices with rank $leq2$ the basis for that space will contain only $6$ of these $9$.
answered Nov 19 '18 at 13:56
ChaiChai
1255
SInce all those matrices have rank $le2$, they all belong to the given set.
– egreg
Nov 19 '18 at 14:23
Is that true? Taking linear combinations of all $9$ will yield all possible matrices, including rank $3$ ones. If we leave out the ones that have 1's in the last column ofr instance, all combinations will have rank $leq2$ , or am I missing something?
– Chai
Nov 19 '18 at 14:28
Well, every matrix belongs to the span of the rank $1$ matrices.
– egreg
Nov 19 '18 at 14:33
I'm not quite sure what you mean by that. As I understand it, OP is looking for a basis of the subspace of $3x3$ matrices that do not include the ones with rank$3$. I believe you could find a basis in this manner. But if I am wrong I'd like to learn as well !
– Chai
Nov 19 '18 at 14:38
I don't really understand why the vector space of all 3x3 matrices has dimension 9? can someone please explain?
– agromek
Nov 19 '18 at 17:45
|
show 3 more comments
The vector space of all $3 x 3$ matrices is not $R^3$. You can verify that the space has dimension $9$ because you will need $9$ vectors for a basis. Probably the most likely ones would be $begin{bmatrix}1 & 0 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$, $begin{bmatrix}0 & 1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$ ... $begin{bmatrix}0 & 0 & 1\0 & 0 & 0\0 & 0 & 0end{bmatrix}$,$begin{bmatrix}0 & 0 & 0\1 & 0 & 0\0 & 0 & 0end{bmatrix}$, $begin{bmatrix}0 & 0 & 0\0 & 1 & 0\0 & 0 & 0end{bmatrix}$, $begin{bmatrix}0 & 0 & 0\0 & 0 & 1\0 & 0 & 0end{bmatrix}$,$begin{bmatrix}0 & 0 & 0\0 & 0 & 0\1 & 0 & 0end{bmatrix}$, $begin{bmatrix}0 & 0 & 0\0 & 0 & 0\0 & 1 & 0end{bmatrix}$, $begin{bmatrix}0 & 0 & 0\0 & 0 & 0\0 & 0 & 1end{bmatrix}$
Now considering that your subspace only contains matrices with rank $leq2$ the basis for that space will contain only $6$ of these $9$.
answered Nov 19 '18 at 13:56
ChaiChai
1255
The vector space of all $3 x 3$ matrices is not $R^3$. You can verify that the space has dimension $9$ because you will need $9$ vectors for a basis. Probably the most likely ones would be $begin{bmatrix}1 & 0 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$, $begin{bmatrix}0 & 1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$ ... $begin{bmatrix}0 & 0 & 1\0 & 0 & 0\0 & 0 & 0end{bmatrix}$,$begin{bmatrix}0 & 0 & 0\1 & 0 & 0\0 & 0 & 0end{bmatrix}$, $begin{bmatrix}0 & 0 & 0\0 & 1 & 0\0 & 0 & 0end{bmatrix}$, $begin{bmatrix}0 & 0 & 0\0 & 0 & 1\0 & 0 & 0end{bmatrix}$,$begin{bmatrix}0 & 0 & 0\0 & 0 & 0\1 & 0 & 0end{bmatrix}$, $begin{bmatrix}0 & 0 & 0\0 & 0 & 0\0 & 1 & 0end{bmatrix}$, $begin{bmatrix}0 & 0 & 0\0 & 0 & 0\0 & 0 & 1end{bmatrix}$
Now considering that your subspace only contains matrices with rank $leq2$ the basis for that space will contain only $6$ of these $9$.
answered Nov 19 '18 at 13:56
ChaiChai
1255
edited Nov 19 '18 at 21:04
answered Nov 19 '18 at 13:56
ChaiChai
1255
answered Nov 19 '18 at 13:56
ChaiChai
1255
answered Nov 19 '18 at 13:56
ChaiChai
1255
1255
SInce all those matrices have rank $le2$, they all belong to the given set.
– egreg
Nov 19 '18 at 14:23
Is that true? Taking linear combinations of all $9$ will yield all possible matrices, including rank $3$ ones. If we leave out the ones that have 1's in the last column ofr instance, all combinations will have rank $leq2$ , or am I missing something?
– Chai
Nov 19 '18 at 14:28
Well, every matrix belongs to the span of the rank $1$ matrices.
– egreg
Nov 19 '18 at 14:33
I'm not quite sure what you mean by that. As I understand it, OP is looking for a basis of the subspace of $3x3$ matrices that do not include the ones with rank$3$. I believe you could find a basis in this manner. But if I am wrong I'd like to learn as well !
– Chai
Nov 19 '18 at 14:38
I don't really understand why the vector space of all 3x3 matrices has dimension 9? can someone please explain?
– agromek
Nov 19 '18 at 17:45
|
show 3 more comments
SInce all those matrices have rank $le2$, they all belong to the given set.
– egreg
Nov 19 '18 at 14:23
Is that true? Taking linear combinations of all $9$ will yield all possible matrices, including rank $3$ ones. If we leave out the ones that have 1's in the last column ofr instance, all combinations will have rank $leq2$ , or am I missing something?
– Chai
Nov 19 '18 at 14:28
Well, every matrix belongs to the span of the rank $1$ matrices.
– egreg
Nov 19 '18 at 14:33
I'm not quite sure what you mean by that. As I understand it, OP is looking for a basis of the subspace of $3x3$ matrices that do not include the ones with rank$3$. I believe you could find a basis in this manner. But if I am wrong I'd like to learn as well !
– Chai
Nov 19 '18 at 14:38
I don't really understand why the vector space of all 3x3 matrices has dimension 9? can someone please explain?
– agromek
Nov 19 '18 at 17:45
SInce all those matrices have rank $le2$, they all belong to the given set.
– egreg
Nov 19 '18 at 14:23
SInce all those matrices have rank $le2$, they all belong to the given set.
– egreg
Nov 19 '18 at 14:23
Is that true? Taking linear combinations of all $9$ will yield all possible matrices, including rank $3$ ones. If we leave out the ones that have 1's in the last column ofr instance, all combinations will have rank $leq2$ , or am I missing something?
– Chai
Nov 19 '18 at 14:28
Is that true? Taking linear combinations of all $9$ will yield all possible matrices, including rank $3$ ones. If we leave out the ones that have 1's in the last column ofr instance, all combinations will have rank $leq2$ , or am I missing something?
– Chai
Nov 19 '18 at 14:28
Well, every matrix belongs to the span of the rank $1$ matrices.
– egreg
Nov 19 '18 at 14:33
Well, every matrix belongs to the span of the rank $1$ matrices.
– egreg
Nov 19 '18 at 14:33
I'm not quite sure what you mean by that. As I understand it, OP is looking for a basis of the subspace of $3x3$ matrices that do not include the ones with rank$3$. I believe you could find a basis in this manner. But if I am wrong I'd like to learn as well !
– Chai
Nov 19 '18 at 14:38
I'm not quite sure what you mean by that. As I understand it, OP is looking for a basis of the subspace of $3x3$ matrices that do not include the ones with rank$3$. I believe you could find a basis in this manner. But if I am wrong I'd like to learn as well !
– Chai
Nov 19 '18 at 14:38
I don't really understand why the vector space of all 3x3 matrices has dimension 9? can someone please explain?
– agromek
Nov 19 '18 at 17:45
I don't really understand why the vector space of all 3x3 matrices has dimension 9? can someone please explain?
– agromek
Nov 19 '18 at 17:45
|
show 3 more comments
It's unclear which space you're talking about---subspace of what? If you are viewing $3times3$ matrices as elements of $V=mathbb R^{3,3}$ i.e. $3times3$ matrices under addition, then the answer is all of $V$.
answered Nov 19 '18 at 12:21
Richard MartinRichard Martin
1,61118
the set of B matrices is the subspace of R^3
– agromek
Nov 19 '18 at 12:31
No, I don't think so.
– Richard Martin
Nov 19 '18 at 12:33
the whole task was that I have two 3x3 matrices: A and C - A has rank 2 and C can be any 3x3 matrix and the question is: what is the basis of a subspace of the space of all 3x3 matrices that is generated by the set of 3x3 matrices B that satisfy the equation A*C=B (I'm trying to translate it into english as exact as I can)
– agromek
Nov 19 '18 at 12:34
maybe I'm interpreting the phrase "the space of all 3x3 matrices" wrongly, I thought it is just R^3 but maybe it's something else?
– agromek
Nov 19 '18 at 12:37
The space of all 3x3 matrices acts on $mathbb R^3$ but it is not of itself $mathbb R^3$.
– Richard Martin
Nov 19 '18 at 12:38
|
show 2 more comments
It's unclear which space you're talking about---subspace of what? If you are viewing $3times3$ matrices as elements of $V=mathbb R^{3,3}$ i.e. $3times3$ matrices under addition, then the answer is all of $V$.
answered Nov 19 '18 at 12:21
Richard MartinRichard Martin
1,61118
the set of B matrices is the subspace of R^3
– agromek
Nov 19 '18 at 12:31
No, I don't think so.
– Richard Martin
Nov 19 '18 at 12:33
the whole task was that I have two 3x3 matrices: A and C - A has rank 2 and C can be any 3x3 matrix and the question is: what is the basis of a subspace of the space of all 3x3 matrices that is generated by the set of 3x3 matrices B that satisfy the equation A*C=B (I'm trying to translate it into english as exact as I can)
– agromek
Nov 19 '18 at 12:34
maybe I'm interpreting the phrase "the space of all 3x3 matrices" wrongly, I thought it is just R^3 but maybe it's something else?
– agromek
Nov 19 '18 at 12:37
The space of all 3x3 matrices acts on $mathbb R^3$ but it is not of itself $mathbb R^3$.
– Richard Martin
Nov 19 '18 at 12:38
|
show 2 more comments
It's unclear which space you're talking about---subspace of what? If you are viewing $3times3$ matrices as elements of $V=mathbb R^{3,3}$ i.e. $3times3$ matrices under addition, then the answer is all of $V$.
answered Nov 19 '18 at 12:21
Richard MartinRichard Martin
1,61118
It's unclear which space you're talking about---subspace of what? If you are viewing $3times3$ matrices as elements of $V=mathbb R^{3,3}$ i.e. $3times3$ matrices under addition, then the answer is all of $V$.
answered Nov 19 '18 at 12:21
Richard MartinRichard Martin
1,61118
answered Nov 19 '18 at 12:21
Richard MartinRichard Martin
1,61118
answered Nov 19 '18 at 12:21
Richard MartinRichard Martin
1,61118
answered Nov 19 '18 at 12:21
Richard MartinRichard Martin
1,61118
1,61118
the set of B matrices is the subspace of R^3
– agromek
Nov 19 '18 at 12:31
No, I don't think so.
– Richard Martin
Nov 19 '18 at 12:33
the whole task was that I have two 3x3 matrices: A and C - A has rank 2 and C can be any 3x3 matrix and the question is: what is the basis of a subspace of the space of all 3x3 matrices that is generated by the set of 3x3 matrices B that satisfy the equation A*C=B (I'm trying to translate it into english as exact as I can)
– agromek
Nov 19 '18 at 12:34
maybe I'm interpreting the phrase "the space of all 3x3 matrices" wrongly, I thought it is just R^3 but maybe it's something else?
– agromek
Nov 19 '18 at 12:37
The space of all 3x3 matrices acts on $mathbb R^3$ but it is not of itself $mathbb R^3$.
– Richard Martin
Nov 19 '18 at 12:38
|
show 2 more comments
the set of B matrices is the subspace of R^3
– agromek
Nov 19 '18 at 12:31
No, I don't think so.
– Richard Martin
Nov 19 '18 at 12:33
the whole task was that I have two 3x3 matrices: A and C - A has rank 2 and C can be any 3x3 matrix and the question is: what is the basis of a subspace of the space of all 3x3 matrices that is generated by the set of 3x3 matrices B that satisfy the equation A*C=B (I'm trying to translate it into english as exact as I can)
– agromek
Nov 19 '18 at 12:34
maybe I'm interpreting the phrase "the space of all 3x3 matrices" wrongly, I thought it is just R^3 but maybe it's something else?
– agromek
Nov 19 '18 at 12:37
The space of all 3x3 matrices acts on $mathbb R^3$ but it is not of itself $mathbb R^3$.
– Richard Martin
Nov 19 '18 at 12:38
the set of B matrices is the subspace of R^3
– agromek
Nov 19 '18 at 12:31
the set of B matrices is the subspace of R^3
– agromek
Nov 19 '18 at 12:31
No, I don't think so.
– Richard Martin
Nov 19 '18 at 12:33
No, I don't think so.
– Richard Martin
Nov 19 '18 at 12:33
the whole task was that I have two 3x3 matrices: A and C - A has rank 2 and C can be any 3x3 matrix and the question is: what is the basis of a subspace of the space of all 3x3 matrices that is generated by the set of 3x3 matrices B that satisfy the equation A*C=B (I'm trying to translate it into english as exact as I can)
– agromek
Nov 19 '18 at 12:34
the whole task was that I have two 3x3 matrices: A and C - A has rank 2 and C can be any 3x3 matrix and the question is: what is the basis of a subspace of the space of all 3x3 matrices that is generated by the set of 3x3 matrices B that satisfy the equation A*C=B (I'm trying to translate it into english as exact as I can)
– agromek
Nov 19 '18 at 12:34
maybe I'm interpreting the phrase "the space of all 3x3 matrices" wrongly, I thought it is just R^3 but maybe it's something else?
– agromek
Nov 19 '18 at 12:37
maybe I'm interpreting the phrase "the space of all 3x3 matrices" wrongly, I thought it is just R^3 but maybe it's something else?
– agromek
Nov 19 '18 at 12:37
The space of all 3x3 matrices acts on $mathbb R^3$ but it is not of itself $mathbb R^3$.
– Richard Martin
Nov 19 '18 at 12:38
The space of all 3x3 matrices acts on $mathbb R^3$ but it is not of itself $mathbb R^3$.
– Richard Martin
Nov 19 '18 at 12:38
|
show 2 more comments
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 19 '18 at 12:06