Mixing
Proof that $X = left{ A in mathbb R^{6,5} : V subset ker(A) right}$ is linear subspace and find its $dim$
$begingroup$
Proof that if $dim V = 3 $ then $X = left{ A in mathbb R^{6,5} : V subset ker(A) right}$ is linear subspace and find its $dim$
What I've done
I think that I have done proof.
$ forall vec{v} in V. Avec{v} = 0 $ and $Bvec{v}=0$
but from property of the matrix product it follows that:
$$ Avec{v} + Bvec{v} = (A+B)vec{v} $$
Moreover if $Avec{x}=0$ then $(alpha A)vec{x} =0$ and in to other side too. So ok.
My problem
But how in use of given informations find its dimension? My scribbles do not lead to anything.
linear-algebra vector-spaces
edited Dec 31 '18 at 17:21
Bernard
119k639112
asked Dec 31 '18 at 17:12
VirtualUserVirtualUser
58912
$endgroup$
add a comment |
$begingroup$
Proof that if $dim V = 3 $ then $X = left{ A in mathbb R^{6,5} : V subset ker(A) right}$ is linear subspace and find its $dim$
What I've done
I think that I have done proof.
$ forall vec{v} in V. Avec{v} = 0 $ and $Bvec{v}=0$
but from property of the matrix product it follows that:
$$ Avec{v} + Bvec{v} = (A+B)vec{v} $$
Moreover if $Avec{x}=0$ then $(alpha A)vec{x} =0$ and in to other side too. So ok.
My problem
But how in use of given informations find its dimension? My scribbles do not lead to anything.
linear-algebra vector-spaces
edited Dec 31 '18 at 17:21
Bernard
119k639112
asked Dec 31 '18 at 17:12
VirtualUserVirtualUser
58912
$endgroup$
$begingroup$
What do you denote $Bbb R^{6,5}$?
$endgroup$
– Bernard
Dec 31 '18 at 17:23
$begingroup$
Sorry, I used to writeRRbut there it does not work I didn't know how to write it. But now I will remember, thanks
$endgroup$
– VirtualUser
Dec 31 '18 at 17:33
add a comment |
$begingroup$
Proof that if $dim V = 3 $ then $X = left{ A in mathbb R^{6,5} : V subset ker(A) right}$ is linear subspace and find its $dim$
What I've done
I think that I have done proof.
$ forall vec{v} in V. Avec{v} = 0 $ and $Bvec{v}=0$
but from property of the matrix product it follows that:
$$ Avec{v} + Bvec{v} = (A+B)vec{v} $$
Moreover if $Avec{x}=0$ then $(alpha A)vec{x} =0$ and in to other side too. So ok.
My problem
But how in use of given informations find its dimension? My scribbles do not lead to anything.
linear-algebra vector-spaces
edited Dec 31 '18 at 17:21
Bernard
119k639112
asked Dec 31 '18 at 17:12
VirtualUserVirtualUser
58912
$endgroup$
Proof that if $dim V = 3 $ then $X = left{ A in mathbb R^{6,5} : V subset ker(A) right}$ is linear subspace and find its $dim$
What I've done
I think that I have done proof.
$ forall vec{v} in V. Avec{v} = 0 $ and $Bvec{v}=0$
but from property of the matrix product it follows that:
$$ Avec{v} + Bvec{v} = (A+B)vec{v} $$
Moreover if $Avec{x}=0$ then $(alpha A)vec{x} =0$ and in to other side too. So ok.
My problem
But how in use of given informations find its dimension? My scribbles do not lead to anything.
linear-algebra vector-spaces
linear-algebra vector-spaces
edited Dec 31 '18 at 17:21
Bernard
119k639112
asked Dec 31 '18 at 17:12
VirtualUserVirtualUser
58912
edited Dec 31 '18 at 17:21
Bernard
119k639112
asked Dec 31 '18 at 17:12
VirtualUserVirtualUser
58912
edited Dec 31 '18 at 17:21
Bernard
119k639112
edited Dec 31 '18 at 17:21
Bernard
119k639112
edited Dec 31 '18 at 17:21
Bernard
119k639112
119k639112
asked Dec 31 '18 at 17:12
VirtualUserVirtualUser
58912
asked Dec 31 '18 at 17:12
VirtualUserVirtualUser
58912
asked Dec 31 '18 at 17:12
VirtualUserVirtualUser
58912
58912
$begingroup$
What do you denote $Bbb R^{6,5}$?
$endgroup$
– Bernard
Dec 31 '18 at 17:23
$begingroup$
Sorry, I used to writeRRbut there it does not work I didn't know how to write it. But now I will remember, thanks
$endgroup$
– VirtualUser
Dec 31 '18 at 17:33
add a comment |
$begingroup$
What do you denote $Bbb R^{6,5}$?
$endgroup$
– Bernard
Dec 31 '18 at 17:23
$begingroup$
Sorry, I used to writeRRbut there it does not work I didn't know how to write it. But now I will remember, thanks
$endgroup$
– VirtualUser
Dec 31 '18 at 17:33
$begingroup$
What do you denote $Bbb R^{6,5}$?
$endgroup$
– Bernard
Dec 31 '18 at 17:23
$begingroup$
What do you denote $Bbb R^{6,5}$?
$endgroup$
– Bernard
Dec 31 '18 at 17:23
$begingroup$
Sorry, I used to write
RR but there it does not work I didn't know how to write it. But now I will remember, thanks$endgroup$
– VirtualUser
Dec 31 '18 at 17:33
$begingroup$
Sorry, I used to write
RR but there it does not work I didn't know how to write it. But now I will remember, thanks$endgroup$
– VirtualUser
Dec 31 '18 at 17:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$mathbb R^{6,5}$ is a linear space of dimension $6 times 5=30$.
If $(v_1, v_2, v_3)$ (with $v_i in mathbb R^5$) is a basis of $V$. You have
$$X = {A in mathbb R^{6,5} ; A.v_1=A.v_2=A.v_3=0}.$$
It means that you're imposing $3 times 6 = 18$ independent conditions in the dual space $left(mathbb R^{6,5}right)^vee$. Hence $dim X = 30 -18=12$.
Another proof not using the dual space
Without loss of generality, you can suppose that $(v_1, dots, v_5)$ is a basis of $mathcal V = mathbb R^5$ where $(v_1, v_2,v_3)$ is a basis of $V$. And that $(v_1^prime, dots, v_6^prime)$ is a basis of $mathcal V^prime=mathbb R^6$.
For $A =(a_{ij}) in X$ expressed in those basis, you must have $$a_{11} =dots =a_{61}=a_{12}=dots=a_{62}=a_{13}=dots=a_{63}=0$$ as $V subseteq ker A$.
If we denote $E_{ij}$ the matrix with coefficient $i,j$ equal to $1$ and the others vanishing, $(E_{14}, dots, E_{64}, E_{15}, dots, E_{65})$ is a basis of $X$. Proving that the dimension of that linear subspace is equal to $12$.
answered Dec 31 '18 at 17:28
mathcounterexamples.netmathcounterexamples.net
25.3k21953
$endgroup$
$begingroup$
I don't understand this notation: $A.v_1=A.v_2=A.v_3=0$. Does $A.v_{i} $ means $i$ column? I suspect that these are not the same vectors as the vectors from base $V$?
$endgroup$
– VirtualUser
Dec 31 '18 at 17:47
$begingroup$
For a vector $v in V$ (with 5 coordinates) $A.v$ means the image by $A$ of vector $v$. $A.v$ is a vector with 6 coordinates.
$endgroup$
– mathcounterexamples.net
Dec 31 '18 at 17:49
$begingroup$
Can it be done without dual space? I haven't got this on lecture and probably I can't use that
$endgroup$
– VirtualUser
Jan 2 at 19:12
$begingroup$
What are the key theorems of your lecture?
$endgroup$
– mathcounterexamples.net
Jan 2 at 20:35
$begingroup$
image, kernel, rank, basis of linear space, sum of linear spaces linear independence and linear combinations
$endgroup$
– VirtualUser
Jan 2 at 20:39
|
show 4 more comments
$begingroup$
Since $Vsubsetker(A)forall Ain X$, the rows of $A$ are orthogonal to all vectors in $V$ and thus belong to $V^perp$, which has dimension $2$. Each row vector of $A$ is the linear combination of the two basis vectors $v_1,v_2$ of $V^perp$.
$$A=begin{bmatrix}a_{11}v_1^T+a_{12}v_2^T\a_{21}v_1^T+a_{22}v_2^T\vdots\a_{61}v_1^T+a_{62}v_2^Tend{bmatrix}=a_{11}begin{bmatrix}v_1^T\0\vdots\0end{bmatrix}+a_{12}begin{bmatrix}v_2^T\0\vdots\0end{bmatrix}+a_{21}begin{bmatrix}0\v_1^T\vdots\0end{bmatrix}+...+a_{62}begin{bmatrix}0\0\vdots\v_2^Tend{bmatrix}$$
Thus, $X$ has $12$ basis vectors.
answered Jan 1 at 12:40
Shubham JohriShubham Johri
4,666717
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$mathbb R^{6,5}$ is a linear space of dimension $6 times 5=30$.
If $(v_1, v_2, v_3)$ (with $v_i in mathbb R^5$) is a basis of $V$. You have
$$X = {A in mathbb R^{6,5} ; A.v_1=A.v_2=A.v_3=0}.$$
It means that you're imposing $3 times 6 = 18$ independent conditions in the dual space $left(mathbb R^{6,5}right)^vee$. Hence $dim X = 30 -18=12$.
Another proof not using the dual space
Without loss of generality, you can suppose that $(v_1, dots, v_5)$ is a basis of $mathcal V = mathbb R^5$ where $(v_1, v_2,v_3)$ is a basis of $V$. And that $(v_1^prime, dots, v_6^prime)$ is a basis of $mathcal V^prime=mathbb R^6$.
For $A =(a_{ij}) in X$ expressed in those basis, you must have $$a_{11} =dots =a_{61}=a_{12}=dots=a_{62}=a_{13}=dots=a_{63}=0$$ as $V subseteq ker A$.
If we denote $E_{ij}$ the matrix with coefficient $i,j$ equal to $1$ and the others vanishing, $(E_{14}, dots, E_{64}, E_{15}, dots, E_{65})$ is a basis of $X$. Proving that the dimension of that linear subspace is equal to $12$.
answered Dec 31 '18 at 17:28
mathcounterexamples.netmathcounterexamples.net
25.3k21953
$endgroup$
$begingroup$
I don't understand this notation: $A.v_1=A.v_2=A.v_3=0$. Does $A.v_{i} $ means $i$ column? I suspect that these are not the same vectors as the vectors from base $V$?
$endgroup$
– VirtualUser
Dec 31 '18 at 17:47
$begingroup$
For a vector $v in V$ (with 5 coordinates) $A.v$ means the image by $A$ of vector $v$. $A.v$ is a vector with 6 coordinates.
$endgroup$
– mathcounterexamples.net
Dec 31 '18 at 17:49
$begingroup$
Can it be done without dual space? I haven't got this on lecture and probably I can't use that
$endgroup$
– VirtualUser
Jan 2 at 19:12
$begingroup$
What are the key theorems of your lecture?
$endgroup$
– mathcounterexamples.net
Jan 2 at 20:35
$begingroup$
image, kernel, rank, basis of linear space, sum of linear spaces linear independence and linear combinations
$endgroup$
– VirtualUser
Jan 2 at 20:39
|
show 4 more comments
$begingroup$
$mathbb R^{6,5}$ is a linear space of dimension $6 times 5=30$.
If $(v_1, v_2, v_3)$ (with $v_i in mathbb R^5$) is a basis of $V$. You have
$$X = {A in mathbb R^{6,5} ; A.v_1=A.v_2=A.v_3=0}.$$
It means that you're imposing $3 times 6 = 18$ independent conditions in the dual space $left(mathbb R^{6,5}right)^vee$. Hence $dim X = 30 -18=12$.
Another proof not using the dual space
Without loss of generality, you can suppose that $(v_1, dots, v_5)$ is a basis of $mathcal V = mathbb R^5$ where $(v_1, v_2,v_3)$ is a basis of $V$. And that $(v_1^prime, dots, v_6^prime)$ is a basis of $mathcal V^prime=mathbb R^6$.
For $A =(a_{ij}) in X$ expressed in those basis, you must have $$a_{11} =dots =a_{61}=a_{12}=dots=a_{62}=a_{13}=dots=a_{63}=0$$ as $V subseteq ker A$.
If we denote $E_{ij}$ the matrix with coefficient $i,j$ equal to $1$ and the others vanishing, $(E_{14}, dots, E_{64}, E_{15}, dots, E_{65})$ is a basis of $X$. Proving that the dimension of that linear subspace is equal to $12$.
answered Dec 31 '18 at 17:28
mathcounterexamples.netmathcounterexamples.net
25.3k21953
$endgroup$
$begingroup$
I don't understand this notation: $A.v_1=A.v_2=A.v_3=0$. Does $A.v_{i} $ means $i$ column? I suspect that these are not the same vectors as the vectors from base $V$?
$endgroup$
– VirtualUser
Dec 31 '18 at 17:47
$begingroup$
For a vector $v in V$ (with 5 coordinates) $A.v$ means the image by $A$ of vector $v$. $A.v$ is a vector with 6 coordinates.
$endgroup$
– mathcounterexamples.net
Dec 31 '18 at 17:49
$begingroup$
Can it be done without dual space? I haven't got this on lecture and probably I can't use that
$endgroup$
– VirtualUser
Jan 2 at 19:12
$begingroup$
What are the key theorems of your lecture?
$endgroup$
– mathcounterexamples.net
Jan 2 at 20:35
$begingroup$
image, kernel, rank, basis of linear space, sum of linear spaces linear independence and linear combinations
$endgroup$
– VirtualUser
Jan 2 at 20:39
|
show 4 more comments
$begingroup$
$mathbb R^{6,5}$ is a linear space of dimension $6 times 5=30$.
If $(v_1, v_2, v_3)$ (with $v_i in mathbb R^5$) is a basis of $V$. You have
$$X = {A in mathbb R^{6,5} ; A.v_1=A.v_2=A.v_3=0}.$$
It means that you're imposing $3 times 6 = 18$ independent conditions in the dual space $left(mathbb R^{6,5}right)^vee$. Hence $dim X = 30 -18=12$.
Another proof not using the dual space
Without loss of generality, you can suppose that $(v_1, dots, v_5)$ is a basis of $mathcal V = mathbb R^5$ where $(v_1, v_2,v_3)$ is a basis of $V$. And that $(v_1^prime, dots, v_6^prime)$ is a basis of $mathcal V^prime=mathbb R^6$.
For $A =(a_{ij}) in X$ expressed in those basis, you must have $$a_{11} =dots =a_{61}=a_{12}=dots=a_{62}=a_{13}=dots=a_{63}=0$$ as $V subseteq ker A$.
If we denote $E_{ij}$ the matrix with coefficient $i,j$ equal to $1$ and the others vanishing, $(E_{14}, dots, E_{64}, E_{15}, dots, E_{65})$ is a basis of $X$. Proving that the dimension of that linear subspace is equal to $12$.
answered Dec 31 '18 at 17:28
mathcounterexamples.netmathcounterexamples.net
25.3k21953
$endgroup$
$mathbb R^{6,5}$ is a linear space of dimension $6 times 5=30$.
If $(v_1, v_2, v_3)$ (with $v_i in mathbb R^5$) is a basis of $V$. You have
$$X = {A in mathbb R^{6,5} ; A.v_1=A.v_2=A.v_3=0}.$$
It means that you're imposing $3 times 6 = 18$ independent conditions in the dual space $left(mathbb R^{6,5}right)^vee$. Hence $dim X = 30 -18=12$.
Another proof not using the dual space
Without loss of generality, you can suppose that $(v_1, dots, v_5)$ is a basis of $mathcal V = mathbb R^5$ where $(v_1, v_2,v_3)$ is a basis of $V$. And that $(v_1^prime, dots, v_6^prime)$ is a basis of $mathcal V^prime=mathbb R^6$.
For $A =(a_{ij}) in X$ expressed in those basis, you must have $$a_{11} =dots =a_{61}=a_{12}=dots=a_{62}=a_{13}=dots=a_{63}=0$$ as $V subseteq ker A$.
If we denote $E_{ij}$ the matrix with coefficient $i,j$ equal to $1$ and the others vanishing, $(E_{14}, dots, E_{64}, E_{15}, dots, E_{65})$ is a basis of $X$. Proving that the dimension of that linear subspace is equal to $12$.
answered Dec 31 '18 at 17:28
mathcounterexamples.netmathcounterexamples.net
25.3k21953
edited Jan 3 at 5:51
answered Dec 31 '18 at 17:28
mathcounterexamples.netmathcounterexamples.net
25.3k21953
answered Dec 31 '18 at 17:28
mathcounterexamples.netmathcounterexamples.net
25.3k21953
answered Dec 31 '18 at 17:28
mathcounterexamples.netmathcounterexamples.net
25.3k21953
25.3k21953
$begingroup$
I don't understand this notation: $A.v_1=A.v_2=A.v_3=0$. Does $A.v_{i} $ means $i$ column? I suspect that these are not the same vectors as the vectors from base $V$?
$endgroup$
– VirtualUser
Dec 31 '18 at 17:47
$begingroup$
For a vector $v in V$ (with 5 coordinates) $A.v$ means the image by $A$ of vector $v$. $A.v$ is a vector with 6 coordinates.
$endgroup$
– mathcounterexamples.net
Dec 31 '18 at 17:49
$begingroup$
Can it be done without dual space? I haven't got this on lecture and probably I can't use that
$endgroup$
– VirtualUser
Jan 2 at 19:12
$begingroup$
What are the key theorems of your lecture?
$endgroup$
– mathcounterexamples.net
Jan 2 at 20:35
$begingroup$
image, kernel, rank, basis of linear space, sum of linear spaces linear independence and linear combinations
$endgroup$
– VirtualUser
Jan 2 at 20:39
|
show 4 more comments
$begingroup$
I don't understand this notation: $A.v_1=A.v_2=A.v_3=0$. Does $A.v_{i} $ means $i$ column? I suspect that these are not the same vectors as the vectors from base $V$?
$endgroup$
– VirtualUser
Dec 31 '18 at 17:47
$begingroup$
For a vector $v in V$ (with 5 coordinates) $A.v$ means the image by $A$ of vector $v$. $A.v$ is a vector with 6 coordinates.
$endgroup$
– mathcounterexamples.net
Dec 31 '18 at 17:49
$begingroup$
Can it be done without dual space? I haven't got this on lecture and probably I can't use that
$endgroup$
– VirtualUser
Jan 2 at 19:12
$begingroup$
What are the key theorems of your lecture?
$endgroup$
– mathcounterexamples.net
Jan 2 at 20:35
$begingroup$
image, kernel, rank, basis of linear space, sum of linear spaces linear independence and linear combinations
$endgroup$
– VirtualUser
Jan 2 at 20:39
$begingroup$
I don't understand this notation: $A.v_1=A.v_2=A.v_3=0$. Does $A.v_{i} $ means $i$ column? I suspect that these are not the same vectors as the vectors from base $V$?
$endgroup$
– VirtualUser
Dec 31 '18 at 17:47
$begingroup$
I don't understand this notation: $A.v_1=A.v_2=A.v_3=0$. Does $A.v_{i} $ means $i$ column? I suspect that these are not the same vectors as the vectors from base $V$?
$endgroup$
– VirtualUser
Dec 31 '18 at 17:47
$begingroup$
For a vector $v in V$ (with 5 coordinates) $A.v$ means the image by $A$ of vector $v$. $A.v$ is a vector with 6 coordinates.
$endgroup$
– mathcounterexamples.net
Dec 31 '18 at 17:49
$begingroup$
For a vector $v in V$ (with 5 coordinates) $A.v$ means the image by $A$ of vector $v$. $A.v$ is a vector with 6 coordinates.
$endgroup$
– mathcounterexamples.net
Dec 31 '18 at 17:49
$begingroup$
Can it be done without dual space? I haven't got this on lecture and probably I can't use that
$endgroup$
– VirtualUser
Jan 2 at 19:12
$begingroup$
Can it be done without dual space? I haven't got this on lecture and probably I can't use that
$endgroup$
– VirtualUser
Jan 2 at 19:12
$begingroup$
What are the key theorems of your lecture?
$endgroup$
– mathcounterexamples.net
Jan 2 at 20:35
$begingroup$
What are the key theorems of your lecture?
$endgroup$
– mathcounterexamples.net
Jan 2 at 20:35
$begingroup$
image, kernel, rank, basis of linear space, sum of linear spaces linear independence and linear combinations
$endgroup$
– VirtualUser
Jan 2 at 20:39
$begingroup$
image, kernel, rank, basis of linear space, sum of linear spaces linear independence and linear combinations
$endgroup$
– VirtualUser
Jan 2 at 20:39
|
show 4 more comments
$begingroup$
Since $Vsubsetker(A)forall Ain X$, the rows of $A$ are orthogonal to all vectors in $V$ and thus belong to $V^perp$, which has dimension $2$. Each row vector of $A$ is the linear combination of the two basis vectors $v_1,v_2$ of $V^perp$.
$$A=begin{bmatrix}a_{11}v_1^T+a_{12}v_2^T\a_{21}v_1^T+a_{22}v_2^T\vdots\a_{61}v_1^T+a_{62}v_2^Tend{bmatrix}=a_{11}begin{bmatrix}v_1^T\0\vdots\0end{bmatrix}+a_{12}begin{bmatrix}v_2^T\0\vdots\0end{bmatrix}+a_{21}begin{bmatrix}0\v_1^T\vdots\0end{bmatrix}+...+a_{62}begin{bmatrix}0\0\vdots\v_2^Tend{bmatrix}$$
Thus, $X$ has $12$ basis vectors.
answered Jan 1 at 12:40
Shubham JohriShubham Johri
4,666717
$endgroup$
add a comment |
$begingroup$
Since $Vsubsetker(A)forall Ain X$, the rows of $A$ are orthogonal to all vectors in $V$ and thus belong to $V^perp$, which has dimension $2$. Each row vector of $A$ is the linear combination of the two basis vectors $v_1,v_2$ of $V^perp$.
$$A=begin{bmatrix}a_{11}v_1^T+a_{12}v_2^T\a_{21}v_1^T+a_{22}v_2^T\vdots\a_{61}v_1^T+a_{62}v_2^Tend{bmatrix}=a_{11}begin{bmatrix}v_1^T\0\vdots\0end{bmatrix}+a_{12}begin{bmatrix}v_2^T\0\vdots\0end{bmatrix}+a_{21}begin{bmatrix}0\v_1^T\vdots\0end{bmatrix}+...+a_{62}begin{bmatrix}0\0\vdots\v_2^Tend{bmatrix}$$
Thus, $X$ has $12$ basis vectors.
answered Jan 1 at 12:40
Shubham JohriShubham Johri
4,666717
$endgroup$
add a comment |
$begingroup$
Since $Vsubsetker(A)forall Ain X$, the rows of $A$ are orthogonal to all vectors in $V$ and thus belong to $V^perp$, which has dimension $2$. Each row vector of $A$ is the linear combination of the two basis vectors $v_1,v_2$ of $V^perp$.
$$A=begin{bmatrix}a_{11}v_1^T+a_{12}v_2^T\a_{21}v_1^T+a_{22}v_2^T\vdots\a_{61}v_1^T+a_{62}v_2^Tend{bmatrix}=a_{11}begin{bmatrix}v_1^T\0\vdots\0end{bmatrix}+a_{12}begin{bmatrix}v_2^T\0\vdots\0end{bmatrix}+a_{21}begin{bmatrix}0\v_1^T\vdots\0end{bmatrix}+...+a_{62}begin{bmatrix}0\0\vdots\v_2^Tend{bmatrix}$$
Thus, $X$ has $12$ basis vectors.
answered Jan 1 at 12:40
Shubham JohriShubham Johri
4,666717
$endgroup$
Since $Vsubsetker(A)forall Ain X$, the rows of $A$ are orthogonal to all vectors in $V$ and thus belong to $V^perp$, which has dimension $2$. Each row vector of $A$ is the linear combination of the two basis vectors $v_1,v_2$ of $V^perp$.
$$A=begin{bmatrix}a_{11}v_1^T+a_{12}v_2^T\a_{21}v_1^T+a_{22}v_2^T\vdots\a_{61}v_1^T+a_{62}v_2^Tend{bmatrix}=a_{11}begin{bmatrix}v_1^T\0\vdots\0end{bmatrix}+a_{12}begin{bmatrix}v_2^T\0\vdots\0end{bmatrix}+a_{21}begin{bmatrix}0\v_1^T\vdots\0end{bmatrix}+...+a_{62}begin{bmatrix}0\0\vdots\v_2^Tend{bmatrix}$$
Thus, $X$ has $12$ basis vectors.
answered Jan 1 at 12:40
Shubham JohriShubham Johri
4,666717
answered Jan 1 at 12:40
Shubham JohriShubham Johri
4,666717
answered Jan 1 at 12:40
Shubham JohriShubham Johri
4,666717
answered Jan 1 at 12:40
Shubham JohriShubham Johri
4,666717
4,666717
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$begingroup$
What do you denote $Bbb R^{6,5}$?
$endgroup$
– Bernard
Dec 31 '18 at 17:23
$begingroup$
Sorry, I used to write
RRbut there it does not work I didn't know how to write it. But now I will remember, thanks$endgroup$
– VirtualUser
Dec 31 '18 at 17:33