Find condition that some vector is base of fundamental subspace
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If you have some vector $r_1,r_2,n_1,n_2,c_1,c_2,l_1,l_2 in R^4$
a) Under what condition it will be $r_1,r_2,n_1,n_2,c_1,c_2,l_1,l_2$ base of fundamental subspace $R(A^T),N(A),R(A),N(A^T)$ in that order, of some matrices $Ain M_4(mathbb R)$?
b) Under condition that you get for a) give an example of such a matrix.
$mathbb R^4=N(A)⊕R(A)^T$ and $mathbb R^4=R(A)⊕N(A)^T$ so $N(A)cap R(A)^T=0$ if we take some vector $xin N(A)cap R(A)^T$ since $xin N(A)$ and $xin R(A)^T$ then $x=alpha_1 r_1+alpha_2 n2, x=beta1 n_1+beta2 n_2$ $alpha_1 r_1+alpha_2 n2=beta1 n_1+beta2 n_2$ since $alpha_1 n_1-beta_1n_1-beta_2 n_2+alpha_2 n_2=0$ they must be linear independent and they must be orthogonal, that is the same for $c_1,c_2,l_1,l_2$,
I try to find matrix but I have no idea,do you have some idea?
linear-algebra matrices linear-transformations
asked Sep 6 at 9:03
Marko Škorić
4008
add a comment |Â
up vote
1
down vote
favorite
If you have some vector $r_1,r_2,n_1,n_2,c_1,c_2,l_1,l_2 in R^4$
a) Under what condition it will be $r_1,r_2,n_1,n_2,c_1,c_2,l_1,l_2$ base of fundamental subspace $R(A^T),N(A),R(A),N(A^T)$ in that order, of some matrices $Ain M_4(mathbb R)$?
b) Under condition that you get for a) give an example of such a matrix.
$mathbb R^4=N(A)⊕R(A)^T$ and $mathbb R^4=R(A)⊕N(A)^T$ so $N(A)cap R(A)^T=0$ if we take some vector $xin N(A)cap R(A)^T$ since $xin N(A)$ and $xin R(A)^T$ then $x=alpha_1 r_1+alpha_2 n2, x=beta1 n_1+beta2 n_2$ $alpha_1 r_1+alpha_2 n2=beta1 n_1+beta2 n_2$ since $alpha_1 n_1-beta_1n_1-beta_2 n_2+alpha_2 n_2=0$ they must be linear independent and they must be orthogonal, that is the same for $c_1,c_2,l_1,l_2$,
I try to find matrix but I have no idea,do you have some idea?
linear-algebra matrices linear-transformations
asked Sep 6 at 9:03
Marko Škorić
4008
There is no reason to assume that $dim (R(A))=2$ for an arbitrary matrix $Ain mathbb R^4times 4$ (I assume that $R(A)$ denotes the range or image of $A$). In b), the matrix $A=beginpmatrix1&0&0&0\0&1&0&0\0&0&1&0\0&0&0&1endpmatrix$ would be an example, since $A=A^T$ and $N(A) = 0$ and $R(A)=mathbb R^4$. Therefore all sums are direct.
– Babelfish
Sep 6 at 14:46
add a comment |Â
up vote
1
down vote
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up vote
1
down vote
favorite
If you have some vector $r_1,r_2,n_1,n_2,c_1,c_2,l_1,l_2 in R^4$
a) Under what condition it will be $r_1,r_2,n_1,n_2,c_1,c_2,l_1,l_2$ base of fundamental subspace $R(A^T),N(A),R(A),N(A^T)$ in that order, of some matrices $Ain M_4(mathbb R)$?
b) Under condition that you get for a) give an example of such a matrix.
$mathbb R^4=N(A)⊕R(A)^T$ and $mathbb R^4=R(A)⊕N(A)^T$ so $N(A)cap R(A)^T=0$ if we take some vector $xin N(A)cap R(A)^T$ since $xin N(A)$ and $xin R(A)^T$ then $x=alpha_1 r_1+alpha_2 n2, x=beta1 n_1+beta2 n_2$ $alpha_1 r_1+alpha_2 n2=beta1 n_1+beta2 n_2$ since $alpha_1 n_1-beta_1n_1-beta_2 n_2+alpha_2 n_2=0$ they must be linear independent and they must be orthogonal, that is the same for $c_1,c_2,l_1,l_2$,
I try to find matrix but I have no idea,do you have some idea?
linear-algebra matrices linear-transformations
asked Sep 6 at 9:03
Marko Škorić
4008
If you have some vector $r_1,r_2,n_1,n_2,c_1,c_2,l_1,l_2 in R^4$
a) Under what condition it will be $r_1,r_2,n_1,n_2,c_1,c_2,l_1,l_2$ base of fundamental subspace $R(A^T),N(A),R(A),N(A^T)$ in that order, of some matrices $Ain M_4(mathbb R)$?
b) Under condition that you get for a) give an example of such a matrix.
$mathbb R^4=N(A)⊕R(A)^T$ and $mathbb R^4=R(A)⊕N(A)^T$ so $N(A)cap R(A)^T=0$ if we take some vector $xin N(A)cap R(A)^T$ since $xin N(A)$ and $xin R(A)^T$ then $x=alpha_1 r_1+alpha_2 n2, x=beta1 n_1+beta2 n_2$ $alpha_1 r_1+alpha_2 n2=beta1 n_1+beta2 n_2$ since $alpha_1 n_1-beta_1n_1-beta_2 n_2+alpha_2 n_2=0$ they must be linear independent and they must be orthogonal, that is the same for $c_1,c_2,l_1,l_2$,
I try to find matrix but I have no idea,do you have some idea?
linear-algebra matrices linear-transformations
linear-algebra matrices linear-transformations
asked Sep 6 at 9:03
Marko Škorić
4008
asked Sep 6 at 9:03
Marko Škorić
4008
asked Sep 6 at 9:03
Marko Škorić
4008
asked Sep 6 at 9:03
Marko Škorić
4008
asked Sep 6 at 9:03
Marko Škorić
4008
4008
There is no reason to assume that $dim (R(A))=2$ for an arbitrary matrix $Ain mathbb R^4times 4$ (I assume that $R(A)$ denotes the range or image of $A$). In b), the matrix $A=beginpmatrix1&0&0&0\0&1&0&0\0&0&1&0\0&0&0&1endpmatrix$ would be an example, since $A=A^T$ and $N(A) = 0$ and $R(A)=mathbb R^4$. Therefore all sums are direct.
– Babelfish
Sep 6 at 14:46
add a comment |Â
There is no reason to assume that $dim (R(A))=2$ for an arbitrary matrix $Ain mathbb R^4times 4$ (I assume that $R(A)$ denotes the range or image of $A$). In b), the matrix $A=beginpmatrix1&0&0&0\0&1&0&0\0&0&1&0\0&0&0&1endpmatrix$ would be an example, since $A=A^T$ and $N(A) = 0$ and $R(A)=mathbb R^4$. Therefore all sums are direct.
– Babelfish
Sep 6 at 14:46
There is no reason to assume that $dim (R(A))=2$ for an arbitrary matrix $Ain mathbb R^4times 4$ (I assume that $R(A)$ denotes the range or image of $A$). In b), the matrix $A=beginpmatrix1&0&0&0\0&1&0&0\0&0&1&0\0&0&0&1endpmatrix$ would be an example, since $A=A^T$ and $N(A) = 0$ and $R(A)=mathbb R^4$. Therefore all sums are direct.
– Babelfish
Sep 6 at 14:46
There is no reason to assume that $dim (R(A))=2$ for an arbitrary matrix $Ain mathbb R^4times 4$ (I assume that $R(A)$ denotes the range or image of $A$). In b), the matrix $A=beginpmatrix1&0&0&0\0&1&0&0\0&0&1&0\0&0&0&1endpmatrix$ would be an example, since $A=A^T$ and $N(A) = 0$ and $R(A)=mathbb R^4$. Therefore all sums are direct.
– Babelfish
Sep 6 at 14:46
add a comment |Â
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There is no reason to assume that $dim (R(A))=2$ for an arbitrary matrix $Ain mathbb R^4times 4$ (I assume that $R(A)$ denotes the range or image of $A$). In b), the matrix $A=beginpmatrix1&0&0&0\0&1&0&0\0&0&1&0\0&0&0&1endpmatrix$ would be an example, since $A=A^T$ and $N(A) = 0$ and $R(A)=mathbb R^4$. Therefore all sums are direct.
– Babelfish
Sep 6 at 14:46