direct sum of null space and range space The 2019 Stack Overflow Developer Survey Results Are InVisulizing column/row space and null/left null space, A and xleft shift operator null-spaceFinding a linear transformation with a given null spaceUnderstanding the definition of the direct sum of subspaces of a vector spaceProve or give a counterexample: $V=textnullT+textrangeT$Finding basis for Null Space of matrixDirect Sum: Span of Basis VectorsWhat is the difference between orthogonal subspaces and orthogonal complements?Prove that $V$ is a direct sum of subspacesFinding matrices which P is null space
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direct sum of null space and range space
The 2019 Stack Overflow Developer Survey Results Are InVisulizing column/row space and null/left null space, A and xleft shift operator null-spaceFinding a linear transformation with a given null spaceUnderstanding the definition of the direct sum of subspaces of a vector spaceProve or give a counterexample: $V=textnullT+textrangeT$Finding basis for Null Space of matrixDirect Sum: Span of Basis VectorsWhat is the difference between orthogonal subspaces and orthogonal complements?Prove that $V$ is a direct sum of subspacesFinding matrices which P is null space
$begingroup$
Suppose A $in$ L ($mathcalR^4$) such that A(x) is given by the following vector in $mathcalR^4$ beginbmatrix
x_2 + x_3 \
x_3 \
0\
0
endbmatrix
Can I get $mathcalR^4$ = N$_A$ $bigoplus$ R$_A$ ? If not, any suggestion for choosing subspaces X and Y of $mathcalR^4$ to satisfy this direct sum formula ?
What I get is: N$_A$ is the set of vectors such as [x$_1$,0,0,x$_4$] in which x$_1$,x$_4$ $in$ $mathcalR$; and R$_A$ is the set of vectors such as [x$_2$+x$_3$, x$_3$, 0, 0] in which x$_2$,x$_3$ $in$ $mathcalR$. But I find it is a bit impossible to write $mathcalR^4$ as a direct sum of null space and range space.
linear-algebra
asked Mar 23 at 1:47
EricEric
577
$endgroup$
|
show 2 more comments
$begingroup$
Suppose A $in$ L ($mathcalR^4$) such that A(x) is given by the following vector in $mathcalR^4$ beginbmatrix
x_2 + x_3 \
x_3 \
0\
0
endbmatrix
Can I get $mathcalR^4$ = N$_A$ $bigoplus$ R$_A$ ? If not, any suggestion for choosing subspaces X and Y of $mathcalR^4$ to satisfy this direct sum formula ?
What I get is: N$_A$ is the set of vectors such as [x$_1$,0,0,x$_4$] in which x$_1$,x$_4$ $in$ $mathcalR$; and R$_A$ is the set of vectors such as [x$_2$+x$_3$, x$_3$, 0, 0] in which x$_2$,x$_3$ $in$ $mathcalR$. But I find it is a bit impossible to write $mathcalR^4$ as a direct sum of null space and range space.
linear-algebra
asked Mar 23 at 1:47
EricEric
577
$endgroup$
1
$begingroup$
Check that $(1,0,0,0)^t$ lies in both the null space and the range so you cannot get $mathbbR^4 = mathbfN_Aoplus mathbfR_A$. Asto “choosing subspaces to satisfy the direct sum formula”, there’s lots and lots; in fact, given any subspace for $X$ you can find infinitely many choices for $Y$. You presumably, though, want subspaces satisfying some conditions.
$endgroup$
– Arturo Magidin
Mar 23 at 1:53
$begingroup$
That's because $mathbb R^4$ is not the internal direct sum of those two spaces, because in fact those two spaces have a non-trivial intersection (which is $operatornamespan(beginbmatrix1 & 0 & 0 & 0endbmatrix^T)$. But $mathbb R^4$ is isomorphic to the (external) direct sum of the two spaces, and this is obvious from their dimensions.
$endgroup$
– M. Vinay
Mar 23 at 1:53
$begingroup$
Other conditions are (i) two subspaces are both invariant subspaces (ii) dimension are both two.
$endgroup$
– Eric
Mar 23 at 1:55
$begingroup$
First, you should put all the information in the question. Were you hoping we would guess these conditions? Sorry; the government really doesn’t like it when I read minds without a warrant. Second, “invariant” doesn’t mean anything by itself. It should be invariant relative to something... (presumably... the action of $A$?)
$endgroup$
– Arturo Magidin
Mar 23 at 1:59
$begingroup$
Given A $in$ L(X), Y $subset$ X subspace of X s.t. A(Y) $subset$ Y, then we say Y is an invariant subspace of X.
$endgroup$
– Eric
Mar 23 at 3:49
|
show 2 more comments
$begingroup$
Suppose A $in$ L ($mathcalR^4$) such that A(x) is given by the following vector in $mathcalR^4$ beginbmatrix
x_2 + x_3 \
x_3 \
0\
0
endbmatrix
Can I get $mathcalR^4$ = N$_A$ $bigoplus$ R$_A$ ? If not, any suggestion for choosing subspaces X and Y of $mathcalR^4$ to satisfy this direct sum formula ?
What I get is: N$_A$ is the set of vectors such as [x$_1$,0,0,x$_4$] in which x$_1$,x$_4$ $in$ $mathcalR$; and R$_A$ is the set of vectors such as [x$_2$+x$_3$, x$_3$, 0, 0] in which x$_2$,x$_3$ $in$ $mathcalR$. But I find it is a bit impossible to write $mathcalR^4$ as a direct sum of null space and range space.
linear-algebra
asked Mar 23 at 1:47
EricEric
577
$endgroup$
Suppose A $in$ L ($mathcalR^4$) such that A(x) is given by the following vector in $mathcalR^4$ beginbmatrix
x_2 + x_3 \
x_3 \
0\
0
endbmatrix
Can I get $mathcalR^4$ = N$_A$ $bigoplus$ R$_A$ ? If not, any suggestion for choosing subspaces X and Y of $mathcalR^4$ to satisfy this direct sum formula ?
What I get is: N$_A$ is the set of vectors such as [x$_1$,0,0,x$_4$] in which x$_1$,x$_4$ $in$ $mathcalR$; and R$_A$ is the set of vectors such as [x$_2$+x$_3$, x$_3$, 0, 0] in which x$_2$,x$_3$ $in$ $mathcalR$. But I find it is a bit impossible to write $mathcalR^4$ as a direct sum of null space and range space.
linear-algebra
linear-algebra
asked Mar 23 at 1:47
EricEric
577
asked Mar 23 at 1:47
EricEric
577
asked Mar 23 at 1:47
EricEric
577
asked Mar 23 at 1:47
EricEric
577
asked Mar 23 at 1:47
EricEric
577
577
1
$begingroup$
Check that $(1,0,0,0)^t$ lies in both the null space and the range so you cannot get $mathbbR^4 = mathbfN_Aoplus mathbfR_A$. Asto “choosing subspaces to satisfy the direct sum formula”, there’s lots and lots; in fact, given any subspace for $X$ you can find infinitely many choices for $Y$. You presumably, though, want subspaces satisfying some conditions.
$endgroup$
– Arturo Magidin
Mar 23 at 1:53
$begingroup$
That's because $mathbb R^4$ is not the internal direct sum of those two spaces, because in fact those two spaces have a non-trivial intersection (which is $operatornamespan(beginbmatrix1 & 0 & 0 & 0endbmatrix^T)$. But $mathbb R^4$ is isomorphic to the (external) direct sum of the two spaces, and this is obvious from their dimensions.
$endgroup$
– M. Vinay
Mar 23 at 1:53
$begingroup$
Other conditions are (i) two subspaces are both invariant subspaces (ii) dimension are both two.
$endgroup$
– Eric
Mar 23 at 1:55
$begingroup$
First, you should put all the information in the question. Were you hoping we would guess these conditions? Sorry; the government really doesn’t like it when I read minds without a warrant. Second, “invariant” doesn’t mean anything by itself. It should be invariant relative to something... (presumably... the action of $A$?)
$endgroup$
– Arturo Magidin
Mar 23 at 1:59
$begingroup$
Given A $in$ L(X), Y $subset$ X subspace of X s.t. A(Y) $subset$ Y, then we say Y is an invariant subspace of X.
$endgroup$
– Eric
Mar 23 at 3:49
|
show 2 more comments
1
$begingroup$
Check that $(1,0,0,0)^t$ lies in both the null space and the range so you cannot get $mathbbR^4 = mathbfN_Aoplus mathbfR_A$. Asto “choosing subspaces to satisfy the direct sum formula”, there’s lots and lots; in fact, given any subspace for $X$ you can find infinitely many choices for $Y$. You presumably, though, want subspaces satisfying some conditions.
$endgroup$
– Arturo Magidin
Mar 23 at 1:53
$begingroup$
That's because $mathbb R^4$ is not the internal direct sum of those two spaces, because in fact those two spaces have a non-trivial intersection (which is $operatornamespan(beginbmatrix1 & 0 & 0 & 0endbmatrix^T)$. But $mathbb R^4$ is isomorphic to the (external) direct sum of the two spaces, and this is obvious from their dimensions.
$endgroup$
– M. Vinay
Mar 23 at 1:53
$begingroup$
Other conditions are (i) two subspaces are both invariant subspaces (ii) dimension are both two.
$endgroup$
– Eric
Mar 23 at 1:55
$begingroup$
First, you should put all the information in the question. Were you hoping we would guess these conditions? Sorry; the government really doesn’t like it when I read minds without a warrant. Second, “invariant” doesn’t mean anything by itself. It should be invariant relative to something... (presumably... the action of $A$?)
$endgroup$
– Arturo Magidin
Mar 23 at 1:59
$begingroup$
Given A $in$ L(X), Y $subset$ X subspace of X s.t. A(Y) $subset$ Y, then we say Y is an invariant subspace of X.
$endgroup$
– Eric
Mar 23 at 3:49
1
1
$begingroup$
Check that $(1,0,0,0)^t$ lies in both the null space and the range so you cannot get $mathbbR^4 = mathbfN_Aoplus mathbfR_A$. Asto “choosing subspaces to satisfy the direct sum formula”, there’s lots and lots; in fact, given any subspace for $X$ you can find infinitely many choices for $Y$. You presumably, though, want subspaces satisfying some conditions.
$endgroup$
– Arturo Magidin
Mar 23 at 1:53
$begingroup$
Check that $(1,0,0,0)^t$ lies in both the null space and the range so you cannot get $mathbbR^4 = mathbfN_Aoplus mathbfR_A$. Asto “choosing subspaces to satisfy the direct sum formula”, there’s lots and lots; in fact, given any subspace for $X$ you can find infinitely many choices for $Y$. You presumably, though, want subspaces satisfying some conditions.
$endgroup$
– Arturo Magidin
Mar 23 at 1:53
$begingroup$
That's because $mathbb R^4$ is not the internal direct sum of those two spaces, because in fact those two spaces have a non-trivial intersection (which is $operatornamespan(beginbmatrix1 & 0 & 0 & 0endbmatrix^T)$. But $mathbb R^4$ is isomorphic to the (external) direct sum of the two spaces, and this is obvious from their dimensions.
$endgroup$
– M. Vinay
Mar 23 at 1:53
$begingroup$
That's because $mathbb R^4$ is not the internal direct sum of those two spaces, because in fact those two spaces have a non-trivial intersection (which is $operatornamespan(beginbmatrix1 & 0 & 0 & 0endbmatrix^T)$. But $mathbb R^4$ is isomorphic to the (external) direct sum of the two spaces, and this is obvious from their dimensions.
$endgroup$
– M. Vinay
Mar 23 at 1:53
$begingroup$
Other conditions are (i) two subspaces are both invariant subspaces (ii) dimension are both two.
$endgroup$
– Eric
Mar 23 at 1:55
$begingroup$
Other conditions are (i) two subspaces are both invariant subspaces (ii) dimension are both two.
$endgroup$
– Eric
Mar 23 at 1:55
$begingroup$
First, you should put all the information in the question. Were you hoping we would guess these conditions? Sorry; the government really doesn’t like it when I read minds without a warrant. Second, “invariant” doesn’t mean anything by itself. It should be invariant relative to something... (presumably... the action of $A$?)
$endgroup$
– Arturo Magidin
Mar 23 at 1:59
$begingroup$
First, you should put all the information in the question. Were you hoping we would guess these conditions? Sorry; the government really doesn’t like it when I read minds without a warrant. Second, “invariant” doesn’t mean anything by itself. It should be invariant relative to something... (presumably... the action of $A$?)
$endgroup$
– Arturo Magidin
Mar 23 at 1:59
$begingroup$
Given A $in$ L(X), Y $subset$ X subspace of X s.t. A(Y) $subset$ Y, then we say Y is an invariant subspace of X.
$endgroup$
– Eric
Mar 23 at 3:49
$begingroup$
Given A $in$ L(X), Y $subset$ X subspace of X s.t. A(Y) $subset$ Y, then we say Y is an invariant subspace of X.
$endgroup$
– Eric
Mar 23 at 3:49
|
show 2 more comments
0
active
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$begingroup$
Check that $(1,0,0,0)^t$ lies in both the null space and the range so you cannot get $mathbbR^4 = mathbfN_Aoplus mathbfR_A$. Asto “choosing subspaces to satisfy the direct sum formula”, there’s lots and lots; in fact, given any subspace for $X$ you can find infinitely many choices for $Y$. You presumably, though, want subspaces satisfying some conditions.
$endgroup$
– Arturo Magidin
Mar 23 at 1:53
$begingroup$
That's because $mathbb R^4$ is not the internal direct sum of those two spaces, because in fact those two spaces have a non-trivial intersection (which is $operatornamespan(beginbmatrix1 & 0 & 0 & 0endbmatrix^T)$. But $mathbb R^4$ is isomorphic to the (external) direct sum of the two spaces, and this is obvious from their dimensions.
$endgroup$
– M. Vinay
Mar 23 at 1:53
$begingroup$
Other conditions are (i) two subspaces are both invariant subspaces (ii) dimension are both two.
$endgroup$
– Eric
Mar 23 at 1:55
$begingroup$
First, you should put all the information in the question. Were you hoping we would guess these conditions? Sorry; the government really doesn’t like it when I read minds without a warrant. Second, “invariant” doesn’t mean anything by itself. It should be invariant relative to something... (presumably... the action of $A$?)
$endgroup$
– Arturo Magidin
Mar 23 at 1:59
$begingroup$
Given A $in$ L(X), Y $subset$ X subspace of X s.t. A(Y) $subset$ Y, then we say Y is an invariant subspace of X.
$endgroup$
– Eric
Mar 23 at 3:49