Is the following statement is True false regarding inner product
$begingroup$
Is the following statement is True false
Let $V = mathbb{R}^5$ be equipped with the usual euclidean inner-product.
If $W$ and $Z$ are subspaces of $V$ such that both of them are of dimension
$3,$ then there exists $z in Z$ such that $z neq 0$ and $z ⊥ W.$
My attempt : i thinks this statement is True take $W= (1,-1,1)$ and $Z=(-1,1,-1)$ the $Z ⊥ W$ that is $langle Z. Wrangle=0$
Is my thinking is correct or not ?
Any hints/solution will be appreciated
linear-algebra
asked Jan 9 at 10:38
jasminejasmine
1,796418
$endgroup$
add a comment |
$begingroup$
Is the following statement is True false
Let $V = mathbb{R}^5$ be equipped with the usual euclidean inner-product.
If $W$ and $Z$ are subspaces of $V$ such that both of them are of dimension
$3,$ then there exists $z in Z$ such that $z neq 0$ and $z ⊥ W.$
My attempt : i thinks this statement is True take $W= (1,-1,1)$ and $Z=(-1,1,-1)$ the $Z ⊥ W$ that is $langle Z. Wrangle=0$
Is my thinking is correct or not ?
Any hints/solution will be appreciated
linear-algebra
asked Jan 9 at 10:38
jasminejasmine
1,796418
$endgroup$
3
$begingroup$
What if W=Z i.e. W and Z are the same subspace ? The conditions of the question don't seem to rule this out.
$endgroup$
– gandalf61
Jan 9 at 10:46
2
$begingroup$
The question is not clear. Your $W$ and $Z$ seem vectors of $mathbb{R}^3$, so their are not subspaces of $mathbb{R}^5$ nor elements of this vector space.
$endgroup$
– Emilio Novati
Jan 9 at 10:48
1
$begingroup$
exactly! by the way, I tried to incorporate your comments in my answer, i hope that is ok
$endgroup$
– Enkidu
Jan 9 at 10:49
add a comment |
$begingroup$
Is the following statement is True false
Let $V = mathbb{R}^5$ be equipped with the usual euclidean inner-product.
If $W$ and $Z$ are subspaces of $V$ such that both of them are of dimension
$3,$ then there exists $z in Z$ such that $z neq 0$ and $z ⊥ W.$
My attempt : i thinks this statement is True take $W= (1,-1,1)$ and $Z=(-1,1,-1)$ the $Z ⊥ W$ that is $langle Z. Wrangle=0$
Is my thinking is correct or not ?
Any hints/solution will be appreciated
linear-algebra
asked Jan 9 at 10:38
jasminejasmine
1,796418
$endgroup$
Is the following statement is True false
Let $V = mathbb{R}^5$ be equipped with the usual euclidean inner-product.
If $W$ and $Z$ are subspaces of $V$ such that both of them are of dimension
$3,$ then there exists $z in Z$ such that $z neq 0$ and $z ⊥ W.$
My attempt : i thinks this statement is True take $W= (1,-1,1)$ and $Z=(-1,1,-1)$ the $Z ⊥ W$ that is $langle Z. Wrangle=0$
Is my thinking is correct or not ?
Any hints/solution will be appreciated
linear-algebra
linear-algebra
asked Jan 9 at 10:38
jasminejasmine
1,796418
asked Jan 9 at 10:38
jasminejasmine
1,796418
asked Jan 9 at 10:38
jasminejasmine
1,796418
asked Jan 9 at 10:38
jasminejasmine
1,796418
asked Jan 9 at 10:38
jasminejasmine
1,796418
1,796418
3
$begingroup$
What if W=Z i.e. W and Z are the same subspace ? The conditions of the question don't seem to rule this out.
$endgroup$
– gandalf61
Jan 9 at 10:46
2
$begingroup$
The question is not clear. Your $W$ and $Z$ seem vectors of $mathbb{R}^3$, so their are not subspaces of $mathbb{R}^5$ nor elements of this vector space.
$endgroup$
– Emilio Novati
Jan 9 at 10:48
1
$begingroup$
exactly! by the way, I tried to incorporate your comments in my answer, i hope that is ok
$endgroup$
– Enkidu
Jan 9 at 10:49
add a comment |
3
$begingroup$
What if W=Z i.e. W and Z are the same subspace ? The conditions of the question don't seem to rule this out.
$endgroup$
– gandalf61
Jan 9 at 10:46
2
$begingroup$
The question is not clear. Your $W$ and $Z$ seem vectors of $mathbb{R}^3$, so their are not subspaces of $mathbb{R}^5$ nor elements of this vector space.
$endgroup$
– Emilio Novati
Jan 9 at 10:48
1
$begingroup$
exactly! by the way, I tried to incorporate your comments in my answer, i hope that is ok
$endgroup$
– Enkidu
Jan 9 at 10:49
3
3
$begingroup$
What if W=Z i.e. W and Z are the same subspace ? The conditions of the question don't seem to rule this out.
$endgroup$
– gandalf61
Jan 9 at 10:46
$begingroup$
What if W=Z i.e. W and Z are the same subspace ? The conditions of the question don't seem to rule this out.
$endgroup$
– gandalf61
Jan 9 at 10:46
2
2
$begingroup$
The question is not clear. Your $W$ and $Z$ seem vectors of $mathbb{R}^3$, so their are not subspaces of $mathbb{R}^5$ nor elements of this vector space.
$endgroup$
– Emilio Novati
Jan 9 at 10:48
$begingroup$
The question is not clear. Your $W$ and $Z$ seem vectors of $mathbb{R}^3$, so their are not subspaces of $mathbb{R}^5$ nor elements of this vector space.
$endgroup$
– Emilio Novati
Jan 9 at 10:48
1
1
$begingroup$
exactly! by the way, I tried to incorporate your comments in my answer, i hope that is ok
$endgroup$
– Enkidu
Jan 9 at 10:49
$begingroup$
exactly! by the way, I tried to incorporate your comments in my answer, i hope that is ok
$endgroup$
– Enkidu
Jan 9 at 10:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, it is false, it needs further assumptions, also as stated in the comments, be aware that we are talking subspaces here, not vectors! for example (as gandalf61 said) take $Z=Wsubset V$ then we have, since $langle_,_ rangle$ is non degenerate, for all $0 neq z in Z$ a $w in W=Z$ such that $langle z , w rangle neq 0$ in particular (to give a very concrete example) you could pick $w=z$ and then we already know that $$langle z, wrangle = langle z, zrangle = 0 iff z=0$$
Also, be aware that only the concrete counterexample with $z=w$ needs an inner product, for the first one it suffices to have bilinear form $langle _ ,_ rangle$ inducing a non degenerate bilinearform on $Z=W$.
answered Jan 9 at 10:48
EnkiduEnkidu
1,36719
$endgroup$
$begingroup$
thanks u @Enkidu
$endgroup$
– jasmine
Jan 9 at 10:53
add a comment |
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1 Answer
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$begingroup$
No, it is false, it needs further assumptions, also as stated in the comments, be aware that we are talking subspaces here, not vectors! for example (as gandalf61 said) take $Z=Wsubset V$ then we have, since $langle_,_ rangle$ is non degenerate, for all $0 neq z in Z$ a $w in W=Z$ such that $langle z , w rangle neq 0$ in particular (to give a very concrete example) you could pick $w=z$ and then we already know that $$langle z, wrangle = langle z, zrangle = 0 iff z=0$$
Also, be aware that only the concrete counterexample with $z=w$ needs an inner product, for the first one it suffices to have bilinear form $langle _ ,_ rangle$ inducing a non degenerate bilinearform on $Z=W$.
answered Jan 9 at 10:48
EnkiduEnkidu
1,36719
$endgroup$
$begingroup$
thanks u @Enkidu
$endgroup$
– jasmine
Jan 9 at 10:53
add a comment |
$begingroup$
No, it is false, it needs further assumptions, also as stated in the comments, be aware that we are talking subspaces here, not vectors! for example (as gandalf61 said) take $Z=Wsubset V$ then we have, since $langle_,_ rangle$ is non degenerate, for all $0 neq z in Z$ a $w in W=Z$ such that $langle z , w rangle neq 0$ in particular (to give a very concrete example) you could pick $w=z$ and then we already know that $$langle z, wrangle = langle z, zrangle = 0 iff z=0$$
Also, be aware that only the concrete counterexample with $z=w$ needs an inner product, for the first one it suffices to have bilinear form $langle _ ,_ rangle$ inducing a non degenerate bilinearform on $Z=W$.
answered Jan 9 at 10:48
EnkiduEnkidu
1,36719
$endgroup$
$begingroup$
thanks u @Enkidu
$endgroup$
– jasmine
Jan 9 at 10:53
add a comment |
$begingroup$
No, it is false, it needs further assumptions, also as stated in the comments, be aware that we are talking subspaces here, not vectors! for example (as gandalf61 said) take $Z=Wsubset V$ then we have, since $langle_,_ rangle$ is non degenerate, for all $0 neq z in Z$ a $w in W=Z$ such that $langle z , w rangle neq 0$ in particular (to give a very concrete example) you could pick $w=z$ and then we already know that $$langle z, wrangle = langle z, zrangle = 0 iff z=0$$
Also, be aware that only the concrete counterexample with $z=w$ needs an inner product, for the first one it suffices to have bilinear form $langle _ ,_ rangle$ inducing a non degenerate bilinearform on $Z=W$.
answered Jan 9 at 10:48
EnkiduEnkidu
1,36719
$endgroup$
No, it is false, it needs further assumptions, also as stated in the comments, be aware that we are talking subspaces here, not vectors! for example (as gandalf61 said) take $Z=Wsubset V$ then we have, since $langle_,_ rangle$ is non degenerate, for all $0 neq z in Z$ a $w in W=Z$ such that $langle z , w rangle neq 0$ in particular (to give a very concrete example) you could pick $w=z$ and then we already know that $$langle z, wrangle = langle z, zrangle = 0 iff z=0$$
Also, be aware that only the concrete counterexample with $z=w$ needs an inner product, for the first one it suffices to have bilinear form $langle _ ,_ rangle$ inducing a non degenerate bilinearform on $Z=W$.
answered Jan 9 at 10:48
EnkiduEnkidu
1,36719
edited Jan 9 at 10:53
answered Jan 9 at 10:48
EnkiduEnkidu
1,36719
answered Jan 9 at 10:48
EnkiduEnkidu
1,36719
answered Jan 9 at 10:48
EnkiduEnkidu
1,36719
1,36719
$begingroup$
thanks u @Enkidu
$endgroup$
– jasmine
Jan 9 at 10:53
add a comment |
$begingroup$
thanks u @Enkidu
$endgroup$
– jasmine
Jan 9 at 10:53
$begingroup$
thanks u @Enkidu
$endgroup$
– jasmine
Jan 9 at 10:53
$begingroup$
thanks u @Enkidu
$endgroup$
– jasmine
Jan 9 at 10:53
add a comment |
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3
$begingroup$
What if W=Z i.e. W and Z are the same subspace ? The conditions of the question don't seem to rule this out.
$endgroup$
– gandalf61
Jan 9 at 10:46
2
$begingroup$
The question is not clear. Your $W$ and $Z$ seem vectors of $mathbb{R}^3$, so their are not subspaces of $mathbb{R}^5$ nor elements of this vector space.
$endgroup$
– Emilio Novati
Jan 9 at 10:48
1
$begingroup$
exactly! by the way, I tried to incorporate your comments in my answer, i hope that is ok
$endgroup$
– Enkidu
Jan 9 at 10:49