Mixing
Proof about double orthogonal complements
$begingroup$
What I had to prove was:
Let $W$ be a subspace of an inner product space $V$. Show that $Wsubset W^{perpperp}$ and that $W=W^{perpperp}$ if dim$V$ is finite.
There are two areas where I'm having trouble labelled with **.
My attempt is:
If $win W$, then $forall vin W^{perp}$, $langle w,vrangle=0$. **Then somehow $win W^{perpperp}$ which would show that $Wsubset W^{perpperp}$ but I'm not sure why $win W^{perpperp}$.
Let $V$ have finite dimension $n$, and let dim$W=m$. Then dim$W^{perp}=$dim$V-$dim$W=n-m$, and dim$W^{perpperp}=$dim$V-$dim$W^{perp}=n-(n-m)=m=$dim$W$. **Then I'm not sure how this shows or helps show that $W=W^{perpperp}$.
Could someone help me with these two areas of the proof that I'm struggling with?
inner-product-space orthogonality
$endgroup$
add a comment |
$begingroup$
What I had to prove was:
Let $W$ be a subspace of an inner product space $V$. Show that $Wsubset W^{perpperp}$ and that $W=W^{perpperp}$ if dim$V$ is finite.
There are two areas where I'm having trouble labelled with **.
My attempt is:
If $win W$, then $forall vin W^{perp}$, $langle w,vrangle=0$. **Then somehow $win W^{perpperp}$ which would show that $Wsubset W^{perpperp}$ but I'm not sure why $win W^{perpperp}$.
Let $V$ have finite dimension $n$, and let dim$W=m$. Then dim$W^{perp}=$dim$V-$dim$W=n-m$, and dim$W^{perpperp}=$dim$V-$dim$W^{perp}=n-(n-m)=m=$dim$W$. **Then I'm not sure how this shows or helps show that $W=W^{perpperp}$.
Could someone help me with these two areas of the proof that I'm struggling with?
inner-product-space orthogonality
$endgroup$
$begingroup$
If you can show that $Wsubset W^{perpperp}$ and $dim W = dim W^{perpperp}$, does that tell you anything about the relationship between them?
$endgroup$
– Mark
Oct 31 '16 at 8:34
$begingroup$
I'm guessing that $W=W^{perpperp}$ but how do you know that these two conditions is enough to conclude this?
$endgroup$
– user342661
Oct 31 '16 at 9:48
add a comment |
$begingroup$
What I had to prove was:
Let $W$ be a subspace of an inner product space $V$. Show that $Wsubset W^{perpperp}$ and that $W=W^{perpperp}$ if dim$V$ is finite.
There are two areas where I'm having trouble labelled with **.
My attempt is:
If $win W$, then $forall vin W^{perp}$, $langle w,vrangle=0$. **Then somehow $win W^{perpperp}$ which would show that $Wsubset W^{perpperp}$ but I'm not sure why $win W^{perpperp}$.
Let $V$ have finite dimension $n$, and let dim$W=m$. Then dim$W^{perp}=$dim$V-$dim$W=n-m$, and dim$W^{perpperp}=$dim$V-$dim$W^{perp}=n-(n-m)=m=$dim$W$. **Then I'm not sure how this shows or helps show that $W=W^{perpperp}$.
Could someone help me with these two areas of the proof that I'm struggling with?
inner-product-space orthogonality
$endgroup$
What I had to prove was:
Let $W$ be a subspace of an inner product space $V$. Show that $Wsubset W^{perpperp}$ and that $W=W^{perpperp}$ if dim$V$ is finite.
There are two areas where I'm having trouble labelled with **.
My attempt is:
If $win W$, then $forall vin W^{perp}$, $langle w,vrangle=0$. **Then somehow $win W^{perpperp}$ which would show that $Wsubset W^{perpperp}$ but I'm not sure why $win W^{perpperp}$.
Let $V$ have finite dimension $n$, and let dim$W=m$. Then dim$W^{perp}=$dim$V-$dim$W=n-m$, and dim$W^{perpperp}=$dim$V-$dim$W^{perp}=n-(n-m)=m=$dim$W$. **Then I'm not sure how this shows or helps show that $W=W^{perpperp}$.
Could someone help me with these two areas of the proof that I'm struggling with?
inner-product-space orthogonality
inner-product-space orthogonality
asked Oct 31 '16 at 8:27
user342661
$begingroup$
If you can show that $Wsubset W^{perpperp}$ and $dim W = dim W^{perpperp}$, does that tell you anything about the relationship between them?
$endgroup$
– Mark
Oct 31 '16 at 8:34
$begingroup$
I'm guessing that $W=W^{perpperp}$ but how do you know that these two conditions is enough to conclude this?
$endgroup$
– user342661
Oct 31 '16 at 9:48
add a comment |
$begingroup$
If you can show that $Wsubset W^{perpperp}$ and $dim W = dim W^{perpperp}$, does that tell you anything about the relationship between them?
$endgroup$
– Mark
Oct 31 '16 at 8:34
$begingroup$
I'm guessing that $W=W^{perpperp}$ but how do you know that these two conditions is enough to conclude this?
$endgroup$
– user342661
Oct 31 '16 at 9:48
$begingroup$
If you can show that $Wsubset W^{perpperp}$ and $dim W = dim W^{perpperp}$, does that tell you anything about the relationship between them?
$endgroup$
– Mark
Oct 31 '16 at 8:34
$begingroup$
If you can show that $Wsubset W^{perpperp}$ and $dim W = dim W^{perpperp}$, does that tell you anything about the relationship between them?
$endgroup$
– Mark
Oct 31 '16 at 8:34
$begingroup$
I'm guessing that $W=W^{perpperp}$ but how do you know that these two conditions is enough to conclude this?
$endgroup$
– user342661
Oct 31 '16 at 9:48
$begingroup$
I'm guessing that $W=W^{perpperp}$ but how do you know that these two conditions is enough to conclude this?
$endgroup$
– user342661
Oct 31 '16 at 9:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It might help to rename, say $F=W^perp$
Then for all $win W$ and $fin F$ you have: $langle w,frangle =0$
which shows that $W$ and $F$ are orthogonal to each other. In particular, $win F^perp = (W^perp)^perp$. In other words $Wsubset (W^perp)^perp$.
If $V$ has finite dimension, then there are different ways to go: Either construct orthonormal bases for $W$ and $W^perp$ and show that $V=Woplus W^perp$ (from which the conclusion follows) or use that the orthogonal projection $P:Vrightarrow W$ has $W^perp$ as kernel.
answered Oct 31 '16 at 9:00
H. H. RughH. H. Rugh
23.2k11134
$endgroup$
add a comment |
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$begingroup$
It might help to rename, say $F=W^perp$
Then for all $win W$ and $fin F$ you have: $langle w,frangle =0$
which shows that $W$ and $F$ are orthogonal to each other. In particular, $win F^perp = (W^perp)^perp$. In other words $Wsubset (W^perp)^perp$.
If $V$ has finite dimension, then there are different ways to go: Either construct orthonormal bases for $W$ and $W^perp$ and show that $V=Woplus W^perp$ (from which the conclusion follows) or use that the orthogonal projection $P:Vrightarrow W$ has $W^perp$ as kernel.
answered Oct 31 '16 at 9:00
H. H. RughH. H. Rugh
23.2k11134
$endgroup$
add a comment |
$begingroup$
It might help to rename, say $F=W^perp$
Then for all $win W$ and $fin F$ you have: $langle w,frangle =0$
which shows that $W$ and $F$ are orthogonal to each other. In particular, $win F^perp = (W^perp)^perp$. In other words $Wsubset (W^perp)^perp$.
If $V$ has finite dimension, then there are different ways to go: Either construct orthonormal bases for $W$ and $W^perp$ and show that $V=Woplus W^perp$ (from which the conclusion follows) or use that the orthogonal projection $P:Vrightarrow W$ has $W^perp$ as kernel.
answered Oct 31 '16 at 9:00
H. H. RughH. H. Rugh
23.2k11134
$endgroup$
add a comment |
$begingroup$
It might help to rename, say $F=W^perp$
Then for all $win W$ and $fin F$ you have: $langle w,frangle =0$
which shows that $W$ and $F$ are orthogonal to each other. In particular, $win F^perp = (W^perp)^perp$. In other words $Wsubset (W^perp)^perp$.
If $V$ has finite dimension, then there are different ways to go: Either construct orthonormal bases for $W$ and $W^perp$ and show that $V=Woplus W^perp$ (from which the conclusion follows) or use that the orthogonal projection $P:Vrightarrow W$ has $W^perp$ as kernel.
answered Oct 31 '16 at 9:00
H. H. RughH. H. Rugh
23.2k11134
$endgroup$
It might help to rename, say $F=W^perp$
Then for all $win W$ and $fin F$ you have: $langle w,frangle =0$
which shows that $W$ and $F$ are orthogonal to each other. In particular, $win F^perp = (W^perp)^perp$. In other words $Wsubset (W^perp)^perp$.
If $V$ has finite dimension, then there are different ways to go: Either construct orthonormal bases for $W$ and $W^perp$ and show that $V=Woplus W^perp$ (from which the conclusion follows) or use that the orthogonal projection $P:Vrightarrow W$ has $W^perp$ as kernel.
answered Oct 31 '16 at 9:00
H. H. RughH. H. Rugh
23.2k11134
answered Oct 31 '16 at 9:00
H. H. RughH. H. Rugh
23.2k11134
answered Oct 31 '16 at 9:00
H. H. RughH. H. Rugh
23.2k11134
answered Oct 31 '16 at 9:00
H. H. RughH. H. Rugh
23.2k11134
23.2k11134
add a comment |
add a comment |
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$begingroup$
If you can show that $Wsubset W^{perpperp}$ and $dim W = dim W^{perpperp}$, does that tell you anything about the relationship between them?
$endgroup$
– Mark
Oct 31 '16 at 8:34
$begingroup$
I'm guessing that $W=W^{perpperp}$ but how do you know that these two conditions is enough to conclude this?
$endgroup$
– user342661
Oct 31 '16 at 9:48