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Show that $W_1+W_2$ is one of the subspaces and $W_1cap W_2 $ is equal to other

2

$begingroup$

Question: given $W_1$, $W_2$ be finite dimensional subspaces of vector space such that $$dim(W_1+W_2)=1+dim(W_1cap W_2)$$

Show that $W_1+W_2$ is equal to one of the subspaces and $W_1cap W_2$ is equal to the other.

I know that,

begin{align}dim(W_1cap W_2)&≤dim W_1\
dim(W_1cap W_2)&≤dim W_2 \
dim(W_1+W_2)&=dim W_1+ dim W_2-dim(W_1cap W_2)end{align}

With this information,

$$1+dim(W_1cap W_2)=dim W_1 +dim W_2-dim(W_1cap W_2)$$

Hence $$2dim(W_1cap W_2)= dim W_1+dim W_2-1$$

Still unable to prove, please help.

linear-algebra vector-spaces

share|cite|improve this question

edited Mar 22 at 13:23

Jimmy R.

33.3k42257

asked Mar 22 at 13:08

Akash PatalwanshiAkash Patalwanshi

1,0171820

$endgroup$

add a comment | 

2

$begingroup$

Question: given $W_1$, $W_2$ be finite dimensional subspaces of vector space such that $$dim(W_1+W_2)=1+dim(W_1cap W_2)$$

Show that $W_1+W_2$ is equal to one of the subspaces and $W_1cap W_2$ is equal to the other.

I know that,

begin{align}dim(W_1cap W_2)&≤dim W_1\
dim(W_1cap W_2)&≤dim W_2 \
dim(W_1+W_2)&=dim W_1+ dim W_2-dim(W_1cap W_2)end{align}

With this information,

$$1+dim(W_1cap W_2)=dim W_1 +dim W_2-dim(W_1cap W_2)$$

Hence $$2dim(W_1cap W_2)= dim W_1+dim W_2-1$$

Still unable to prove, please help.

linear-algebra vector-spaces

share|cite|improve this question

edited Mar 22 at 13:23

Jimmy R.

33.3k42257

asked Mar 22 at 13:08

Akash PatalwanshiAkash Patalwanshi

1,0171820

$endgroup$

add a comment | 

$begingroup$

Question: given $W_1$, $W_2$ be finite dimensional subspaces of vector space such that $$dim(W_1+W_2)=1+dim(W_1cap W_2)$$

Show that $W_1+W_2$ is equal to one of the subspaces and $W_1cap W_2$ is equal to the other.

I know that,

begin{align}dim(W_1cap W_2)&≤dim W_1\
dim(W_1cap W_2)&≤dim W_2 \
dim(W_1+W_2)&=dim W_1+ dim W_2-dim(W_1cap W_2)end{align}

With this information,

$$1+dim(W_1cap W_2)=dim W_1 +dim W_2-dim(W_1cap W_2)$$

Hence $$2dim(W_1cap W_2)= dim W_1+dim W_2-1$$

Still unable to prove, please help.

linear-algebra vector-spaces

share|cite|improve this question

edited Mar 22 at 13:23

Jimmy R.

33.3k42257

asked Mar 22 at 13:08

Akash PatalwanshiAkash Patalwanshi

1,0171820

$endgroup$

share|cite|improve this question

edited Mar 22 at 13:23

Jimmy R.

33.3k42257

asked Mar 22 at 13:08

Akash PatalwanshiAkash Patalwanshi

1,0171820

share|cite|improve this question

edited Mar 22 at 13:23

Jimmy R.

33.3k42257

asked Mar 22 at 13:08

Akash PatalwanshiAkash Patalwanshi

1,0171820

edited Mar 22 at 13:23

Jimmy R.

33.3k42257

edited Mar 22 at 13:23

Jimmy R.

33.3k42257

asked Mar 22 at 13:08

Akash PatalwanshiAkash Patalwanshi

1,0171820

asked Mar 22 at 13:08

Akash PatalwanshiAkash Patalwanshi

1,0171820

3 Answers
3

active

oldest

votes

2

$begingroup$

So far so good with your solution. Assume now that both $W_1$ and $W_2$ are bigger than $W_1cap W_2$, then you have that $dim(W_1cap W_2)+1le dim (W_1)$ and the same for $W_2$ and your last equation becomes
begin{align}2dim(W_1cap W_2)&=dim W_1+dim W_2-1\&ge (1+dim(W_1cap W_2))+(1+dim(W_1cap W_2))-1\&=1+2dim(W_1cap W_2)end{align} which is a contradiction. So, either $W_1=W_1cap W_2$ or $W_2=W_1cap W_2$. Say without loss of generality that this is $W_1$. Then $W_1subseteq W_2$ and it follows that $W_1+W_2=W_2$.

share|cite|improve this answer

edited Mar 22 at 13:38

answered Mar 22 at 13:17

Jimmy R.Jimmy R.

33.3k42257

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sir, $dim(W_1cap W_2)-1≤dimW_1$ but sir, while using this in equation which I had obtained, you written, $dimW_1≥1+dim(W_1cap W_2)$
  $endgroup$
  – Akash Patalwanshi
  Mar 22 at 13:33
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, there was a typo. Should be correct now. Thanks for noticing
  $endgroup$
  – Jimmy R.
  Mar 22 at 13:38
  

  
  
  
  

add a comment | 

3

$begingroup$

The spaces $W_1$ and $W_2$ both contain $W_1 cap W_2$, and are both contained in $W_1 + W_2$. Now the factor spaces $F_1 := W_1/(W_1 cap W_2)$ and $F_2 := W_2/(W_1cap W_2)$ are both subspaces of $(W_1+W_2)/(W_1cap W_2) := F$. But this space $F$ has dimension one by what you are given, so there aren't that many subspaces.

share|cite|improve this answer

answered Mar 22 at 13:17

DirkDirk

4,658219

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sir, please elaborate last line "so there aren't many subspaces.
  $endgroup$
  – Akash Patalwanshi
  Mar 22 at 13:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $F$ is a space of dimension one. Can you write down all subspaces of a space of dimension one? Which one of them can be $F_1$, which one can be $F_2$?
  $endgroup$
  – Dirk
  Mar 22 at 14:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sir, lf $F$ is one dimensional vector space then $F$ and ${0}$ are only subspaces of $F$.
  $endgroup$
  – Akash Patalwanshi
  Mar 22 at 16:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so ...much sir, I think you want to say, either $F_1=F$ and $F_2={0}$ or $F_2=F$ and $F_1={0}$. In the first case, $W_1=W_1+W_2$ and $W_1cap W_2=W_2$ and in second case $W_1+W_2=W_2$ and $W_1cap W_2=W_1$. Beautiful answer.
  $endgroup$
  – Akash Patalwanshi
  Mar 22 at 16:55
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

$newcommand{Span}[1]{leftlangle #1 rightrangle}$Hint 1

Suppose $W_{1} + W_{2} ne W_{1}$. Then there is $u in W_{2} setminus W_{1}$.

Hint 2

Then $W_{1} + W_{2} = (W_{1} cap W_{2}) + Span{u}$, by the dimension condition.

Hint 3

Then $W_{1} + W_{2} = (W_{1} cap W_{2}) + Span{u} subseteq W_{2}$, so that $W_{1} + W_{2} = W_{2}$.

Hint 4

$W_{1} + W_{2} supsetneq W_{1} supseteq W_{1} cap W_{2}$, so that by the dimension condition $W_{1} = W_{1} cap W_{2}$.

share|cite|improve this answer

edited Mar 22 at 13:37

answered Mar 22 at 13:22

Andreas CarantiAndreas Caranti

57.2k34497

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much. Sir
  $endgroup$
  – Akash Patalwanshi
  Mar 22 at 13:48
  

  
  
  
  

add a comment | 

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2

$begingroup$

So far so good with your solution. Assume now that both $W_1$ and $W_2$ are bigger than $W_1cap W_2$, then you have that $dim(W_1cap W_2)+1le dim (W_1)$ and the same for $W_2$ and your last equation becomes
begin{align}2dim(W_1cap W_2)&=dim W_1+dim W_2-1\&ge (1+dim(W_1cap W_2))+(1+dim(W_1cap W_2))-1\&=1+2dim(W_1cap W_2)end{align} which is a contradiction. So, either $W_1=W_1cap W_2$ or $W_2=W_1cap W_2$. Say without loss of generality that this is $W_1$. Then $W_1subseteq W_2$ and it follows that $W_1+W_2=W_2$.

share|cite|improve this answer

edited Mar 22 at 13:38

answered Mar 22 at 13:17

Jimmy R.Jimmy R.

33.3k42257

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sir, $dim(W_1cap W_2)-1≤dimW_1$ but sir, while using this in equation which I had obtained, you written, $dimW_1≥1+dim(W_1cap W_2)$
  $endgroup$
  – Akash Patalwanshi
  Mar 22 at 13:33
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, there was a typo. Should be correct now. Thanks for noticing
  $endgroup$
  – Jimmy R.
  Mar 22 at 13:38
  

  
  
  
  

add a comment | 

2

$begingroup$

So far so good with your solution. Assume now that both $W_1$ and $W_2$ are bigger than $W_1cap W_2$, then you have that $dim(W_1cap W_2)+1le dim (W_1)$ and the same for $W_2$ and your last equation becomes
begin{align}2dim(W_1cap W_2)&=dim W_1+dim W_2-1\&ge (1+dim(W_1cap W_2))+(1+dim(W_1cap W_2))-1\&=1+2dim(W_1cap W_2)end{align} which is a contradiction. So, either $W_1=W_1cap W_2$ or $W_2=W_1cap W_2$. Say without loss of generality that this is $W_1$. Then $W_1subseteq W_2$ and it follows that $W_1+W_2=W_2$.

share|cite|improve this answer

edited Mar 22 at 13:38

answered Mar 22 at 13:17

Jimmy R.Jimmy R.

33.3k42257

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sir, $dim(W_1cap W_2)-1≤dimW_1$ but sir, while using this in equation which I had obtained, you written, $dimW_1≥1+dim(W_1cap W_2)$
  $endgroup$
  – Akash Patalwanshi
  Mar 22 at 13:33
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, there was a typo. Should be correct now. Thanks for noticing
  $endgroup$
  – Jimmy R.
  Mar 22 at 13:38
  

  
  
  
  

add a comment | 

$begingroup$

So far so good with your solution. Assume now that both $W_1$ and $W_2$ are bigger than $W_1cap W_2$, then you have that $dim(W_1cap W_2)+1le dim (W_1)$ and the same for $W_2$ and your last equation becomes
begin{align}2dim(W_1cap W_2)&=dim W_1+dim W_2-1\&ge (1+dim(W_1cap W_2))+(1+dim(W_1cap W_2))-1\&=1+2dim(W_1cap W_2)end{align} which is a contradiction. So, either $W_1=W_1cap W_2$ or $W_2=W_1cap W_2$. Say without loss of generality that this is $W_1$. Then $W_1subseteq W_2$ and it follows that $W_1+W_2=W_2$.

share|cite|improve this answer

edited Mar 22 at 13:38

answered Mar 22 at 13:17

Jimmy R.Jimmy R.

33.3k42257

$endgroup$

share|cite|improve this answer

edited Mar 22 at 13:38

answered Mar 22 at 13:17

Jimmy R.Jimmy R.

33.3k42257

answered Mar 22 at 13:17

Jimmy R.Jimmy R.

33.3k42257

answered Mar 22 at 13:17

Jimmy R.Jimmy R.

33.3k42257

3

$begingroup$

The spaces $W_1$ and $W_2$ both contain $W_1 cap W_2$, and are both contained in $W_1 + W_2$. Now the factor spaces $F_1 := W_1/(W_1 cap W_2)$ and $F_2 := W_2/(W_1cap W_2)$ are both subspaces of $(W_1+W_2)/(W_1cap W_2) := F$. But this space $F$ has dimension one by what you are given, so there aren't that many subspaces.

share|cite|improve this answer

answered Mar 22 at 13:17

DirkDirk

4,658219

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sir, please elaborate last line "so there aren't many subspaces.
  $endgroup$
  – Akash Patalwanshi
  Mar 22 at 13:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $F$ is a space of dimension one. Can you write down all subspaces of a space of dimension one? Which one of them can be $F_1$, which one can be $F_2$?
  $endgroup$
  – Dirk
  Mar 22 at 14:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sir, lf $F$ is one dimensional vector space then $F$ and ${0}$ are only subspaces of $F$.
  $endgroup$
  – Akash Patalwanshi
  Mar 22 at 16:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so ...much sir, I think you want to say, either $F_1=F$ and $F_2={0}$ or $F_2=F$ and $F_1={0}$. In the first case, $W_1=W_1+W_2$ and $W_1cap W_2=W_2$ and in second case $W_1+W_2=W_2$ and $W_1cap W_2=W_1$. Beautiful answer.
  $endgroup$
  – Akash Patalwanshi
  Mar 22 at 16:55
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

The spaces $W_1$ and $W_2$ both contain $W_1 cap W_2$, and are both contained in $W_1 + W_2$. Now the factor spaces $F_1 := W_1/(W_1 cap W_2)$ and $F_2 := W_2/(W_1cap W_2)$ are both subspaces of $(W_1+W_2)/(W_1cap W_2) := F$. But this space $F$ has dimension one by what you are given, so there aren't that many subspaces.

share|cite|improve this answer

answered Mar 22 at 13:17

DirkDirk

4,658219

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sir, please elaborate last line "so there aren't many subspaces.
  $endgroup$
  – Akash Patalwanshi
  Mar 22 at 13:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $F$ is a space of dimension one. Can you write down all subspaces of a space of dimension one? Which one of them can be $F_1$, which one can be $F_2$?
  $endgroup$
  – Dirk
  Mar 22 at 14:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sir, lf $F$ is one dimensional vector space then $F$ and ${0}$ are only subspaces of $F$.
  $endgroup$
  – Akash Patalwanshi
  Mar 22 at 16:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so ...much sir, I think you want to say, either $F_1=F$ and $F_2={0}$ or $F_2=F$ and $F_1={0}$. In the first case, $W_1=W_1+W_2$ and $W_1cap W_2=W_2$ and in second case $W_1+W_2=W_2$ and $W_1cap W_2=W_1$. Beautiful answer.
  $endgroup$
  – Akash Patalwanshi
  Mar 22 at 16:55
  
  
  

  
  
  
  

add a comment | 

$begingroup$

The spaces $W_1$ and $W_2$ both contain $W_1 cap W_2$, and are both contained in $W_1 + W_2$. Now the factor spaces $F_1 := W_1/(W_1 cap W_2)$ and $F_2 := W_2/(W_1cap W_2)$ are both subspaces of $(W_1+W_2)/(W_1cap W_2) := F$. But this space $F$ has dimension one by what you are given, so there aren't that many subspaces.

share|cite|improve this answer

answered Mar 22 at 13:17

DirkDirk

4,658219

$endgroup$

share|cite|improve this answer

answered Mar 22 at 13:17

DirkDirk

4,658219

answered Mar 22 at 13:17

DirkDirk

4,658219

answered Mar 22 at 13:17

DirkDirk

4,658219

3

$begingroup$

$newcommand{Span}[1]{leftlangle #1 rightrangle}$Hint 1

Suppose $W_{1} + W_{2} ne W_{1}$. Then there is $u in W_{2} setminus W_{1}$.

Hint 2

Then $W_{1} + W_{2} = (W_{1} cap W_{2}) + Span{u}$, by the dimension condition.

Hint 3

Then $W_{1} + W_{2} = (W_{1} cap W_{2}) + Span{u} subseteq W_{2}$, so that $W_{1} + W_{2} = W_{2}$.

Hint 4

$W_{1} + W_{2} supsetneq W_{1} supseteq W_{1} cap W_{2}$, so that by the dimension condition $W_{1} = W_{1} cap W_{2}$.

share|cite|improve this answer

edited Mar 22 at 13:37

answered Mar 22 at 13:22

Andreas CarantiAndreas Caranti

57.2k34497

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much. Sir
  $endgroup$
  – Akash Patalwanshi
  Mar 22 at 13:48
  

  
  
  
  

add a comment | 

3

$begingroup$

$newcommand{Span}[1]{leftlangle #1 rightrangle}$Hint 1

Suppose $W_{1} + W_{2} ne W_{1}$. Then there is $u in W_{2} setminus W_{1}$.

Hint 2

Then $W_{1} + W_{2} = (W_{1} cap W_{2}) + Span{u}$, by the dimension condition.

Hint 3

Then $W_{1} + W_{2} = (W_{1} cap W_{2}) + Span{u} subseteq W_{2}$, so that $W_{1} + W_{2} = W_{2}$.

Hint 4

$W_{1} + W_{2} supsetneq W_{1} supseteq W_{1} cap W_{2}$, so that by the dimension condition $W_{1} = W_{1} cap W_{2}$.

share|cite|improve this answer

edited Mar 22 at 13:37

answered Mar 22 at 13:22

Andreas CarantiAndreas Caranti

57.2k34497

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much. Sir
  $endgroup$
  – Akash Patalwanshi
  Mar 22 at 13:48
  

  
  
  
  

add a comment | 

$begingroup$

$newcommand{Span}[1]{leftlangle #1 rightrangle}$Hint 1

Suppose $W_{1} + W_{2} ne W_{1}$. Then there is $u in W_{2} setminus W_{1}$.

Hint 2

Then $W_{1} + W_{2} = (W_{1} cap W_{2}) + Span{u}$, by the dimension condition.

Hint 3

Then $W_{1} + W_{2} = (W_{1} cap W_{2}) + Span{u} subseteq W_{2}$, so that $W_{1} + W_{2} = W_{2}$.

Hint 4

$W_{1} + W_{2} supsetneq W_{1} supseteq W_{1} cap W_{2}$, so that by the dimension condition $W_{1} = W_{1} cap W_{2}$.

share|cite|improve this answer

edited Mar 22 at 13:37

answered Mar 22 at 13:22

Andreas CarantiAndreas Caranti

57.2k34497

$endgroup$

share|cite|improve this answer

edited Mar 22 at 13:37

answered Mar 22 at 13:22

Andreas CarantiAndreas Caranti

57.2k34497

answered Mar 22 at 13:22

Andreas CarantiAndreas Caranti

57.2k34497

answered Mar 22 at 13:22

Andreas CarantiAndreas Caranti

57.2k34497