Basic linear algebra doubt on Dimension of Vector space












0












$begingroup$


Let $V(R)=M$ of order $2$



$$W_1=left{
begin{pmatrix}
a & b \
0 & 0 \
end{pmatrix}
: a ,b in Rright}$$



$$W_2=left{
begin{pmatrix}
a & 0 \
c & 0 \
end{pmatrix}
: a ,c in R right}$$



$$W_1+W_2=left{begin{pmatrix}
2a & b \
c & 0 \
end{pmatrix}
: a ,b,c in Rright}$$



I am trying to find out the dimension of $W_1+W_2$ using the definition: The number of elements in the basis for $V$ is called dimension of $V$.



Now we have vectors $v_1=(2a,c),v_2=(b,0)$



both are linearly independent and each have two elements in it so dimension is $2$. Is this the correct way to find dimension ?






linear-algebra matrices vector-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    For $W_i$, do you intend to type them in set notation but having difficulty displaying them?
    $endgroup$
    – Siong Thye Goh
    Jan 13 at 11:24










  • $begingroup$
    What is $W_2 $?
    $endgroup$
    – Bernard
    Jan 13 at 11:26










  • $begingroup$
    @SiongThyeGoh I didnt 't get you. I am just unable to find dimension.
    $endgroup$
    – Daman deep
    Jan 13 at 11:28










  • $begingroup$
    @Bernard typo check now
    $endgroup$
    – Daman deep
    Jan 13 at 11:29










  • $begingroup$
    is $W_1$ a subspace? or a particular matrix? what do you mean by dimension of a particular matrix if it is a particular matrix?
    $endgroup$
    – Siong Thye Goh
    Jan 13 at 11:32


















0












$begingroup$


Let $V(R)=M$ of order $2$



$$W_1=left{
begin{pmatrix}
a & b \
0 & 0 \
end{pmatrix}
: a ,b in Rright}$$



$$W_2=left{
begin{pmatrix}
a & 0 \
c & 0 \
end{pmatrix}
: a ,c in R right}$$



$$W_1+W_2=left{begin{pmatrix}
2a & b \
c & 0 \
end{pmatrix}
: a ,b,c in Rright}$$



I am trying to find out the dimension of $W_1+W_2$ using the definition: The number of elements in the basis for $V$ is called dimension of $V$.



Now we have vectors $v_1=(2a,c),v_2=(b,0)$



both are linearly independent and each have two elements in it so dimension is $2$. Is this the correct way to find dimension ?






linear-algebra matrices vector-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    For $W_i$, do you intend to type them in set notation but having difficulty displaying them?
    $endgroup$
    – Siong Thye Goh
    Jan 13 at 11:24










  • $begingroup$
    What is $W_2 $?
    $endgroup$
    – Bernard
    Jan 13 at 11:26










  • $begingroup$
    @SiongThyeGoh I didnt 't get you. I am just unable to find dimension.
    $endgroup$
    – Daman deep
    Jan 13 at 11:28










  • $begingroup$
    @Bernard typo check now
    $endgroup$
    – Daman deep
    Jan 13 at 11:29










  • $begingroup$
    is $W_1$ a subspace? or a particular matrix? what do you mean by dimension of a particular matrix if it is a particular matrix?
    $endgroup$
    – Siong Thye Goh
    Jan 13 at 11:32
















0












0








0





$begingroup$


Let $V(R)=M$ of order $2$



$$W_1=left{
begin{pmatrix}
a & b \
0 & 0 \
end{pmatrix}
: a ,b in Rright}$$



$$W_2=left{
begin{pmatrix}
a & 0 \
c & 0 \
end{pmatrix}
: a ,c in R right}$$



$$W_1+W_2=left{begin{pmatrix}
2a & b \
c & 0 \
end{pmatrix}
: a ,b,c in Rright}$$



I am trying to find out the dimension of $W_1+W_2$ using the definition: The number of elements in the basis for $V$ is called dimension of $V$.



Now we have vectors $v_1=(2a,c),v_2=(b,0)$



both are linearly independent and each have two elements in it so dimension is $2$. Is this the correct way to find dimension ?






linear-algebra matrices vector-spaces







share|cite|improve this question











$endgroup$




Let $V(R)=M$ of order $2$



$$W_1=left{
begin{pmatrix}
a & b \
0 & 0 \
end{pmatrix}
: a ,b in Rright}$$



$$W_2=left{
begin{pmatrix}
a & 0 \
c & 0 \
end{pmatrix}
: a ,c in R right}$$



$$W_1+W_2=left{begin{pmatrix}
2a & b \
c & 0 \
end{pmatrix}
: a ,b,c in Rright}$$



I am trying to find out the dimension of $W_1+W_2$ using the definition: The number of elements in the basis for $V$ is called dimension of $V$.



Now we have vectors $v_1=(2a,c),v_2=(b,0)$



both are linearly independent and each have two elements in it so dimension is $2$. Is this the correct way to find dimension ?






linear-algebra matrices vector-spaces




linear-algebra matrices vector-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 11:50









José Carlos Santos

175k24134243




175k24134243










asked Jan 13 at 11:16









Daman deepDaman deep

756420




756420












  • $begingroup$
    For $W_i$, do you intend to type them in set notation but having difficulty displaying them?
    $endgroup$
    – Siong Thye Goh
    Jan 13 at 11:24










  • $begingroup$
    What is $W_2 $?
    $endgroup$
    – Bernard
    Jan 13 at 11:26










  • $begingroup$
    @SiongThyeGoh I didnt 't get you. I am just unable to find dimension.
    $endgroup$
    – Daman deep
    Jan 13 at 11:28










  • $begingroup$
    @Bernard typo check now
    $endgroup$
    – Daman deep
    Jan 13 at 11:29










  • $begingroup$
    is $W_1$ a subspace? or a particular matrix? what do you mean by dimension of a particular matrix if it is a particular matrix?
    $endgroup$
    – Siong Thye Goh
    Jan 13 at 11:32




















  • $begingroup$
    For $W_i$, do you intend to type them in set notation but having difficulty displaying them?
    $endgroup$
    – Siong Thye Goh
    Jan 13 at 11:24










  • $begingroup$
    What is $W_2 $?
    $endgroup$
    – Bernard
    Jan 13 at 11:26










  • $begingroup$
    @SiongThyeGoh I didnt 't get you. I am just unable to find dimension.
    $endgroup$
    – Daman deep
    Jan 13 at 11:28










  • $begingroup$
    @Bernard typo check now
    $endgroup$
    – Daman deep
    Jan 13 at 11:29










  • $begingroup$
    is $W_1$ a subspace? or a particular matrix? what do you mean by dimension of a particular matrix if it is a particular matrix?
    $endgroup$
    – Siong Thye Goh
    Jan 13 at 11:32


















$begingroup$
For $W_i$, do you intend to type them in set notation but having difficulty displaying them?
$endgroup$
– Siong Thye Goh
Jan 13 at 11:24




$begingroup$
For $W_i$, do you intend to type them in set notation but having difficulty displaying them?
$endgroup$
– Siong Thye Goh
Jan 13 at 11:24












$begingroup$
What is $W_2 $?
$endgroup$
– Bernard
Jan 13 at 11:26




$begingroup$
What is $W_2 $?
$endgroup$
– Bernard
Jan 13 at 11:26












$begingroup$
@SiongThyeGoh I didnt 't get you. I am just unable to find dimension.
$endgroup$
– Daman deep
Jan 13 at 11:28




$begingroup$
@SiongThyeGoh I didnt 't get you. I am just unable to find dimension.
$endgroup$
– Daman deep
Jan 13 at 11:28












$begingroup$
@Bernard typo check now
$endgroup$
– Daman deep
Jan 13 at 11:29




$begingroup$
@Bernard typo check now
$endgroup$
– Daman deep
Jan 13 at 11:29












$begingroup$
is $W_1$ a subspace? or a particular matrix? what do you mean by dimension of a particular matrix if it is a particular matrix?
$endgroup$
– Siong Thye Goh
Jan 13 at 11:32






$begingroup$
is $W_1$ a subspace? or a particular matrix? what do you mean by dimension of a particular matrix if it is a particular matrix?
$endgroup$
– Siong Thye Goh
Jan 13 at 11:32












2 Answers
2






active

oldest

votes


















1












$begingroup$

Check that $$left{begin{pmatrix} 1 & 0 \ 0 & 0end{pmatrix}, begin{pmatrix} 0 & 1 \ 0 & 0end{pmatrix}, begin{pmatrix} 0 & 0 \ 1 & 0end{pmatrix} right}$$



is a basis for $W_1+W_2$.



Hopefully you can state the dimension correctly.



The question is not talking about column space of a particular matrix.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    oh so I was calculated column space which $2$ here.
    $endgroup$
    – Daman deep
    Jan 13 at 11:39



















2












$begingroup$

Note that$$W_1+W_2=left{begin{pmatrix}a+a'&b\c&0end{pmatrix},middle|,a,a',b,cinmathbb{R}right}.$$It's dimension is three, because$$W_1+W_2=operatorname{span}left{begin{pmatrix}1&0\0&0end{pmatrix},begin{pmatrix}0&1\0&0end{pmatrix},begin{pmatrix}0&0\1&0end{pmatrix}right}$$and the previous set is linearly independent.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What does previous set is linearly independent means ?
    $endgroup$
    – Daman deep
    Jan 13 at 11:40










  • $begingroup$
    Why it is $a+a^{'}$ not $2a$
    $endgroup$
    – Daman deep
    Jan 13 at 11:44












  • $begingroup$
    It means that$$abegin{pmatrix}1&0\0&0end{pmatrix}+bbegin{pmatrix}0&1\0&0end{pmatrix}+cbegin{pmatrix}0&0\1&0end{pmatrix}=begin{pmatrix}0&0\0&0end{pmatrix}iff a=b=c=0.$$
    $endgroup$
    – José Carlos Santos
    Jan 13 at 11:46










  • $begingroup$
    Because the $a$ from an element of $W_1$ doesn't have to be the same $a$ that was used in an element of $W_2$.
    $endgroup$
    – José Carlos Santos
    Jan 13 at 11:49












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Check that $$left{begin{pmatrix} 1 & 0 \ 0 & 0end{pmatrix}, begin{pmatrix} 0 & 1 \ 0 & 0end{pmatrix}, begin{pmatrix} 0 & 0 \ 1 & 0end{pmatrix} right}$$



is a basis for $W_1+W_2$.



Hopefully you can state the dimension correctly.



The question is not talking about column space of a particular matrix.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    oh so I was calculated column space which $2$ here.
    $endgroup$
    – Daman deep
    Jan 13 at 11:39
















1












$begingroup$

Check that $$left{begin{pmatrix} 1 & 0 \ 0 & 0end{pmatrix}, begin{pmatrix} 0 & 1 \ 0 & 0end{pmatrix}, begin{pmatrix} 0 & 0 \ 1 & 0end{pmatrix} right}$$



is a basis for $W_1+W_2$.



Hopefully you can state the dimension correctly.



The question is not talking about column space of a particular matrix.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    oh so I was calculated column space which $2$ here.
    $endgroup$
    – Daman deep
    Jan 13 at 11:39














1












1








1





$begingroup$

Check that $$left{begin{pmatrix} 1 & 0 \ 0 & 0end{pmatrix}, begin{pmatrix} 0 & 1 \ 0 & 0end{pmatrix}, begin{pmatrix} 0 & 0 \ 1 & 0end{pmatrix} right}$$



is a basis for $W_1+W_2$.



Hopefully you can state the dimension correctly.



The question is not talking about column space of a particular matrix.






share|cite|improve this answer









$endgroup$



Check that $$left{begin{pmatrix} 1 & 0 \ 0 & 0end{pmatrix}, begin{pmatrix} 0 & 1 \ 0 & 0end{pmatrix}, begin{pmatrix} 0 & 0 \ 1 & 0end{pmatrix} right}$$



is a basis for $W_1+W_2$.



Hopefully you can state the dimension correctly.



The question is not talking about column space of a particular matrix.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 13 at 11:36









Siong Thye GohSiong Thye Goh

104k1468120




104k1468120












  • $begingroup$
    oh so I was calculated column space which $2$ here.
    $endgroup$
    – Daman deep
    Jan 13 at 11:39


















  • $begingroup$
    oh so I was calculated column space which $2$ here.
    $endgroup$
    – Daman deep
    Jan 13 at 11:39
















$begingroup$
oh so I was calculated column space which $2$ here.
$endgroup$
– Daman deep
Jan 13 at 11:39




$begingroup$
oh so I was calculated column space which $2$ here.
$endgroup$
– Daman deep
Jan 13 at 11:39











2












$begingroup$

Note that$$W_1+W_2=left{begin{pmatrix}a+a'&b\c&0end{pmatrix},middle|,a,a',b,cinmathbb{R}right}.$$It's dimension is three, because$$W_1+W_2=operatorname{span}left{begin{pmatrix}1&0\0&0end{pmatrix},begin{pmatrix}0&1\0&0end{pmatrix},begin{pmatrix}0&0\1&0end{pmatrix}right}$$and the previous set is linearly independent.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What does previous set is linearly independent means ?
    $endgroup$
    – Daman deep
    Jan 13 at 11:40










  • $begingroup$
    Why it is $a+a^{'}$ not $2a$
    $endgroup$
    – Daman deep
    Jan 13 at 11:44












  • $begingroup$
    It means that$$abegin{pmatrix}1&0\0&0end{pmatrix}+bbegin{pmatrix}0&1\0&0end{pmatrix}+cbegin{pmatrix}0&0\1&0end{pmatrix}=begin{pmatrix}0&0\0&0end{pmatrix}iff a=b=c=0.$$
    $endgroup$
    – José Carlos Santos
    Jan 13 at 11:46










  • $begingroup$
    Because the $a$ from an element of $W_1$ doesn't have to be the same $a$ that was used in an element of $W_2$.
    $endgroup$
    – José Carlos Santos
    Jan 13 at 11:49
















2












$begingroup$

Note that$$W_1+W_2=left{begin{pmatrix}a+a'&b\c&0end{pmatrix},middle|,a,a',b,cinmathbb{R}right}.$$It's dimension is three, because$$W_1+W_2=operatorname{span}left{begin{pmatrix}1&0\0&0end{pmatrix},begin{pmatrix}0&1\0&0end{pmatrix},begin{pmatrix}0&0\1&0end{pmatrix}right}$$and the previous set is linearly independent.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What does previous set is linearly independent means ?
    $endgroup$
    – Daman deep
    Jan 13 at 11:40










  • $begingroup$
    Why it is $a+a^{'}$ not $2a$
    $endgroup$
    – Daman deep
    Jan 13 at 11:44












  • $begingroup$
    It means that$$abegin{pmatrix}1&0\0&0end{pmatrix}+bbegin{pmatrix}0&1\0&0end{pmatrix}+cbegin{pmatrix}0&0\1&0end{pmatrix}=begin{pmatrix}0&0\0&0end{pmatrix}iff a=b=c=0.$$
    $endgroup$
    – José Carlos Santos
    Jan 13 at 11:46










  • $begingroup$
    Because the $a$ from an element of $W_1$ doesn't have to be the same $a$ that was used in an element of $W_2$.
    $endgroup$
    – José Carlos Santos
    Jan 13 at 11:49














2












2








2





$begingroup$

Note that$$W_1+W_2=left{begin{pmatrix}a+a'&b\c&0end{pmatrix},middle|,a,a',b,cinmathbb{R}right}.$$It's dimension is three, because$$W_1+W_2=operatorname{span}left{begin{pmatrix}1&0\0&0end{pmatrix},begin{pmatrix}0&1\0&0end{pmatrix},begin{pmatrix}0&0\1&0end{pmatrix}right}$$and the previous set is linearly independent.






share|cite|improve this answer









$endgroup$



Note that$$W_1+W_2=left{begin{pmatrix}a+a'&b\c&0end{pmatrix},middle|,a,a',b,cinmathbb{R}right}.$$It's dimension is three, because$$W_1+W_2=operatorname{span}left{begin{pmatrix}1&0\0&0end{pmatrix},begin{pmatrix}0&1\0&0end{pmatrix},begin{pmatrix}0&0\1&0end{pmatrix}right}$$and the previous set is linearly independent.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 13 at 11:38









José Carlos SantosJosé Carlos Santos

175k24134243




175k24134243












  • $begingroup$
    What does previous set is linearly independent means ?
    $endgroup$
    – Daman deep
    Jan 13 at 11:40










  • $begingroup$
    Why it is $a+a^{'}$ not $2a$
    $endgroup$
    – Daman deep
    Jan 13 at 11:44












  • $begingroup$
    It means that$$abegin{pmatrix}1&0\0&0end{pmatrix}+bbegin{pmatrix}0&1\0&0end{pmatrix}+cbegin{pmatrix}0&0\1&0end{pmatrix}=begin{pmatrix}0&0\0&0end{pmatrix}iff a=b=c=0.$$
    $endgroup$
    – José Carlos Santos
    Jan 13 at 11:46










  • $begingroup$
    Because the $a$ from an element of $W_1$ doesn't have to be the same $a$ that was used in an element of $W_2$.
    $endgroup$
    – José Carlos Santos
    Jan 13 at 11:49


















  • $begingroup$
    What does previous set is linearly independent means ?
    $endgroup$
    – Daman deep
    Jan 13 at 11:40










  • $begingroup$
    Why it is $a+a^{'}$ not $2a$
    $endgroup$
    – Daman deep
    Jan 13 at 11:44












  • $begingroup$
    It means that$$abegin{pmatrix}1&0\0&0end{pmatrix}+bbegin{pmatrix}0&1\0&0end{pmatrix}+cbegin{pmatrix}0&0\1&0end{pmatrix}=begin{pmatrix}0&0\0&0end{pmatrix}iff a=b=c=0.$$
    $endgroup$
    – José Carlos Santos
    Jan 13 at 11:46










  • $begingroup$
    Because the $a$ from an element of $W_1$ doesn't have to be the same $a$ that was used in an element of $W_2$.
    $endgroup$
    – José Carlos Santos
    Jan 13 at 11:49
















$begingroup$
What does previous set is linearly independent means ?
$endgroup$
– Daman deep
Jan 13 at 11:40




$begingroup$
What does previous set is linearly independent means ?
$endgroup$
– Daman deep
Jan 13 at 11:40












$begingroup$
Why it is $a+a^{'}$ not $2a$
$endgroup$
– Daman deep
Jan 13 at 11:44






$begingroup$
Why it is $a+a^{'}$ not $2a$
$endgroup$
– Daman deep
Jan 13 at 11:44














$begingroup$
It means that$$abegin{pmatrix}1&0\0&0end{pmatrix}+bbegin{pmatrix}0&1\0&0end{pmatrix}+cbegin{pmatrix}0&0\1&0end{pmatrix}=begin{pmatrix}0&0\0&0end{pmatrix}iff a=b=c=0.$$
$endgroup$
– José Carlos Santos
Jan 13 at 11:46




$begingroup$
It means that$$abegin{pmatrix}1&0\0&0end{pmatrix}+bbegin{pmatrix}0&1\0&0end{pmatrix}+cbegin{pmatrix}0&0\1&0end{pmatrix}=begin{pmatrix}0&0\0&0end{pmatrix}iff a=b=c=0.$$
$endgroup$
– José Carlos Santos
Jan 13 at 11:46












$begingroup$
Because the $a$ from an element of $W_1$ doesn't have to be the same $a$ that was used in an element of $W_2$.
$endgroup$
– José Carlos Santos
Jan 13 at 11:49




$begingroup$
Because the $a$ from an element of $W_1$ doesn't have to be the same $a$ that was used in an element of $W_2$.
$endgroup$
– José Carlos Santos
Jan 13 at 11:49


















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Cabo Verde

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  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
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Gyllenstierna

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