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If $,v_1,v_2,$ is a basis for a subspace $V$ show that $,v_1+v_2,v_1−v_2,$ is also a basis for $V$ .
I know that we must have to prove that $v_1+v_2$ and $v_1-v_2$ are linearly independent and they also span $V$ because that is a definition for a basis but am unsure how to go about this.
linear-algebra
edited Jul 30 at 8:58
Babelfish
408112
asked Jul 30 at 7:57
J-Dorman
555
add a comment |Â
up vote
0
down vote
favorite
If $,v_1,v_2,$ is a basis for a subspace $V$ show that $,v_1+v_2,v_1−v_2,$ is also a basis for $V$ .
I know that we must have to prove that $v_1+v_2$ and $v_1-v_2$ are linearly independent and they also span $V$ because that is a definition for a basis but am unsure how to go about this.
linear-algebra
edited Jul 30 at 8:58
Babelfish
408112
asked Jul 30 at 7:57
J-Dorman
555
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $,v_1,v_2,$ is a basis for a subspace $V$ show that $,v_1+v_2,v_1−v_2,$ is also a basis for $V$ .
I know that we must have to prove that $v_1+v_2$ and $v_1-v_2$ are linearly independent and they also span $V$ because that is a definition for a basis but am unsure how to go about this.
linear-algebra
edited Jul 30 at 8:58
Babelfish
408112
asked Jul 30 at 7:57
J-Dorman
555
If $,v_1,v_2,$ is a basis for a subspace $V$ show that $,v_1+v_2,v_1−v_2,$ is also a basis for $V$ .
I know that we must have to prove that $v_1+v_2$ and $v_1-v_2$ are linearly independent and they also span $V$ because that is a definition for a basis but am unsure how to go about this.
linear-algebra
edited Jul 30 at 8:58
Babelfish
408112
asked Jul 30 at 7:57
J-Dorman
555
edited Jul 30 at 8:58
Babelfish
408112
edited Jul 30 at 8:58
Babelfish
408112
edited Jul 30 at 8:58
Babelfish
408112
408112
asked Jul 30 at 7:57
J-Dorman
555
asked Jul 30 at 7:57
J-Dorman
555
asked Jul 30 at 7:57
J-Dorman
555
555
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
Let $t(v_1+v_2)+s(v_1-v_2)=0$ , hence $(t+s)v_1+(t-s)v_2=0$. Since $v_1,v_2$ is linearly independent, we get $t+s=0$ and $t-s=0$, hence $t=s=0$.
This shows that $v_1+v_2,v_1-v_2$ is linearly independent.
Since $dim V=2$, we have that $v_1+v_2,v_1-v_2$ span $V$.
answered Jul 30 at 8:03
Fred
37k1237
how does that fact that dim V=2 prove that they span V? can you elaborate?
– J-Dorman
Jul 30 at 8:28
There are several charakterisations for basis. One is, that a basis for $V$ is a maximal linear independent set. The set $, v_1+v_2,$ is linear independent. Every linear independet of $V$ set can be enlarged to a basis of $V$. So there is a basis of size $geq 2$ where those two vectors are a subset. But every basis of $V$ has the same cardinality, which is $2$. So $, v_1+v_2,$ already was a basis.
– Babelfish
Jul 30 at 8:51
@Babelfish I don't understand
– J-Dorman
Jul 30 at 9:23
@J-Dorman Which part do you not understand? It is hard to estimate your linear algebra knowledge
– Babelfish
Jul 30 at 9:28
add a comment |Â
up vote
0
down vote
To show LI, you need to show that the only linear combination of the vectors $v_1+v_2$ and $v_1-v_2$ that gives the zero vector is the zero combination, i.e. the following equation has ONLY the trivial solution.
$$c_1(v_1+v_2)+c_2(v_1-v_2)=0.$$
This amounts to
$$(c_1+c_2)v_1+(c_1-c_2)v_2=0.$$
Now $v_1, v_2$ are LI, therefore these coefficients must be $0$. Thus we have
beginalign*
c_1+c_2 & = 0\
c_1-c_2 & = 0.
endalign*
It is easy to see that it has only the trivial (zero solution). Hence the new set is LI.
Showing span is easy or you can use the fact that the given subspace is 2D.
answered Jul 30 at 8:06
Anurag A
22.2k12243
add a comment |Â
up vote
0
down vote
Here are several possible arguments:
Set $w_1=v_1+v_2$, $;w_2=v_1-v_2$.
- As, by hypothesis, $V$ has dimension $2$, it is enough to prove either that $w_1$ and $w_2$ are linearly independent (which amounts to showing a linear system has only the trivial solution), or that $v_1, v_2in langle w_1,w_2rangle$.
$qquad$This second way is easy: form the equations defining $w_1$ and $w_2$, you deduce readily
$$v_1=tfrac12(w_1+w_2),quad v_2=tfrac12(w_1-w_2).$$
- If you know about determinants, observe the column matrices $;beginbmatrixv_1newline v_2endbmatrix$ and $;beginbmatrixw_1newline w_2endbmatrix$ are linked by the relation
$$beginbmatrixw_1\ w_2endbmatrix=beginbmatrix1&phantom-1\ 1&-1endbmatrixbeginbmatrixv_1newline v_2endbmatrix,$$
and the matrix $;A=beginbmatrix1&phantom-1\ 1&-1endbmatrix$ is invertible since $det A=-2$, so it maps a basis onto a basis.
answered Jul 30 at 8:39
Bernard
110k635102
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Let $t(v_1+v_2)+s(v_1-v_2)=0$ , hence $(t+s)v_1+(t-s)v_2=0$. Since $v_1,v_2$ is linearly independent, we get $t+s=0$ and $t-s=0$, hence $t=s=0$.
This shows that $v_1+v_2,v_1-v_2$ is linearly independent.
Since $dim V=2$, we have that $v_1+v_2,v_1-v_2$ span $V$.
answered Jul 30 at 8:03
Fred
37k1237
how does that fact that dim V=2 prove that they span V? can you elaborate?
– J-Dorman
Jul 30 at 8:28
There are several charakterisations for basis. One is, that a basis for $V$ is a maximal linear independent set. The set $, v_1+v_2,$ is linear independent. Every linear independet of $V$ set can be enlarged to a basis of $V$. So there is a basis of size $geq 2$ where those two vectors are a subset. But every basis of $V$ has the same cardinality, which is $2$. So $, v_1+v_2,$ already was a basis.
– Babelfish
Jul 30 at 8:51
@Babelfish I don't understand
– J-Dorman
Jul 30 at 9:23
@J-Dorman Which part do you not understand? It is hard to estimate your linear algebra knowledge
– Babelfish
Jul 30 at 9:28
add a comment |Â
up vote
2
down vote
Let $t(v_1+v_2)+s(v_1-v_2)=0$ , hence $(t+s)v_1+(t-s)v_2=0$. Since $v_1,v_2$ is linearly independent, we get $t+s=0$ and $t-s=0$, hence $t=s=0$.
This shows that $v_1+v_2,v_1-v_2$ is linearly independent.
Since $dim V=2$, we have that $v_1+v_2,v_1-v_2$ span $V$.
answered Jul 30 at 8:03
Fred
37k1237
how does that fact that dim V=2 prove that they span V? can you elaborate?
– J-Dorman
Jul 30 at 8:28
There are several charakterisations for basis. One is, that a basis for $V$ is a maximal linear independent set. The set $, v_1+v_2,$ is linear independent. Every linear independet of $V$ set can be enlarged to a basis of $V$. So there is a basis of size $geq 2$ where those two vectors are a subset. But every basis of $V$ has the same cardinality, which is $2$. So $, v_1+v_2,$ already was a basis.
– Babelfish
Jul 30 at 8:51
@Babelfish I don't understand
– J-Dorman
Jul 30 at 9:23
@J-Dorman Which part do you not understand? It is hard to estimate your linear algebra knowledge
– Babelfish
Jul 30 at 9:28
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let $t(v_1+v_2)+s(v_1-v_2)=0$ , hence $(t+s)v_1+(t-s)v_2=0$. Since $v_1,v_2$ is linearly independent, we get $t+s=0$ and $t-s=0$, hence $t=s=0$.
This shows that $v_1+v_2,v_1-v_2$ is linearly independent.
Since $dim V=2$, we have that $v_1+v_2,v_1-v_2$ span $V$.
answered Jul 30 at 8:03
Fred
37k1237
Let $t(v_1+v_2)+s(v_1-v_2)=0$ , hence $(t+s)v_1+(t-s)v_2=0$. Since $v_1,v_2$ is linearly independent, we get $t+s=0$ and $t-s=0$, hence $t=s=0$.
This shows that $v_1+v_2,v_1-v_2$ is linearly independent.
Since $dim V=2$, we have that $v_1+v_2,v_1-v_2$ span $V$.
answered Jul 30 at 8:03
Fred
37k1237
answered Jul 30 at 8:03
Fred
37k1237
answered Jul 30 at 8:03
Fred
37k1237
answered Jul 30 at 8:03
Fred
37k1237
37k1237
how does that fact that dim V=2 prove that they span V? can you elaborate?
– J-Dorman
Jul 30 at 8:28
There are several charakterisations for basis. One is, that a basis for $V$ is a maximal linear independent set. The set $, v_1+v_2,$ is linear independent. Every linear independet of $V$ set can be enlarged to a basis of $V$. So there is a basis of size $geq 2$ where those two vectors are a subset. But every basis of $V$ has the same cardinality, which is $2$. So $, v_1+v_2,$ already was a basis.
– Babelfish
Jul 30 at 8:51
@Babelfish I don't understand
– J-Dorman
Jul 30 at 9:23
@J-Dorman Which part do you not understand? It is hard to estimate your linear algebra knowledge
– Babelfish
Jul 30 at 9:28
add a comment |Â
how does that fact that dim V=2 prove that they span V? can you elaborate?
– J-Dorman
Jul 30 at 8:28
There are several charakterisations for basis. One is, that a basis for $V$ is a maximal linear independent set. The set $, v_1+v_2,$ is linear independent. Every linear independet of $V$ set can be enlarged to a basis of $V$. So there is a basis of size $geq 2$ where those two vectors are a subset. But every basis of $V$ has the same cardinality, which is $2$. So $, v_1+v_2,$ already was a basis.
– Babelfish
Jul 30 at 8:51
@Babelfish I don't understand
– J-Dorman
Jul 30 at 9:23
@J-Dorman Which part do you not understand? It is hard to estimate your linear algebra knowledge
– Babelfish
Jul 30 at 9:28
how does that fact that dim V=2 prove that they span V? can you elaborate?
– J-Dorman
Jul 30 at 8:28
how does that fact that dim V=2 prove that they span V? can you elaborate?
– J-Dorman
Jul 30 at 8:28
There are several charakterisations for basis. One is, that a basis for $V$ is a maximal linear independent set. The set $, v_1+v_2,$ is linear independent. Every linear independet of $V$ set can be enlarged to a basis of $V$. So there is a basis of size $geq 2$ where those two vectors are a subset. But every basis of $V$ has the same cardinality, which is $2$. So $, v_1+v_2,$ already was a basis.
– Babelfish
Jul 30 at 8:51
There are several charakterisations for basis. One is, that a basis for $V$ is a maximal linear independent set. The set $, v_1+v_2,$ is linear independent. Every linear independet of $V$ set can be enlarged to a basis of $V$. So there is a basis of size $geq 2$ where those two vectors are a subset. But every basis of $V$ has the same cardinality, which is $2$. So $, v_1+v_2,$ already was a basis.
– Babelfish
Jul 30 at 8:51
@Babelfish I don't understand
– J-Dorman
Jul 30 at 9:23
@Babelfish I don't understand
– J-Dorman
Jul 30 at 9:23
@J-Dorman Which part do you not understand? It is hard to estimate your linear algebra knowledge
– Babelfish
Jul 30 at 9:28
@J-Dorman Which part do you not understand? It is hard to estimate your linear algebra knowledge
– Babelfish
Jul 30 at 9:28
add a comment |Â
up vote
0
down vote
To show LI, you need to show that the only linear combination of the vectors $v_1+v_2$ and $v_1-v_2$ that gives the zero vector is the zero combination, i.e. the following equation has ONLY the trivial solution.
$$c_1(v_1+v_2)+c_2(v_1-v_2)=0.$$
This amounts to
$$(c_1+c_2)v_1+(c_1-c_2)v_2=0.$$
Now $v_1, v_2$ are LI, therefore these coefficients must be $0$. Thus we have
beginalign*
c_1+c_2 & = 0\
c_1-c_2 & = 0.
endalign*
It is easy to see that it has only the trivial (zero solution). Hence the new set is LI.
Showing span is easy or you can use the fact that the given subspace is 2D.
answered Jul 30 at 8:06
Anurag A
22.2k12243
add a comment |Â
up vote
0
down vote
To show LI, you need to show that the only linear combination of the vectors $v_1+v_2$ and $v_1-v_2$ that gives the zero vector is the zero combination, i.e. the following equation has ONLY the trivial solution.
$$c_1(v_1+v_2)+c_2(v_1-v_2)=0.$$
This amounts to
$$(c_1+c_2)v_1+(c_1-c_2)v_2=0.$$
Now $v_1, v_2$ are LI, therefore these coefficients must be $0$. Thus we have
beginalign*
c_1+c_2 & = 0\
c_1-c_2 & = 0.
endalign*
It is easy to see that it has only the trivial (zero solution). Hence the new set is LI.
Showing span is easy or you can use the fact that the given subspace is 2D.
answered Jul 30 at 8:06
Anurag A
22.2k12243
add a comment |Â
up vote
0
down vote
up vote
0
down vote
To show LI, you need to show that the only linear combination of the vectors $v_1+v_2$ and $v_1-v_2$ that gives the zero vector is the zero combination, i.e. the following equation has ONLY the trivial solution.
$$c_1(v_1+v_2)+c_2(v_1-v_2)=0.$$
This amounts to
$$(c_1+c_2)v_1+(c_1-c_2)v_2=0.$$
Now $v_1, v_2$ are LI, therefore these coefficients must be $0$. Thus we have
beginalign*
c_1+c_2 & = 0\
c_1-c_2 & = 0.
endalign*
It is easy to see that it has only the trivial (zero solution). Hence the new set is LI.
Showing span is easy or you can use the fact that the given subspace is 2D.
answered Jul 30 at 8:06
Anurag A
22.2k12243
To show LI, you need to show that the only linear combination of the vectors $v_1+v_2$ and $v_1-v_2$ that gives the zero vector is the zero combination, i.e. the following equation has ONLY the trivial solution.
$$c_1(v_1+v_2)+c_2(v_1-v_2)=0.$$
This amounts to
$$(c_1+c_2)v_1+(c_1-c_2)v_2=0.$$
Now $v_1, v_2$ are LI, therefore these coefficients must be $0$. Thus we have
beginalign*
c_1+c_2 & = 0\
c_1-c_2 & = 0.
endalign*
It is easy to see that it has only the trivial (zero solution). Hence the new set is LI.
Showing span is easy or you can use the fact that the given subspace is 2D.
answered Jul 30 at 8:06
Anurag A
22.2k12243
answered Jul 30 at 8:06
Anurag A
22.2k12243
answered Jul 30 at 8:06
Anurag A
22.2k12243
answered Jul 30 at 8:06
Anurag A
22.2k12243
22.2k12243
add a comment |Â
add a comment |Â
up vote
0
down vote
Here are several possible arguments:
Set $w_1=v_1+v_2$, $;w_2=v_1-v_2$.
- As, by hypothesis, $V$ has dimension $2$, it is enough to prove either that $w_1$ and $w_2$ are linearly independent (which amounts to showing a linear system has only the trivial solution), or that $v_1, v_2in langle w_1,w_2rangle$.
$qquad$This second way is easy: form the equations defining $w_1$ and $w_2$, you deduce readily
$$v_1=tfrac12(w_1+w_2),quad v_2=tfrac12(w_1-w_2).$$
- If you know about determinants, observe the column matrices $;beginbmatrixv_1newline v_2endbmatrix$ and $;beginbmatrixw_1newline w_2endbmatrix$ are linked by the relation
$$beginbmatrixw_1\ w_2endbmatrix=beginbmatrix1&phantom-1\ 1&-1endbmatrixbeginbmatrixv_1newline v_2endbmatrix,$$
and the matrix $;A=beginbmatrix1&phantom-1\ 1&-1endbmatrix$ is invertible since $det A=-2$, so it maps a basis onto a basis.
answered Jul 30 at 8:39
Bernard
110k635102
add a comment |Â
up vote
0
down vote
Here are several possible arguments:
Set $w_1=v_1+v_2$, $;w_2=v_1-v_2$.
- As, by hypothesis, $V$ has dimension $2$, it is enough to prove either that $w_1$ and $w_2$ are linearly independent (which amounts to showing a linear system has only the trivial solution), or that $v_1, v_2in langle w_1,w_2rangle$.
$qquad$This second way is easy: form the equations defining $w_1$ and $w_2$, you deduce readily
$$v_1=tfrac12(w_1+w_2),quad v_2=tfrac12(w_1-w_2).$$
- If you know about determinants, observe the column matrices $;beginbmatrixv_1newline v_2endbmatrix$ and $;beginbmatrixw_1newline w_2endbmatrix$ are linked by the relation
$$beginbmatrixw_1\ w_2endbmatrix=beginbmatrix1&phantom-1\ 1&-1endbmatrixbeginbmatrixv_1newline v_2endbmatrix,$$
and the matrix $;A=beginbmatrix1&phantom-1\ 1&-1endbmatrix$ is invertible since $det A=-2$, so it maps a basis onto a basis.
answered Jul 30 at 8:39
Bernard
110k635102
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Here are several possible arguments:
Set $w_1=v_1+v_2$, $;w_2=v_1-v_2$.
- As, by hypothesis, $V$ has dimension $2$, it is enough to prove either that $w_1$ and $w_2$ are linearly independent (which amounts to showing a linear system has only the trivial solution), or that $v_1, v_2in langle w_1,w_2rangle$.
$qquad$This second way is easy: form the equations defining $w_1$ and $w_2$, you deduce readily
$$v_1=tfrac12(w_1+w_2),quad v_2=tfrac12(w_1-w_2).$$
- If you know about determinants, observe the column matrices $;beginbmatrixv_1newline v_2endbmatrix$ and $;beginbmatrixw_1newline w_2endbmatrix$ are linked by the relation
$$beginbmatrixw_1\ w_2endbmatrix=beginbmatrix1&phantom-1\ 1&-1endbmatrixbeginbmatrixv_1newline v_2endbmatrix,$$
and the matrix $;A=beginbmatrix1&phantom-1\ 1&-1endbmatrix$ is invertible since $det A=-2$, so it maps a basis onto a basis.
answered Jul 30 at 8:39
Bernard
110k635102
Here are several possible arguments:
Set $w_1=v_1+v_2$, $;w_2=v_1-v_2$.
- As, by hypothesis, $V$ has dimension $2$, it is enough to prove either that $w_1$ and $w_2$ are linearly independent (which amounts to showing a linear system has only the trivial solution), or that $v_1, v_2in langle w_1,w_2rangle$.
$qquad$This second way is easy: form the equations defining $w_1$ and $w_2$, you deduce readily
$$v_1=tfrac12(w_1+w_2),quad v_2=tfrac12(w_1-w_2).$$
- If you know about determinants, observe the column matrices $;beginbmatrixv_1newline v_2endbmatrix$ and $;beginbmatrixw_1newline w_2endbmatrix$ are linked by the relation
$$beginbmatrixw_1\ w_2endbmatrix=beginbmatrix1&phantom-1\ 1&-1endbmatrixbeginbmatrixv_1newline v_2endbmatrix,$$
and the matrix $;A=beginbmatrix1&phantom-1\ 1&-1endbmatrix$ is invertible since $det A=-2$, so it maps a basis onto a basis.
answered Jul 30 at 8:39
Bernard
110k635102
answered Jul 30 at 8:39
Bernard
110k635102
answered Jul 30 at 8:39
Bernard
110k635102
answered Jul 30 at 8:39
Bernard
110k635102
110k635102
add a comment |Â
add a comment |Â
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