Mixing
how to prove a spanning set of polynomial
$begingroup$
I am struggling so much understanding this concept of subspace and span.
The question is, Given that
$P2:W={(x+1)(ax+b)| a,b in R}$
show that
${x^2+x, x^2+2x+1}$
is a spanning set of $W$.
I don't know if I got this concept right, but I've tried to do things by letting
$p(x)=x^2+x$
$q(x)=x^2+2x+1$
then multiplying them a coefficient $alpha$ and $beta$ each and adding to fit in with $W$.
but then I got an answer saying $a=alpha + beta$, $a+b = alpha+ 2beta, b=beta$, which means... no solution? I am guessing? so this does not span $W$. Am I right?
linear-algebra vector-spaces
edited Jan 28 at 15:30
Thomas Shelby
4,5142726
asked Jan 28 at 14:09
nokdoonokdoo
161
$endgroup$
add a comment |
$begingroup$
I am struggling so much understanding this concept of subspace and span.
The question is, Given that
$P2:W={(x+1)(ax+b)| a,b in R}$
show that
${x^2+x, x^2+2x+1}$
is a spanning set of $W$.
I don't know if I got this concept right, but I've tried to do things by letting
$p(x)=x^2+x$
$q(x)=x^2+2x+1$
then multiplying them a coefficient $alpha$ and $beta$ each and adding to fit in with $W$.
but then I got an answer saying $a=alpha + beta$, $a+b = alpha+ 2beta, b=beta$, which means... no solution? I am guessing? so this does not span $W$. Am I right?
linear-algebra vector-spaces
edited Jan 28 at 15:30
Thomas Shelby
4,5142726
asked Jan 28 at 14:09
nokdoonokdoo
161
$endgroup$
add a comment |
$begingroup$
I am struggling so much understanding this concept of subspace and span.
The question is, Given that
$P2:W={(x+1)(ax+b)| a,b in R}$
show that
${x^2+x, x^2+2x+1}$
is a spanning set of $W$.
I don't know if I got this concept right, but I've tried to do things by letting
$p(x)=x^2+x$
$q(x)=x^2+2x+1$
then multiplying them a coefficient $alpha$ and $beta$ each and adding to fit in with $W$.
but then I got an answer saying $a=alpha + beta$, $a+b = alpha+ 2beta, b=beta$, which means... no solution? I am guessing? so this does not span $W$. Am I right?
linear-algebra vector-spaces
edited Jan 28 at 15:30
Thomas Shelby
4,5142726
asked Jan 28 at 14:09
nokdoonokdoo
161
$endgroup$
I am struggling so much understanding this concept of subspace and span.
The question is, Given that
$P2:W={(x+1)(ax+b)| a,b in R}$
show that
${x^2+x, x^2+2x+1}$
is a spanning set of $W$.
I don't know if I got this concept right, but I've tried to do things by letting
$p(x)=x^2+x$
$q(x)=x^2+2x+1$
then multiplying them a coefficient $alpha$ and $beta$ each and adding to fit in with $W$.
but then I got an answer saying $a=alpha + beta$, $a+b = alpha+ 2beta, b=beta$, which means... no solution? I am guessing? so this does not span $W$. Am I right?
linear-algebra vector-spaces
linear-algebra vector-spaces
edited Jan 28 at 15:30
Thomas Shelby
4,5142726
asked Jan 28 at 14:09
nokdoonokdoo
161
edited Jan 28 at 15:30
Thomas Shelby
4,5142726
asked Jan 28 at 14:09
nokdoonokdoo
161
edited Jan 28 at 15:30
Thomas Shelby
4,5142726
edited Jan 28 at 15:30
Thomas Shelby
4,5142726
edited Jan 28 at 15:30
Thomas Shelby
4,5142726
4,5142726
asked Jan 28 at 14:09
nokdoonokdoo
161
asked Jan 28 at 14:09
nokdoonokdoo
161
asked Jan 28 at 14:09
nokdoonokdoo
161
161
add a comment |
add a comment |
3 Answers
3
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oldest
votes
$begingroup$
$(x+1)(ax+b)=ax^2+(b+a)x+b.$
Claim: There exists $alpha, betain Bbb R $ such that
$$ax^2+(b+a)x+b=alpha(x^2+x)+beta( x^2+2x+1)tag1$$
$$ax^2+(b+a)x+b=(alpha+beta)x^2+(alpha+2beta)x+beta$$
Comparing coefficients, we get
$$a=alpha+betatag2$$
$$b+a=alpha+2betatag3$$
$$b=beta tag4$$
Substituting $(4) $ in $(1) $ gives $alpha=a-b$ (you can verify the solution using $(3)$).
In other words, $forall a,binBbb R $, there exists $alpha=a-b, beta =b $ such that $(1) $ holds. Hence ${x^2+x,x^2+2x+1} $ is indeed a spanning set.
answered Jan 28 at 14:24
Thomas ShelbyThomas Shelby
4,5142726
$endgroup$
add a comment |
$begingroup$
No, you are wrong. Note that$$x^2+x=(x+1)xtext{ and that }x^2+2x+1=(x+1)(x+1).$$So, both polynomails $x^2+x$ and $x^2+2x+1$ belong indeed to $W$ and therefore they span a subspace of $W$.
On the other hand, if $(x+1)(ax+b)in W$, thenbegin{align}(x+1)(ax+b)&=(a-b)(x+1)x+b(x+1)(x+1)\&=(a-b)(x^2+x)+b(x^2+2x+1)end{align}which belongs, of course, to $operatorname{span}bigl({x^2+x,x^2+2x+1}bigr)$. So, $W=operatorname{span}bigl({x^2+x,x^2+2x+1}bigr)$.
answered Jan 28 at 14:19
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
Hint: First, $W$ is a $2$-dimensional vector space (easy to see).
Now, ${x^2+x,x^2+2x+1}$ is linearly independent (easy to see)
Let $a=1,b=0$. We get $x^2+x$. Now let $a=1,b=1$. We get $x^2+2x+1$.
Thus $W=operatorname{span} {x^2+x,x^2+2x+1}$.
.
answered Jan 28 at 14:38
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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active
oldest
votes
$begingroup$
$(x+1)(ax+b)=ax^2+(b+a)x+b.$
Claim: There exists $alpha, betain Bbb R $ such that
$$ax^2+(b+a)x+b=alpha(x^2+x)+beta( x^2+2x+1)tag1$$
$$ax^2+(b+a)x+b=(alpha+beta)x^2+(alpha+2beta)x+beta$$
Comparing coefficients, we get
$$a=alpha+betatag2$$
$$b+a=alpha+2betatag3$$
$$b=beta tag4$$
Substituting $(4) $ in $(1) $ gives $alpha=a-b$ (you can verify the solution using $(3)$).
In other words, $forall a,binBbb R $, there exists $alpha=a-b, beta =b $ such that $(1) $ holds. Hence ${x^2+x,x^2+2x+1} $ is indeed a spanning set.
answered Jan 28 at 14:24
Thomas ShelbyThomas Shelby
4,5142726
$endgroup$
add a comment |
$begingroup$
$(x+1)(ax+b)=ax^2+(b+a)x+b.$
Claim: There exists $alpha, betain Bbb R $ such that
$$ax^2+(b+a)x+b=alpha(x^2+x)+beta( x^2+2x+1)tag1$$
$$ax^2+(b+a)x+b=(alpha+beta)x^2+(alpha+2beta)x+beta$$
Comparing coefficients, we get
$$a=alpha+betatag2$$
$$b+a=alpha+2betatag3$$
$$b=beta tag4$$
Substituting $(4) $ in $(1) $ gives $alpha=a-b$ (you can verify the solution using $(3)$).
In other words, $forall a,binBbb R $, there exists $alpha=a-b, beta =b $ such that $(1) $ holds. Hence ${x^2+x,x^2+2x+1} $ is indeed a spanning set.
answered Jan 28 at 14:24
Thomas ShelbyThomas Shelby
4,5142726
$endgroup$
add a comment |
$begingroup$
$(x+1)(ax+b)=ax^2+(b+a)x+b.$
Claim: There exists $alpha, betain Bbb R $ such that
$$ax^2+(b+a)x+b=alpha(x^2+x)+beta( x^2+2x+1)tag1$$
$$ax^2+(b+a)x+b=(alpha+beta)x^2+(alpha+2beta)x+beta$$
Comparing coefficients, we get
$$a=alpha+betatag2$$
$$b+a=alpha+2betatag3$$
$$b=beta tag4$$
Substituting $(4) $ in $(1) $ gives $alpha=a-b$ (you can verify the solution using $(3)$).
In other words, $forall a,binBbb R $, there exists $alpha=a-b, beta =b $ such that $(1) $ holds. Hence ${x^2+x,x^2+2x+1} $ is indeed a spanning set.
answered Jan 28 at 14:24
Thomas ShelbyThomas Shelby
4,5142726
$endgroup$
$(x+1)(ax+b)=ax^2+(b+a)x+b.$
Claim: There exists $alpha, betain Bbb R $ such that
$$ax^2+(b+a)x+b=alpha(x^2+x)+beta( x^2+2x+1)tag1$$
$$ax^2+(b+a)x+b=(alpha+beta)x^2+(alpha+2beta)x+beta$$
Comparing coefficients, we get
$$a=alpha+betatag2$$
$$b+a=alpha+2betatag3$$
$$b=beta tag4$$
Substituting $(4) $ in $(1) $ gives $alpha=a-b$ (you can verify the solution using $(3)$).
In other words, $forall a,binBbb R $, there exists $alpha=a-b, beta =b $ such that $(1) $ holds. Hence ${x^2+x,x^2+2x+1} $ is indeed a spanning set.
answered Jan 28 at 14:24
Thomas ShelbyThomas Shelby
4,5142726
answered Jan 28 at 14:24
Thomas ShelbyThomas Shelby
4,5142726
answered Jan 28 at 14:24
Thomas ShelbyThomas Shelby
4,5142726
answered Jan 28 at 14:24
Thomas ShelbyThomas Shelby
4,5142726
4,5142726
add a comment |
add a comment |
$begingroup$
No, you are wrong. Note that$$x^2+x=(x+1)xtext{ and that }x^2+2x+1=(x+1)(x+1).$$So, both polynomails $x^2+x$ and $x^2+2x+1$ belong indeed to $W$ and therefore they span a subspace of $W$.
On the other hand, if $(x+1)(ax+b)in W$, thenbegin{align}(x+1)(ax+b)&=(a-b)(x+1)x+b(x+1)(x+1)\&=(a-b)(x^2+x)+b(x^2+2x+1)end{align}which belongs, of course, to $operatorname{span}bigl({x^2+x,x^2+2x+1}bigr)$. So, $W=operatorname{span}bigl({x^2+x,x^2+2x+1}bigr)$.
answered Jan 28 at 14:19
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
No, you are wrong. Note that$$x^2+x=(x+1)xtext{ and that }x^2+2x+1=(x+1)(x+1).$$So, both polynomails $x^2+x$ and $x^2+2x+1$ belong indeed to $W$ and therefore they span a subspace of $W$.
On the other hand, if $(x+1)(ax+b)in W$, thenbegin{align}(x+1)(ax+b)&=(a-b)(x+1)x+b(x+1)(x+1)\&=(a-b)(x^2+x)+b(x^2+2x+1)end{align}which belongs, of course, to $operatorname{span}bigl({x^2+x,x^2+2x+1}bigr)$. So, $W=operatorname{span}bigl({x^2+x,x^2+2x+1}bigr)$.
answered Jan 28 at 14:19
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
No, you are wrong. Note that$$x^2+x=(x+1)xtext{ and that }x^2+2x+1=(x+1)(x+1).$$So, both polynomails $x^2+x$ and $x^2+2x+1$ belong indeed to $W$ and therefore they span a subspace of $W$.
On the other hand, if $(x+1)(ax+b)in W$, thenbegin{align}(x+1)(ax+b)&=(a-b)(x+1)x+b(x+1)(x+1)\&=(a-b)(x^2+x)+b(x^2+2x+1)end{align}which belongs, of course, to $operatorname{span}bigl({x^2+x,x^2+2x+1}bigr)$. So, $W=operatorname{span}bigl({x^2+x,x^2+2x+1}bigr)$.
answered Jan 28 at 14:19
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
No, you are wrong. Note that$$x^2+x=(x+1)xtext{ and that }x^2+2x+1=(x+1)(x+1).$$So, both polynomails $x^2+x$ and $x^2+2x+1$ belong indeed to $W$ and therefore they span a subspace of $W$.
On the other hand, if $(x+1)(ax+b)in W$, thenbegin{align}(x+1)(ax+b)&=(a-b)(x+1)x+b(x+1)(x+1)\&=(a-b)(x^2+x)+b(x^2+2x+1)end{align}which belongs, of course, to $operatorname{span}bigl({x^2+x,x^2+2x+1}bigr)$. So, $W=operatorname{span}bigl({x^2+x,x^2+2x+1}bigr)$.
answered Jan 28 at 14:19
José Carlos SantosJosé Carlos Santos
171k23132240
edited Jan 28 at 14:26
answered Jan 28 at 14:19
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 28 at 14:19
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 28 at 14:19
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
add a comment |
add a comment |
$begingroup$
Hint: First, $W$ is a $2$-dimensional vector space (easy to see).
Now, ${x^2+x,x^2+2x+1}$ is linearly independent (easy to see)
Let $a=1,b=0$. We get $x^2+x$. Now let $a=1,b=1$. We get $x^2+2x+1$.
Thus $W=operatorname{span} {x^2+x,x^2+2x+1}$.
.
answered Jan 28 at 14:38
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
Hint: First, $W$ is a $2$-dimensional vector space (easy to see).
Now, ${x^2+x,x^2+2x+1}$ is linearly independent (easy to see)
Let $a=1,b=0$. We get $x^2+x$. Now let $a=1,b=1$. We get $x^2+2x+1$.
Thus $W=operatorname{span} {x^2+x,x^2+2x+1}$.
.
answered Jan 28 at 14:38
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
Hint: First, $W$ is a $2$-dimensional vector space (easy to see).
Now, ${x^2+x,x^2+2x+1}$ is linearly independent (easy to see)
Let $a=1,b=0$. We get $x^2+x$. Now let $a=1,b=1$. We get $x^2+2x+1$.
Thus $W=operatorname{span} {x^2+x,x^2+2x+1}$.
.
answered Jan 28 at 14:38
Chris CusterChris Custer
14.2k3827
$endgroup$
Hint: First, $W$ is a $2$-dimensional vector space (easy to see).
Now, ${x^2+x,x^2+2x+1}$ is linearly independent (easy to see)
Let $a=1,b=0$. We get $x^2+x$. Now let $a=1,b=1$. We get $x^2+2x+1$.
Thus $W=operatorname{span} {x^2+x,x^2+2x+1}$.
.
answered Jan 28 at 14:38
Chris CusterChris Custer
14.2k3827
edited Jan 28 at 15:06
answered Jan 28 at 14:38
Chris CusterChris Custer
14.2k3827
answered Jan 28 at 14:38
Chris CusterChris Custer
14.2k3827
answered Jan 28 at 14:38
Chris CusterChris Custer
14.2k3827
14.2k3827
add a comment |
add a comment |
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