MHB Find a Basis for Subspace in P_3(\mathbb{R})

[Petrus]

Hello,
Find a basis for subspace in $$P_3(\mathbb{R})$$ that containrar polynomial $$1+x, -1+x, 2x$$ Also the hole ker T there $$T: P_3(\mathbb{R})-> P_3(\mathbb{R})$$ defines of $$T(a+bx+cx^2+dx^3)=(a+b)x+(c+d)x^2$$

I am unsure how to handle with that ker.. I am aware that My bas determinant $$\neq0$$ well I did try but I have no clue if I did correct.. And Also please tell me if I need to rotate the picture cause I can't se if it is wrong for pc!

I mean basis when I say bas in the picture!
Regards,
$$|\pi\rangle$$

 

[Deveno]

Your question does not make much sense as posted.

If you mean the subspace of $P_3(\Bbb R)$ GENERATED by $\{1+x,-1+x,2x\}$, it should be clear a basis is:

$\{1+x,-1+x\}$, since:

$(1+x) + (-1+x) = 2x$ so this smaller set spans the same space as the larger set, and if:

$c_1(1+x) + c_2(-1+x) = 0$, then:

$c_1 - c_2 = 0$
$c_1 + c_2 = 0$

and the only solution is: $c_1 = c_2 = 0$, so the smaller set is linearly independent.

You do not make it clear how or if this relates to the linear transformation $T$, but I can assure you that since $T$ is not onto, the matrix of $T$ (relative to ANY basis) will NOT have non-zero determinant.

 

[Petrus]

Your question does not make much sense as posted.

If you mean the subspace of $P_3(\Bbb R)$ GENERATED by $\{1+x,-1+x,2x\}$, it should be clear a basis is:

$\{1+x,-1+x\}$, since:

$(1+x) + (-1+x) = 2x$ so this smaller set spans the same space as the larger set, and if:

$c_1(1+x) + c_2(-1+x) = 0$, then:

$c_1 - c_2 = 0$
$c_1 + c_2 = 0$

and the only solution is: $c_1 = c_2 = 0$, so the smaller set is linearly independent.

You do not make it clear how or if this relates to the linear transformation $T$, but I can assure you that since $T$ is not onto, the matrix of $T$ (relative to ANY basis) will NOT have non-zero determinant.

Hello,
Sorry for the bad translate, is this better?
Find a basis for subspace in $$P_3(\mathbb{R})$$ that containrar polynomial $$1+x, -1+x, 2x$$ and whole ker T where $$T: P_3(\mathbb{R})-> P_3(\mathbb{R})$$ defines of $$T(a+bx+cx^2+dx^3)=(a+b)x+(c+d)x^2$$

I understand the progress and WHY I do it, so absolut I am going to put this as column in matrice and Then add unit matrice and do row echelon, Then I am suposed to find all pivot element and that number of that column is My basis from original! Then I can put all My basis as a matrice and confirmed it is a basis with $$det \neq 0$$

 

[Deveno]

OK, I understand there might be some language issues...but clarity is essential in math (otherwise we might wind up answering the wrong question).

If your basis for "some subspace" (which subspace? How is the subspace defined?) must contain the 3 polynomials you list, AND the kernel of $T$, we need to find out what this kernel IS.

It should hopefully be clear to you that if:

$a + bx + cx^2 + dx^3 \in \text{ker}(T)$ that

$a = -b; c = -d$. So we can choose $a$ and $c$ freely, and then $b$ and $d$ are completely determined, our polynomials in $\text{ker}(T)$ are of the form:

$a(1 - x) + c(x^2 - x^3)$.

This shows that a basis for $\text{ker}(T)$ is $\{1 - x,x^2 -x^3\}$ (Prove this is a basis!).

Now one of THESE vectors, is just a scalar multiple of the first 2 we found in my previous post:

$1 - x = (-1)(-1 + x)$.

So if $\{1 + x, -1 + x, x^2 - x^3\}$ is linearly independent, it generates a subspace that contains the three polynomials in your original post, AND all of the kernel of $T$.

Now: how do you show a set is linearly independent?

Again, if you are going to find a basis of a SUBSPACE, the basis vectors for that subspace WILL NOT form a 4x4 matrix with non-zero determinant, UNLESS the subspace is actually ALL of $P_3(\Bbb R)$.

Personally, I think row-reduction is the wrong tool for the job. Use the DEFINITIONS.

 

[Petrus]

Hello,
That basis for ker T i Dont think I understand how you got them, with that method you did not like I did find the basis(which is what we learned and use most when teacher lecture) I got the basis $$1-x,x^2-x^3,1+x,x^3$$ which got $$det=2$$, may I ask if this is correctly? And may I ask if you got any tips for me to understand that ker part? I want to understand that part

Regards,
$$|\pi\rangle$$

 

[Deveno]

Sure.

By definition:

$\text{ker}(T) = \{p(x) \in P_3(\Bbb R) : T(p(x)) = 0\}$

This means that:

$T(p(x)) = T(a + bx + cx^2 + dx^3) = (a+b)x + (c+d)x^2 = 0 = 0 + 0x + 0x^2 + 0x^3$.

Equating coefficients, we have:

$a+b = 0$
$c+d = 0$.

Hence $b = -a$ and $d = -c$.

Alternatively, in the basis $B = \{1,x,x^2,x^3\}, T$ has the matrix:

$[T]_B = \begin{bmatrix}0&0&0&0\\1&1&0&0\\0&0&1&1\\0&0&0&0 \end{bmatrix}$

By inspection, it is clear that rank(T) is at most 2 (it has 2 zero rows). It is also clear that the 2 non-zero rows are linearly independent, since they act on disjoint sets of coordinates. Hence we conclude that rank(T) = 2. By the rank-nullity theorem, the kernel has dimension 2. Can we find two polynomials that are LI in the kernel? The fact that the 2nd row only uses $a$ and $b$ of the input domain column vector:

$(a,b,c,d)^T$

suggests we find a polynomial of the form:

$a + bx$ in the kernel. Since we know $b = -a$ any polynomial of the form $a - ax = a(1 - x)$ will do (these are vectors of the form $(a,-a,0,0)^T$ in the basis $B$).

Similar reasoning leads to the other polynomial of the form $cx^2 - cx^3 = c(x^2 - x^3)$.

Now these ARE linearly independent, since if:

$r_1(1 - x) + r_2(x^2 - x^3) = r_1 - r_1x + r_2x^2 - r_3x^3 = 0 = 0 + 0x + 0x^2 + 0x^3$

We have:

$r_1 = -r_1 = 0$
$r_2 = -r_2 = 0$.

So $\{1 - x, x^2 - x^3\}$ forms a basis for $\text{ker}(T)$, since we already know that:

$\text{dim }_{\Bbb R}(\text{ker}(T)) = 2$.

Now $T(2x) = 2x \neq 0$ and $T(1+x) = 2x \neq 0$, so to get either one of THESE vectors in our subspace, we need some other vector NOT in $\text{ker}(T)$ (since the set $\{1 - x, x^2 - x^3\}$ only spans the kernel).

It really doesn't matter which one we pick, since:

$2x = (-1)(1 - x) + (1 + x)$ and
$1 + x = (1 - x) + (\frac{1}{2})(2x)$

so we get the same spanned space in either case. I chose $1 + x$, but BASES ARE NOT UNIQUE, and choosing $2x$ would also be correct. Again, we have to show that our 3-element set is now linearly independent (whichever 3rd vector we choose), but that is not difficult (and I urge you to do so).

*********

Now...IF you actually want to extend this to a basis of ALL of $P_3(\Bbb R)$, we will need a 4th vector. All we require is that this 4th vector not be in:

$\text{span}(\{1 - x,1 + x, x^2 - x^3\})$.

Either of the polynomials $x^2$ or $x^3$ will do (other choices are also possible).

*********

Now, you say you prove a set is a basis by forming a matrix out of your chosen vectors, and evaluating this matrix's determinant to show it is non-zero. Yes, this works, HOWEVER:

it is usually easier to prove linear independence DIRECTLY FROM THE DEFINITION. A 4x4 matrix determinant is often rather time-consuming to calculate (and it only gets worse with larger matrices).

 

[Petrus]

Hello,
I want to first thank you for your Verry well explain and for taking your time!
If we use the basis I did early type for the whole $$P_3(\mathbb{R})$$
By deffination we se that
$$a(1-x)+b(x^2-x^3)+c(1+x)+dx^3=0$$
we se that we only got one x^2 and the only way to get it zero is that $$b=0$$ that means that $$d=0$$ now we got
$$a(1-x)+c(1+x)=0$$ and the only way is that $$a,b=0$$ hence it's linear independent hence it is our basis.
Is this correctly?
I am Really gratefull for that you take your time to make such explain,post!

Regards,
$$|\pi\rangle$$

 

[Deveno]

Exactly! And you can see that takes MUCH less time than calculating a determinant.

It is possible, however, that the basis your original problem (which you did not post word for word, perhaps because your book is in another language) is wanting is this one:

$\{1 - x, 2x, x^2 - x^3, 2x^2\}$.

What makes me think this, is that $\{1 - x, x^2 - x^3\}$ is a basis for $\text{ker}(T)$, and $\{2x,2x^2\}$ is a basis for $\text{im}(T)$, and this partition of the basis makes it obvious that:

$P_3(\Bbb R) = \text{ker}(T) \oplus \text{im}(T)$

which is another way of expressing the rank-nullity theorem for a linear endomorphism.

 

[Petrus]

Exactly! And you can see that takes MUCH less time than calculating a determinant.

It is possible, however, that the basis your original problem (which you did not post word for word, perhaps because your book is in another language) is wanting is this one:

$\{1 - x, 2x, x^2 - x^3, 2x^2\}$.

What makes me think this, is that $\{1 - x, x^2 - x^3\}$ is a basis for $\text{ker}(T)$, and $\{2x,2x^2\}$ is a basis for $\text{im}(T)$, and this partition of the basis makes it obvious that:

$P_3(\Bbb R) = \text{ker}(T) \oplus \text{im}(T)$

which is another way of expressing the rank-nullity theorem for a linear endomorphism.

That Im(T) basis part I am pretty NEW with it, I did not understand how you got it. I Also see after a while that I can choose the basis $$1+x$$ or $$1-x$$ right? $$1+x=1-x+2x$$
Regards,
$$|\pi\rangle$$

 

[Petrus]

Another way (which I like) to prove that $$a(1-x)+b(x^2-x^3)+c(1+x)+dx^3$$ is linear independent is:
$$a-ax+bx^2-bx^3+c+cx+dx^3=0$$
Put $$x=0$$ and we get that $$a=-c$$
now we derivate and put $$x=0$$ and we get $$-a=-c$$ and only soloution for those Two is $$a,c=0$$
now take second derivate and put x=0 and we get $$2b=0=>b=0$$
If we take third derivate and put x=0 we get $$-6b+6d=0$$ and we know that $$b=0$$ we get $$6d=0 => d=0$$ hence it's linear independent! This only works for polynom and $$a-ax+bx^2-bx^3+c+cx+dx^3=0$$ Since this is true for all x we can subsitute $$x=0$$. I likes this way alot:)

Regards,
$$|\pi\rangle$$

 

[Petrus]

Is this correctly understand of I am T
$$L(1,x,x^2,x^3)=1,x,x^2,x^3$$
how do I solve that?

Regards,
$$|\pi\rangle$$

 

[Deveno]

If $T: U \to V$ is ANY linear transformation, then:

$\text{im}(T) = \{v \in V: \exists u \in U\ \text{with } v = T(u)\}$

this is called the IMAGE (or range) of $T$.

With ANY function (and linear transformations are just special kinds of functions) we have 3 things:

1) the domain of definition (or source)
2) the co-domain where the function takes its values (or target)
3) the rule that defines which image (value) we get for each element in the domain

The image is a subset of the co-domain. If a function is surjective (onto), the image equals the co-domain.

In this particular case, $T$ is an endomorphism: that is, its domain and co-domain are the same. The image of $T$, however, is much smaller that the entire space, $T$ has non-trivial kernel, so it is NOT surjective (this isn't true of all linear transformation, just ones that map to the same space as their domain).

Clearly, this particular $T$ only maps to polynomials with $x$ and $x^2$ terms, so nothing with a constant or cubic term is in the image.