Linear Map Input: Solving P'(1-x) | \pi\rangle

In summary, the linear map $T:P_3(\Bbb R) \to P_2(\Bbb R)$ given by $T(p(x)) = p'(1-x)$ can be represented by the matrix:$\begin{bmatrix}0&1&0&0\\0&0&2&0\\0&0&0&3 \end{bmatrix}$in the standard bases for $P_3(\Bbb R)$ and $P_2(\Bbb R)$.
  • #1
Petrus
702
0
Hello!
I have hard to understand this input for this linear map \(\displaystyle T:P_3(R)->P_2(R)\)
\(\displaystyle T(p(x))=P'(1-x)\)
so they get this value when they put in which I have hard understanding
vpyfll.png

I don't understand how they get those, I am totally missing something basic...!
The only logical explain is that in \(\displaystyle p'(x)=3x^2\) where \(\displaystyle x=(1-x)\) I don't get it.

Regards,
\(\displaystyle |\pi\rangle\)
 
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  • #2
Petrus said:
Hello!
I have hard to understand this input for this linear map \(\displaystyle T:P_3(R)->P_2(R)\)
\(\displaystyle T(p(x))=P'(1-x)\)
so they get this value when they put in which I have hard understanding
vpyfll.png

I don't understand how they get those, I am totally missing something basic...!
The only logical explain is that in \(\displaystyle p'(x)=3x^2\) where \(\displaystyle x=(1-x)\) I don't get it.

Regards,
\(\displaystyle |\pi\rangle\)

Hey Petrus!

I suspect you have trouble with the substitution step.

Let's take the last one as an example.
I'm assuming that the step from $p(x)=x^3$ to $p'(x)=3x^2$ is clear.

The latter is equivalent to saying:
$$p'(u)=3u^2$$
That is, $p'$ is a function and if you apply it to $u$ you get $3u^2$.

Now, if we substitute $u=1-x$, we replace all $u$'s by $1-x$ and get:
$$p'(1-x)=3(1-x)^2$$
 
  • #3
I like Serena said:
Hey Petrus!

I suspect you have trouble with the substitution step.

Let's take the last one as an example.
I'm assuming that the step from $p(x)=x^3$ to $p'(x)=3x^2$ is clear.

The latter is equivalent to saying:
$$p'(u)=3u^2$$
That is, $p'$ is a function and if you apply it to $u$ you get $3u^2$.

Now, if we substitute $u=1-x$, we replace all $u$'s by $1-x$ and get:
$$p'(1-x)=3(1-x)^2$$
Thanks a lot for the fast respond! I got it now! Have a nice day!:)

Regards,
\(\displaystyle |\pi\rangle\)
 
  • #4
Although most people know the derivative from calculus, you don't need calculus to define the derivative of a polynomial.

Given a polynomial in $P_n(\Bbb R)$, say:

$p(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n$

We define the derivative $p'(x)$ (I prefer to write this as $Dp$ for reasons we shall see later) as:

$Dp(x) = a_1 + 2a_2x + \cdots + na_nx^{n-1}$

If one uses the basis: $B = {1,x,x^2,\dots,x^n}$ for $P_n(\Bbb R^n)$ one can identify (such an identification is called a linear isomorphism) $P_n(\Bbb R)$ with $\Bbb R^n$ like so:

$a_0 + a_1x + a_2x^2 + \cdots + a_nx^n \mapsto (a_0,a_1,a_2,\dots,a_n)$

In this basis (actually "two bases" but the basis for $P_{n-1}(\Bbb R)$ is "just like" the basis for $P_n(\Bbb R)$), $D$ has the nx(n+1) matrix:

$[D]_B = \begin{bmatrix}0&1&0&\cdots&0\\0&0&2&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&n \end{bmatrix}$

which makes it clear $D$ is a linear mapping.

We also have the linear mapping: $L: P_n(\Bbb R) \to P_n(\Bbb R)$ given by:

$L(p(x)) = p(a - x)$ for $a \in \Bbb R$.

It may be instructive to see the matrix of $L$ with respect to the basis $B$:

$[L]_B = \begin{bmatrix}1&a&a^2&\cdots&a^n\\0&-1&-2a&\cdots&-na^{n-1}\\0&0&1&\cdots&\frac{n(n-1)}{2}a^{n-2}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&(-1)^n \end{bmatrix}$

So the mapping you are given is just $L \circ D$ (with $a = 1$):

$(L \circ D)(p(x)) = L(D(p(x)) = L(p'(x)) = p'(1 - x)$.

For the case $n = 3$, we can see $D$ as having the 3x4 matrix:

$\begin{bmatrix}0&1&0&0\\0&0&2&0\\0&0&0&3 \end{bmatrix}$

and $L$ as being the 3x3 matrix (with $a = 1$):

$\begin{bmatrix}1&1&1\\0&-1&-2\\0&0&1\\ \end{bmatrix}$

Then multiplying these two matrices together, we get the 3x4 matrix:

$\begin{bmatrix}0&1&2&3\\0&0&-2&-6\\0&0&0&3 \end{bmatrix}$ which is the matrix for $T$ in the basis $B$.

Now the vector representing the constant polynomial $p(x) = 1$ in the basis $B$ is just: (1,0,0,0), while the vector representing $x$ is (0,1,0,0), the vector representing $x^2$ is (0,0,1,0), and the vector representing $x^3$ is (0,0,0,1). So applying the matrix of $T$ to these vectors, we get

$T(1) \to [T(1,0,0,0)]_B = [0,0,0]_B \to 0$
$T(x) \to [T(0,1,0,0)]_B = [1,0,0]_B \to 1$
$T(x^2) \to [T(0,0,1,0)]_B = [2,-2,0]_B \to 2-x = 2(1-x)$
$T(x^3) \to [T(0,0,0,1)]_B = [3,-6,3]_B \to 3 - 6x + 3x^2 = 3(1 - x^2)$

If, instead, your book had defined:

$T(p(x)) = (p(x))'$

we would have taken $D \circ L$, giving a different 3x4 matrix (using a 4x4 "L").

I find that the use of derivative notation in linear algebra problems involving polynomial spaces is sort of confusing, and obscures the underlying vector algebra: we just have some linear map $D$ on a vector space $V$, to understand what $D$ does, we need only examine what it does to a basis (and once having CHOSEN a basis, we can do everything in matrices relative to that basis).

Or, perhaps more simply put: everything you want to know about a polynomial vector space is given by the coefficients of the polynomials (the $x$ is just "excess baggage" in terms of the "vector-space-ness", although it does come into play when considering the RING structure of this vector space...but that is a topic beyond the scope of most linear algebra courses).
 
  • #5


Hello! It seems like the input is asking for the solution to the linear map T, which takes a polynomial p(x) and returns the derivative of that polynomial evaluated at 1-x. In other words, T(p(x)) = p'(1-x). The input specifically asks for the solution to T(P'(1-x)), which means the derivative of p'(1-x) evaluated at the input |\pi\rangle. This is a bit confusing because |\pi\rangle is typically used to represent a ket vector in quantum mechanics, but in this context, it could be interpreted as a polynomial with the variable |\pi|. So, the solution to T(P'(1-x)) would be the derivative of P'(1-x) evaluated at |\pi\rangle. I hope this helps clarify the input and its solution.
 

FAQ: Linear Map Input: Solving P'(1-x) | \pi\rangle

1. What is a linear map?

A linear map is a mathematical function that maps a vector from one vector space to another, preserving the vector addition and scalar multiplication properties. In simpler terms, it is a transformation that preserves the shape and orientation of a vector.

2. What is "input" in the context of a linear map?

In the context of a linear map, "input" refers to the vector that is being transformed or mapped to another vector space.

3. How do you solve for P'(1-x) in a linear map?

To solve for P'(1-x) in a linear map, you would first need to know the specific function or transformation represented by P. Then, you would substitute 1-x for the input variable in the function and perform any necessary calculations or simplifications.

4. What is the significance of the notation | \pi\rangle in this linear map?

The notation | \pi\rangle represents a vector in a specific vector space, often referred to as the "ket" notation in linear algebra. In the context of this linear map, it is the vector being used as the input for the function P.

5. Can this linear map be applied to any vector in any vector space?

It depends on the specific function or transformation represented by P. Some linear maps may have restrictions on the types of vectors they can be applied to, while others may be able to be applied to any vector in any vector space. In general, linear maps are versatile and can be applied to a wide range of vectors and vector spaces.

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