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From S.L Linear Algebra:

(1) Let $P_n$ be the vector space of polynomials of degree $ \leq n$. (2) Then the derivative $D: P_n \rightarrow P_n$ is a linear map of $P_n$ into itself. (3) Let $I$ be the identity mapping.

Prove that the following linear maps are invertible:

(a) $I - D^2$

...

My observation (long):

(1)

I find a first sentence very interesting:

Let $P_n$ be the vector space of polynomials of degree $ \leq n$.

But in order for $P_n$ to be a vector space, it must contain a zero vector, which in vector space of polynomials is given by a zero polynomial which has a very confusing degree. But it is argued that it makes most sense for zero polynomial to have $-\infty$ degree.

Hence wouldn't it be more precise to say that $P_n$ is vector space of polynomials that have a degree of $\leq -\infty$? In which case $n= -\infty$?

I don't believe $n$ is associated in any way with dimension of vector space, because in this case this assertion would break rule of cardinality.

(2)

Then the derivative $D: P_n \rightarrow P_n$ is a linear map of $P_n$ into itself.

This is another interesting assertion. We know that exponent property of derivative will change degree of $p_g \in P_n$ to $g-1$.

In this case, if our polynomial is something like $x^{-\infty + 1}$, we would get a derivative $\frac{d}{dx} x^{-\infty + 1} = -\infty x^{-\infty}$ which suggests that kernel is not trivial for $D$ and hence our linear derivative map over $P_n$ to $P_n$ is not injective... And therefore not invertible.

I definitely am wrong with this assertion, since it is assumed that $D$ is invertible over $P_n \rightarrow P_n$

(3)

Let $I$ be the identity mapping.

It is very easy to prove that identity map is both injective and surjective therefore invertible. Since considering that:

$I(v) = v$

We can easily see that $Im(I) = P_n$ and $Ker(I) = {0}$.

Polynomials being scalars:

Considering that polynomials are scalars, isn't basis of vector space zero-dimensional? Therefore by rank-nullity theorem:

$$\textrm{dim} \, P_n = \textrm{dim} \, Im(P_n) + \textrm{dim} \, Ker(P_n)$$

$$0 = \textrm{dim} \, Im(P_n) + \textrm{dim} \, Ker(P_n)$$

$$0 = 0 + 0$$

Therefore by this assumption, kernel is trivial and linear map must be invertible.


Question:

In this case, I'm trying to find that there exists a matrix $A$ such that:

$$(I - D^2)A = I$$

I know that identity map is invertible (as mentioned in my observation), but I'm not so sure about $D^2$. In fact I'm unable to find a matrix associated with $D$ in order to prove my assertion.

What could be the simple solution? Is my observation very incorrect?

Thank you!

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    $\begingroup$ $D$ is certainly not invertible. The claim is that $I - D^2$ is invertible. First, try this for just like $P_2$. What's a basis of $P_2$? $\endgroup$ – user113102 Mar 5 '19 at 20:00
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    $\begingroup$ and the degree(0) = $-\infty$ thing is just a convention to make statements like deg(fg) = deg(f) + deg(g) work. You don't want to actually think of $-\infty$ as anything more than shorthand for 'something which is less than any nonnegative integer, and $-\infty + a = -\infty*a = -\infty$ for any nonnegative integer $a$'. So in particular, 0 \in $P_n$ for any $n$ since $-\infty <n$ for any $n$ by definition. $\endgroup$ – user113102 Mar 5 '19 at 20:04
  • $\begingroup$ @user113102 Well I suppose that basis for $P_2$ vector space would be in a form of a quadratic function $ax^2 + bx + c$ (wouldn't basis of such vector space have cardinality of $0$?). $\endgroup$ – ShellRox Mar 5 '19 at 20:21
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    $\begingroup$ Yeah, so your basis is ${1,x,x^2}$. $\endgroup$ – user113102 Mar 5 '19 at 20:24
  • $\begingroup$ @user113102 I apologize for my confusion, but does that imply that basis is $\{1, x, x^2 \}$ or that there are three possible basis with cardinality of $0$? If former is true, then I guess I have to find $D=AX$ where $X= ax^2 + bx +c $ where $\{a, b, c\}$ is a coordinate vector. In which case, isn't $A \cdot (ax^2 + bx +c) = (2ax + b + 0)$? $\endgroup$ – ShellRox Mar 5 '19 at 20:31
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Invertibility of $I-D^2$ is the same as saying that $1$ is not an eigenvalue of $D^2$.

Suppose $D^2p=p$. Then $p=0$, otherwise the degree of $D^2p$ is strictly less than the degree of $p$.

Technical note: the degree of the zero polynomial is usually taken to be $-\infty$ so as to preserve the properties $$ \deg(p+q)\le\max\{\deg(p),\deg(q)\}, \qquad \deg(pq)=\deg(p)+\deg(q) $$ with the obvious definition $-\infty+n=-\infty$ for every $n$ (finite or $-\infty$).


In my lecture notes I define $P_n$ to be the vector space consisting of polynomials of degree $<n$ (so including the zero polynomial). This way $P_0$ is the trivial vector space and, in general $P_n$ has dimension $n$. But that's not really a problem with your notation. Just remember that your $P_n$ has dimension $n+1$ (for $n\ge0$, of course).


An example with $n=3$ (your notation). If we consider the standard basis $\{1,x,x^2,x^3\}$, then $D^2(1)=0$, $D^2(x)=0$, $D^2(x^2)=2$, $D^2(x^3)=6x$. Hence the matrix of $I-D^2$ is $$ \begin{bmatrix} 1 & 0 & -2 & 0 \\ 0 & 1 & 0 & -6 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$ and the inverse is $$ \begin{bmatrix} 1 & 0 & 2 & 0 \\ 0 & 1 & 0 & 6 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$ The general formula is $$ (I-D^2)^{-1}=I+D^2+D^4+\dots+D^{2k} $$ where $2k\ge n$.

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I'm sorry to say that but there are so many factual errors in what you wrote that I find it hard to collect them all, not even explain. This certainly is beyond what is suitable for a comment, so I will leave an answer.

(1)

You write:

But in order for $P_n$ to be a vector space, it must contain a zero vector, which in vector space of polynomials is given by a zero polynomial which has a very confusing degree. But it is argued that it makes most sense for zero polynomial to have $−\infty$ degree. Hence wouldn't it be more precise to say that $P_n$ is vector space of polynomials that have a degree of $\leq −\infty$? In which case $n=−\infty$?

What is even the problem here? If $n$ is a nonnegative integer, then $-\infty \leq n$, so by definition, $0$ polynomial is an element of any $P_n$. Remember that it's $-\infty$ and not $\infty$.

(2)

This is another interesting assertion. We know that exponent property of derivative will change degree of $p_g \in P_n$ to $g-1$.

This works only for integral $g$. $-\infty$ is not integral. It is illdefined in this context and shoudn't be used. Rules work as long as assumptions are satisfied. Here, they aren't. So the rest of this reasoning is meaningless, :

In this case, if our polynomial is something like $x^{-\infty + 1}$, we would get a derivative $\frac{d}{dx} x^{-\infty + 1} = -\infty x^{-\infty}$

(3)

I definitely am wrong with this assertion, since it is assumed that $D$ is invertible over $P_n \rightarrow P_n$

This is false though, it is not assumed, and it is moreover false. Consider $f(x)=x^2+2$ and $g(x)=x^2-2$. Both derivatives are the same, so $D$ cannot be possibly invertible.

(4)

Polynomials being scalars:

No, polynomials here are vectors. Scalars are real numbers, its coefficients.

Considering that polynomials are scalars, isn't basis of vector space zero-dimensional?

First of all - it's not possible that the basis has dimension zero. The basis is just a set of vectors, for instance $\{(1, 2), (3, 4)\}$. It cannot contain zero vector, so it can't be a vector space and only vector spaces can have some dimension. But I get that you meant to say that the space itself is zero-dimensional (i.e. the basis is empty). It's still not true - the only space with empty basis is the zero space - space where the only element is zero. But there are some polynomials! So the space is non-empty.

(5)

Therefore by rank-nullity theorem: $$ \dim P_n=\dim Im(P_n)+\dim Ker(P_n) $$

$Im$ and $Ker$ act on vector space homomorphisms. $P_n$ is a set of polynomials. What would it even mean? Kernel of a set? Image of a set? Again, this is incomprehensible and meaningless. Rank-nullity is only applicable to linear maps (vector space homomorphisms). You can't plug in anything and hope it will work.

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  • $\begingroup$ Thank you for the effort to finding fundamental mistakes in the answer. I was very confused about relation between dimensionality of vector space and degree of polynomial $n$, which is obviously seen in (1). For (3) I was thinking about linearity, e.g if $D$ is invertible then is $D \cdot D$ invertible as well? (but I forgot $I$, which was probably a main mistake). For (4) and (5) I assumed that polynomials were scalars, and I thought I could prove that kernel of vector space $P_n$ is trivial using rank-nullity. $\endgroup$ – ShellRox Mar 6 '19 at 15:34
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    $\begingroup$ 1. If $D$ is invertible, then $D\cdot D$ is invertible as well, but $D$ is not invertible. 2. I need to ask you again - what do you even mean by "kernel of vector space"? Give some definition please, because I feel otherwise we're not gonna get anywhere. $\endgroup$ – enedil Mar 6 '19 at 16:12
  • $\begingroup$ 1. Yes that's what I was previously thinking, I was wrong. 2. I apologize about "kernel of the vector space", I meant kernel of linear map having domain and codomain as vector space $P_n$, I was trying to find triviality of such kernel to prove injective property of derivative map over vector space of polynomials as mentioned in the question. I guess that's what K.power has displayed in the question for finite-dimensional cases. $\endgroup$ – ShellRox Mar 6 '19 at 16:25
  • $\begingroup$ But there are a lot of linear maps from and to $P_n$. How can you speak about all of them collectively? $\endgroup$ – enedil Mar 6 '19 at 16:35
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    $\begingroup$ But the kernel is not trivial - nonzero polynomial $f(x) = 4$ is in the kernel. $\endgroup$ – enedil Mar 6 '19 at 16:47
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The easiest way (for me at least) to prove invertibility of $I-D^2$ is to show it has a trivial kernel. This will prove that it is injective, and by the rank-nullity theorem we get surjectivity for free.

To that end fix some some polynomial $p=\sum_{i=0}^na_ix^i$. Then $$(I-D^2)p=a_nx^n+a_{n-1}x^{n-1}+\sum_{i=2}^{n}(a_{i-2}-i(i-1)a_i)x^{i-2}.$$ Thus $(I-D^2)p=0$ if and only if $a_n=0,a_{n-1}=0$ and $a_{i-2}=i(i-1)a_i$ for all $i\in I:=\{2,\dots,n\}$. Now choosing $n\in J$ we see that $a_{n-2}=0$ (because $a_n=0$). Similarly $a_{n-3}=0$, and continuing in this manner, using "downward induction", we actually see that for those conditions to hold we require $a_i=0$ for all $i\in\{0,\dots,n\}$. Thus $p\in \operatorname{ker}(I-D^2)$ if and only if $p=0$, so we can conclude that the operator is indeed injective, therefore invertible.

As enedil has pointed out, a trivial kernel only applies invertibility in finite dimensional vector spaces. In infinite dimensional vector spaces we don't have the rank-nullity theorem, so we would also have to check surjectivity of the map.

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  • $\begingroup$ This assumes the space is finitely generated (which is true, but nonetheless worth explaining). $\endgroup$ – enedil Mar 6 '19 at 0:30
  • $\begingroup$ @enedil As the OP was is working through a Lang's linear algebra I doubted they would have been exposed to infinite dimensional vector spaces, but I have added the caveat as suggested. $\endgroup$ – K.Power Mar 6 '19 at 1:10
  • $\begingroup$ @K.Power Thank you very much for the answer. Yes, i don't believe infinite dimensional vector spaces are covered in S.L Linear Algebra, at least not before Chapter 5. But I will definitely consider this for future cases. $\endgroup$ – ShellRox Mar 6 '19 at 16:29
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To show that $I - D^2$ is invertible, one doesn't actually need to compute its inverse.

Hint The derivative operator $D$ maps any (nonzero) polynomial to a polynomial of lesser degree, so with respect to the standard basis $(1, x, \ldots, x^n)$ the matrix representation $[D]$ of $D$ is strictly upper triangular. Thus, so is $[D]^2 = [D^2]$.

So, $[I - D^2] = [I] - [D^2]$ is upper triangular with diagonal entries $1$, hence $\det(I - D^2) = 1$. To produce an explicit expression for $(I - D^2)^{-1}$, notice that $D^{n + 1} p = 0$ for all $p \in P_n$, so $D^{n + 1} = 0$ and thus $(D^2)^m = 0$ for $m \geq \lceil \frac{n + 1}{2} \rceil$. So, the formal series $(I - D^2)^{-1} \sim \sum_{i = 0}^{\infty} (D^2)^i$ terminates; we have: $$(I - D^2)^{-1} = I + D^2 + D^4 + \cdots + D^{\lceil (n + 1) / 2 \rceil}.$$

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