MHB Dimension of Dirac Functionals in $V$: Find the Answer

[A.Magnus]

I am very much struggling with this problem: The set $\{\sin x, \cos x, x \sin x, x \cos x, x+2, x^2-1 \}$ on interval of $[0, \pi]$ is linearly independent and generates vector space $V$. Find the dimension of the kernel of the Dirac functionals in $V$.

Here are what I know of the definitions from my textbook: If $W$ is a vector space of continuous real-valued function on the interval $[0,1]$, and if $\delta: W \rightarrow \mathbb R$ is a map such that $\delta(f) = f(0)$, then $\delta$ is called Dirac functional. (Functional is defined as an element of $W$'s dual space, and dual space is the set of all linear maps of $W$ to $\mathbb R$.

I have been researching online on Dirac functional, but all I got were topics on differential equation and not on linear algebra. I am totally lost, please help. Thank you before hand for your gracious help and time. ~MA

 

[I like Serena]

I am very much struggling with this problem: The set $\{\sin x, \cos x, x \sin x, x \cos x, x+2, x^2-1 \}$ on interval of $[0, \pi]$ is linearly independent and generates vector space $V$. Find the dimension of the kernel of the Dirac functionals in $V$.

Here are what I know of the definitions from my textbook: If $W$ is a vector space of continuous real-valued function on the interval $[0,1]$, and if $\delta: W \rightarrow \mathbb R$ is a map such that $\delta(f) = f(0)$, then $\delta$ is called Dirac functional. (Functional is defined as an element of $W$'s dual space, and dual space is the set of all linear maps of $W$ to $\mathbb R$.

I have been researching online on Dirac functional, but all I got were topics on differential equation and not on linear algebra. I am totally lost, please help. Thank you before hand for your gracious help and time. ~MA

Hi MaryAnn! ;)

So we have:
$$V = \{ f: [0,\pi]\to\mathbb R \mid f(x) \mapsto a\sin x+b\cos x + c x\cos x + d(x+2) +e(x^2-1)\text{ for some }a,b,c,d,e \in \mathbb R \} \simeq \mathbb R^5$$
and we're looking for a kernel in $V$ of $\delta$. That is:
$$\{f \in V\mid \delta(f)=0\}$$
Right? (Wondering)

 

[A.Magnus]

Hi MaryAnn! ;)

So we have:
$$V = \{ f: [0,\pi]\to\mathbb R \mid f(x) \mapsto a\sin x+b\cos x + c x\cos x + d(x+2) +e(x^2-1)\text{ for some }a,b,c,d,e \in \mathbb R \} \simeq \mathbb R^5$$
and we're looking for a kernel in $V$ of $\delta$. That is:
$$\{f \in V\mid \delta(f)=0\}$$
Right? (Wondering)

Thanks, but I am getting more confused than ever. By reading the problem closely, at first I thought that the set of linearly independent functions $\{\sin x, \cos x, ... \}$, let's denote it as $S$, is generating $V.$ But the way you told me is that $S$ is generating $\mathbb R.$ I am not challenging an authoritative guru like you, but could you please explain a little bit. And the dimension is therefore 5? Thanks again for your gracious help! ~MA

 

[I like Serena]

Thanks, but I am getting more confused than ever. By reading the problem closely, at first I thought that the set of linearly independent functions $\{\sin x, \cos x, ... \}$, let's denote it as $S$, is generating $V.$ But the way you told me is that $S$ is generating $\mathbb R.$ I am not challenging an authoritative guru like you, but could you please explain a little bit. And the dimension is therefore 5? Thanks again for your gracious help! ~MA

An element of V is a function f. For instance cos is an element. And any linear combination is also an element of V. For instance 2cos x + 3sin x.
Moreover, each function in V can be uniquely identified by 5 real numbers - the coefficients of the linear combination.

Now consider e.g. f(x)=acos x+bsin x. For which a and b do we have f(0)=0?

 

[Opalg]

I am very much struggling with this problem: The set $\{\sin x, \cos x, x \sin x, x \cos x, x+2, x^2-1 \}$ on interval of $[0, \pi]$ is linearly independent and generates vector space $V$. Find the dimension of the kernel of the Dirac functionals in $V$.

Here are what I know of the definitions from my textbook: If $W$ is a vector space of continuous real-valued function on the interval $[0,1]$, and if $\delta: W \rightarrow \mathbb R$ is a map such that $\delta(f) = f(0)$, then $\delta$ is called Dirac functional. (Functional is defined as an element of $W$'s dual space, and dual space is the set of all linear maps of $W$ to $\mathbb R$.

I have been researching online on Dirac functional, but all I got were topics on differential equation and not on linear algebra. I am totally lost, please help. Thank you before hand for your gracious help and time. ~MA

Maybe it would help to go back to the original question and look at it carefully. You are correct to say that this "Dirac functional" is a linear map from $V$ to $\mathbb R$. I have put "Dirac functional" in quotes because I think that term is not helpful in this context and is likely to put you off. So just think of $\delta$ as a linear map from $V$ to $\mathbb R$. You are asked to find the dimension of its kernel.

You are told that $V$ is generated by a set of six functions (not five: I think ILS missed one!), and you are helpfully told that this set of functions is linearly independent. I have two questions for you.

Q.1. What is the dimension of $V$?

Q.2. (This is really a big hint.) Do you know a theorem about a linear map, giving a connection between the dimensions of its kernel and its range?

 

[I like Serena]

You are told that $V$ is generated by a set of six functions (not five: I think ILS missed one!)

Awwww... there goes my hard won guru status. (Tmi)

 

[A.Magnus]

Maybe it would help to go back to the original question and look at it carefully. You are correct to say that this "Dirac functional" is a linear map from $V$ to $\mathbb R$. I have put "Dirac functional" in quotes because I think that term is not helpful in this context and is likely to put you off. So just think of $\delta$ as a linear map from $V$ to $\mathbb R$. You are asked to find the dimension of its kernel.

You are told that $V$ is generated by a set of six functions (not five: I think ILS missed one!), and you are helpfully told that this set of functions is linearly independent. I have two questions for you.

Q.1. What is the dimension of $V$?

Q.2. (This is really a big hint.) Do you know a theorem about a linear map, giving a connection between the dimensions of its kernel and its range?

Many thanks to Opalg, you came to my rescue just in time! Here are my takes from your big hints:

A1: $dim(V) = 6$, because $V$ is generated by 6 functions.

A2: The dimension formula says $dim(V) = dim(ker (\delta) + dim(Im(\delta)).$

I think that $dim(Im(\delta)) = 3$ because as described by ILS, $\{ f \in V \mid \delta (f) = 0 = f(0)\}$, and
$$\begin{align}
f(0) &= a \sin 0 + b \cos 0 + (c)(0) \sin 0 + (d)(0) \cos 0 + e(0 + 2) + g(0^2 - 1)\\
&=b +2e - g,
\end{align}$$
which indicates that the dimension is indeed $3$.

Hence, after putting the numbers into the dimension formula, we have the $\dim(ker(\delta)) = 3.$ Please let me know if I made yet another dummy mistakes :) Thank you again to both of you for your gracious help! ~MA

 

[Opalg]

Many thanks to Opalg, you came to my rescue just in time! Here are my takes from your big hints:

A1: $dim(V) = 6$, because $V$ is generated by 6 functions.

A2: The dimension formula says $dim(V) = dim(ker (\delta) + dim(Im(\delta)).$

I think that $dim(Im(\delta)) = 3$ because as described by ILS, $\{ f \in V \mid \delta (f) = 0 = f(0)\}$, and
$$\begin{align}
f(0) &= a \sin 0 + b \cos 0 + (c)(0) \sin 0 + (d)(0) \cos 0 + e(0 + 2) + g(0^2 - 1)\\
&=b +2e - g,
\end{align}$$
which indicates that the dimension is indeed $3$.

Hence, after putting the numbers into the dimension formula, we have the $\dim(ker(\delta)) = 3.$ Please let me know if I made yet another dummy mistakes :) Thank you again to both of you for your gracious help! ~MA

Ummm... Not quite. This map $\delta:V\to\mathbb R$ has $\mathbb R$ as its image space. Since $\mathbb R$ is $1$-dimensional, its only subspaces have dimension $0$ (the subspace consisting only of the zero vector) or $1$ (if the subspace is the whole space). So $\mathrm{dim}(\mathrm {Im}(\delta))$ can only be $0$ or $1$.

 

[A.Magnus]

Ummm... Not quite. This map $\delta:V\to\mathbb R$ has $\mathbb R$ as its image space. Since $\mathbb R$ is $1$-dimensional, its only subspaces have dimension $0$ (the subspace consisting only of the zero vector) or $1$ (if the subspace is the whole space). So $\mathrm{dim}(\mathrm {Im}(\delta))$ can only be $0$ or $1$.

A belated thank you! ~MA