Vector space of functions defined by a condition

[Orodruin]

That's exactly the point. Part of my teaching is philosophy is to Not give all the details. Consequently the student also learns to Not assume what I have Not stated.

The thing is you presented it in a way that made it appear as an alternative to checking the vector space axioms. That to me is not ”omitting a few details” but ”omitting details in such a way that students are mislead on purpose”.

Things like commutativity of addition of functions is as easy to show directly in the subspace as it is in the general function space.

 

[nuuskur]

The thing is you presented it in a way that made it appear as an alternative to checking the vector space axioms. That to me is not ”omitting a few details” but ”omitting details in such a way that students are mislead on purpose”.

The goal is to incite questions/protest. That goal is evidently achieved.

 

[Orodruin]

The goal is to incite questions/protest. That goal is evidently achieved.

Protest from me, not the OP… that can hardly be interpreted as a sign of the OP realising the subtextual.

 

[nuuskur]

Protest from me, not the OP… that can hardly be interpreted as a sign of the OP realising the subtextual.

Indeed. Perhaps, next time the OP will recall this exchange.

You don't need to worry about what-ifs in the event you hadn't said anything. I would have.

 

[mathwonk]

recall that if there is a surjective linear transformation T from V to W, then the dimension of V equals the sum of the dimension of W plus the dimension of the kernel of T. In your case there is a nice surjective linear transformation from your space V to the 2 dimensional space of functions spanned by x and x^2 on [0,1], namely restriction to [0,1]. So what is the kernel? It should be pretty clear that kernel is 3 dimensional, and hence V is 5 dimensional.

Then to get a basis of V, take a basis of the kernel, and add in any pair of functions in V that restrict respectively to x and x^2 on [0,1]. (There is an obvious choice.)

Apologies if I have missed this hint in earlier discussions. This is of course inspired by what is said earlier.

 

[Hall]

recall that if there is a surjective linear transformation T from V to W, then the dimension of V equals the sum of the dimension of W plus the dimension of the kernel of T. In your case there is a nice surjective linear transformation from your space V to the 2 dimensional space of functions spanned by x and x^2 on [0,1], namely restriction to [0,1]. So what is the kernel? It should be pretty clear that kernel is 3 dimensional, and hence V is 5 dimensional.

Then to get a basis of V, take a basis of the kernel, and add in any pair of functions in V that restrict respectively to x and x^2 on [0,1]. (There is an obvious choice.)

Apologies if I have missed this hint in earlier discussions. This is of course inspired by what is said earlier.

As w have to maintain continuity at 1, we can write our $f$ as
$$
f(x)=
\begin{cases}
ax^2 + bx & x \in [0,1] \\
A(x^3 -1) + B(x^2 -1) + C(x-1) + a+b & x \in[1,2] \\
\end{cases}$$
(credit: @Steve4Physics)

Let $T$ be the operator which restricts the function to $[0,1]$, therefore, $T : V \to W$, where $V$ is the space given in the question, and $W$ is all polynomials with degree less than or equal to 2 and $p(0)=0$.
$T(f) = ax^2 +bx$

The following function will always yield $0$
$$
f(x) =
\begin{cases}
0 & x \in [0,1] \\
A(x^3-1) + B(x^2 -1) +C(x-1) & x \in[1,2]\\
\end{cases}
$$
(the continuity is maintained).

So, the null space is the space of all those functions. Its basis is
$f_1 = \begin{cases} 0 & x \in[0,1]\\ x^3-1 & x \in [1,2] \\ \end{cases}$

$f_2 = \begin{cases} 0 & x \in[0,1] \\ x^2 -1 & x \in [1,2] \\ \end{cases}$

$f_3 = \begin{cases} 0 & x \in[0,1]\\ x-1 & x \in [1,2]\\ \end{cases}$

The basis for range of $T$ is $\{x^2, x\}$.

$dim V = 3 + 2 =5$

Basis for $V$ is $\{f_1, f_2, f_3, f_4, f_5\}$
$$
f_4 =
\begin{cases}
x^2 & x \in [0,1] \\
1 & x \in [1,2]\\
\end{cases}
$$
$$
f_5 =
\begin{cases}
x & x \in[0,1] \\
1 & x \in[1,2]\\
\end{cases}
$$

Your hint was really unique.

 

[Hall]

This solution is for ("attributed to" wouldn't be right because I don't know if he really wanted me to do like this) @Steve4Physics

$$f(x) = \begin{cases} ax^2 + bx & x \in [0,1] \\ Ax^3 + Bx^2 + Cx + (a+b -A-B-C) & x \in [1,2] \end{cases}$$
Can be written as $(ax^2 + bx, A (x^3 -1) + B(x^2 -1) + C(x-1) +a+b)$ where the first coordinate shows the function when $x \in [0,1]$ and second coordinate when $x \in [1,2]$. Now, it's quite easy to find the basis:
$$
a(x^2,1) + b(x, 1) + A(0, x^3 -1) +B(0, x^2 -1) + C(0, x-1)$$

 

[PeroK]

I would have transformed this problem into $\mathbb R^6$ by mapping the polynomial to $(a,b, A,B,C,D)$ with the condition that $a+b = A+B+C +D$.

Then, for a basis you just take each of $a,b,A,B,C$ equal to $1$ in turn and set $D$ accordingly.

 

[Steve4Physics]

This solution is for ("attributed to" wouldn't be right because I don't know if he really wanted me to do like this) @Steve4Physics

For information, @Hall is referring to a suggested approach (not a solution) to a similar, simpler problem, in a PM.