# Find a basis for this vector space

1. Nov 20, 2015

### Steve Turchin

1. The problem statement, all variables and given/known data
Find a basis for the following vector space:

$V = \{ p \in \mathbb C_{\leq4} ^{[z]} | \ p(1)=p(i)$ and $p(2)=0 \}$
(Where $\mathbb C_{\leq4} ^{[z]}$ denotes the polynomials of degree at most 4)

2. Relevant equations
N/A

3. The attempt at a solution
I tried to find bi-terms with all possible degree combinations. such as:
$8z-z^4$ and $2z^3-z^4$
The $p(2)=0$ part is easy, but I can't seem to find any bi-terms that pass $p(1)=p(i)$
I'm afraid that randomly trying out tri-term and quad-term combinations can get messy.

And a side question: Is it true that, suppose there are no polynomials for which $p(1)=p(i)$, or more generally, a vector space that is the trivial one which contains only the zero vector. Then the basis of that vector space is the empty set?

Last edited: Nov 20, 2015
2. Nov 20, 2015

### Staff: Mentor

Your polynomial has the form $p(z) = a_4z^4 + a_3z^3 + a_2z^2 + a_1z + a_0$. Since p(2) = 0, then z - 2 must be a factor, leaving the other factor a cubic.
So $p(z) = (z - 2)(c_3z^3 + c_2z^2 + c_1z + c_0)$
Use the given conditions that p(1) = p(i) to see if you can solve for the coefficients. Having one equation with four constants might be a clue for the size of the basis.

3. Nov 20, 2015

### Steve Turchin

Thank you so much for the thorough reply and sorry for posting this on the wrong forum.
Solving $p(1)=p(i)$
$(1-2)(c_3+c_2+c_1+c_0)=(i-2)(c_3 i^3+c_2 i^2+c_1 i+c_0)$
I get: $-(c_3+c_2+c_1+c_0)+i(c_0-c_2+2c_3-2c_1)+c_3-c_1+2c_2-2c_0$
$c_0-c_2+2c_3-2c_1=0$

$2c_3+3c_2-c_0=0$
$2c_3+c_2-c_1=0$
Arbitrary values: $c_3=1,c_2=1,c_1=3,c_0=5$
So there's one polynomial: $p_1(z)= (z-2)(z^3+z^2+3z+5)$

Please correct me if I'm wrong: If there are four constants then I need four linear independent polynomial vectors. (or is it four linear independent vectors of constants?)

Here's another three:
$p_2(z)= (z-2)(2z^3+4z^2+8z+16)$
$p_3(z)= (z-2)(5z^3-9z^2+z-17)$
$p_4(z)= (z-2)(3z^3+z^2-7z+9)$

Got this far. How do I know if these polynomials are linearly independent, and $\mathsf {span}(p_1(z),p_2(z),p_3(z),p_4(z)) = V$

To check for linear independence, do I need to reduce a matrix, of the vectors of constants, to row echelon form, and make sure that there is only a single answer? If true, should I use the $c_3,c_2,c_1,c_0$ vectors or the $a_4,a_3,a_2,a_1,a_0$ vectors?
Is it true that if $\mathsf {dim}(p_1(z),p_2(z),p_3(z),p_4(z)) = 4$ and the the vectors are linearly independent, then that is the basis? By size of the basis, did you mean the dimension, that is equal to the amount of constants, in this case?

Last edited: Nov 20, 2015
4. Nov 20, 2015

### vela

Staff Emeritus
I can guess where you got $c_0-c_2+2c_3-2c_1=0$, but the origin of the following two equations is unclear. In any case, you can't claim $c_0-c_2+2c_3-2c_1=0$ because the coefficients are complex.

Like Mark, I initially thought to factor (z-2) out of the polynomial, but I think the problem is actually easier to do if you stick with the original version of f(z) in terms of the $a$'s. Write down the two equations you get from $f(2)=0$ and $f(1)-f(i)=0$. You'll have a system of equations, so you can solve for two coefficients in terms of the others. Perhaps once you see that, it'll jog your memory on how to find a basis.

5. Nov 21, 2015

### Steve Turchin

Thanks a lot vela!

$p(2)=16a_4+8a_3+4a_2+2a_1+a_0=0$

$p(1)-p(i)=0=a_4+a_3+a_2+a_1+a_0-(a_4-ia_3-a_2+ia_1+a_0)$

$(1+i)a_3+2a_2+(1-i)a_1=0 \ \ \Rightarrow \ a_3=\frac{(i-1)a_1-2a_2}{1+i}$

Solving the first equation:
$16a_4+8(\frac{i-1}{1+i}a_1-\frac{2}{1+i}a_2)+4a_2+2a_1+a_0 = 0$

Arbitrary values for $a_3,a_2,a_1:$
$a_1=1, \ a_2=-1, \ a_3=1$

$16a_4+8\cdot1+4\cdot(-1)+2\cdot1+a_0=0$
$16a_4+6+a_0 = 0$

Arbitrary values for $a_4,a_0:$
$a_4=-1 , a_0=0$

I get: $-z^4,z^3,-z^2,z,10$
The coefficients are real (subspace of complex). Do the coefficients have to be NOT real?
Is this a basis, or do I need several polynomials?

edit: Just noticed, the arbitrary values I chose are not linearly independent...

edit 2 : Did some more calculations and came up with these three vectors:
$v_1= z^4-16$
$v_2= 4-8i+z^2-\frac{2}{1+i}z^3$
$v_3=-6+z-z^2+z^3$
Since in the system of coefficients, two coefficients can be solved in terms of the others.
I believe the basis should consist of three vectors.

Last edited: Nov 21, 2015
6. Nov 21, 2015

### pasmith

You need at least a quadratic, since if $p(1) = p(i)$ for linear $p$ then $p$ is constant, and in view of $p(2) = 0$ must be zero.

Therefore you can start with $$f(z) = a + (z - 1)(z - i)(Bz^2 + Cz + D)$$ as the most general polynomial of degree at most 4 which satisfies $f(1) = f(i)$.

But since you want $f(2) = 0$ it makes sense to write $$Bz^2 + Cz + D = \frac{b(z-2)^2 + c(z - 2) + d}{(2 - 1)(2 - i)} = \frac{b(z-2)^2 + c(z - 2) + d}{2 - i},$$ to obtain $$f(z) = a + \frac{(z - 1)(z - i)}{2 - i} \left( b(z - 2)^2 + c(z - 2) - a \right)$$ as the most general polynomial of degree at most 4 which satisfies $f(1) = f(i)$ and $f(2) = 0$.

7. Nov 21, 2015

### vela

Staff Emeritus
No, they can be real.

Looks good. It sounds like you figured it out correctly.