Question

Let $P_{11}\left(x\right)$ be the real vector space of polynomials, in the variable $x$ with real coefficients and having degree at most 11, together with the zero polynomial. Let

$E=\{s_0\left(x\right),s_1\left(x\right),\mathrm{.}\mathrm{.}\mathrm{.},s_{11}\left(x\right)\},F=\{r_0\left(x\right),r_1\left(x\right),\mathrm{.}\mathrm{.}\mathrm{.},r_{11}\left(x\right)\}$

be subsets of $P_{11}\left(x\right)$ having 12 elements each and satisfying

$s_0\left(3\right)=s_1\left(3\right),=\mathrm{.}\mathrm{.}\mathrm{.}=s_{11}\left(3\right)=0,r_0\left(4\right)=r_1\left(4\right)=\mathrm{.}\mathrm{.}\mathrm{.}=r_{11}\left(4\right)=1.$

Then, which one of the following is TRUE?

1. Any such 𝐸 is not necessarily linearly dependent and any such 𝐹 is not necessarily linearly dependent
2. Any such 𝐸 is necessarily linearly dependent but any such 𝐹 is not necessarily linearly dependent
3. Any such 𝐸 is not necessarily linearly dependent but any such 𝐹 is necessarily linearly dependent
4. Any such 𝐸 is necessarily linearly dependent and any such 𝐹 is necessarily linearly dependent

Answer: Any such 𝐸 is necessarily linearly dependent but any such 𝐹 is not necessarily linearly dependent

Solution :-

We have $E=\{s_0\left(x\right),s_1\left(x\right),\mathrm{.}\mathrm{.}\mathrm{.},s_{11}\left(x\right)\}\subseteq P_{11}\left(x\right)$ , satisfying $s_0\left(3\right)=s_1\left(3\right)=\mathrm{.}\mathrm{.}\mathrm{.}=s_{11}\left(3\right)=0$ .
Here, each polynomials in $E$ gives value 0 as we put $x=3$ . So, $x=3$ is a root of each polynomials in $E$ . So each polynomials in $E$ is of the form $s_i\left(x\right)=\left(x-3\right)q_i\left(x\right)$ , where $q_i\left(x\right)\in P_{10}\left(x\right)$ , $i\in \{0,1,\cdots ,11\}$ . This suggest that each polynomials in $E$ is divisible by $\left(x-3\right)$ .

According to queation given that $P_{11}\left(x\right)=\{a_0+a_{1}x+\cdots +a_{11}{x}^{11}:a_0,a_1,\cdots ,a_{11}\in ℝ\}$ is a vector space. Now, we form a set W of the collection of all polynomials of degree 11 which are divisible by $\left(x-3\right)$ , then the set W will be a subset of $P_{11}\left(x\right)$ .

So, W is of the form $W=\{\left(x-3\right)q\left(x\right):q\left(x\right)\in P_{10}\left(x\right)\}$ . Here, W is not just a subset of $P_{11}\left(x\right)$ but W forms a subspace of $P_{11}\left(x\right)$ .

Let us verify whether W forms a subspace of $P_{11}\left(x\right)$ or not.
Let $\left(x-3\right)q_k,\left(x-3\right)q_m\in W\text{and}\alpha ,\beta \in ℝ$ here $q_k,q_m\in P_{10}\left(x\right)$ , then $\alpha \left(x-3\right)q_k+\beta \left(x-3\right)q_m=\left(x-3\right)\left(\alpha q_k+\beta q_m\right)\in W$ because $\left(\alpha q_k+\beta q_m\right)\in P_{10}\left(x\right)$ , since $P_{10}\left(x\right)$ is a vector space.

Hence, W is a subspace of $P_{11}\left(x\right)$ . And we know that basis for $P_{10}\left(x\right)$ is the set $\{1,x,\cdots ,x^{10}\}$ , so basis for W is the set $\{\left(x-3\right)\cdot 1,\left(x-3\right)\cdot x,\cdots ,\left(x-3\right)\cdot x^{10}\}$ . Because of W maked by multiplying each vectors of $P_{10}\left(x\right)$ through $\left(x-3\right)$ .

So, the dimension of subspace W and the dimension of $P_{10}\left(x\right)$ are the same, which is 11.

We established above all elements of $E$ look like $s_i\left(x\right)=\left(x-3\right)q_i\left(x\right)$ , where $q_i\left(x\right)\in P_{10}\left(x\right)$ , $i\in \{0,1,\cdots ,11\}$ , and all elements of $s_i\left(x\right)$ type belong to W. So, all elements of $E$ are elements of subspace W. Therefore, all such $E$ is also a subset of subspace W.
Now, we know that if any subset of a vector space contains more elements than the dimension of that vector space, then that subset is linearly dependent.

Here, the total vectors in $E$ are 12, and the dimension of W is 11; hence, any such $E$ is necessarily linearly dependent on W, and W is a subspace of $P_{11}\left(x\right)$ , so $E$ is also necessarily linearly dependent on $P_{11}\left(x\right)$ .

Now, We have $F=\{r_0\left(x\right),r_1\left(x\right),\mathrm{.}\mathrm{.}\mathrm{.},r_{11}\left(x\right)\}\subseteq P_{11}\left(x\right)$ , satisfying $r_0\left(4\right)=r_1\left(4\right)=\mathrm{.}\mathrm{.}\mathrm{.}=r_{11}\left(4\right)=1$ .

Let us consider a set of vectors, $r_0\left(x\right)=1,\text{and}{r}_{i+1}\left(x\right)=\left(x-4\right)x^i+1,\text{where}i\in \{0,1,\cdots ,10\}\mathrm{.}$ Here, all polynomials $r_0\left(x\right),r_1\left(x\right),\cdots ,r_{11}\left(x\right)$ attains 1 at $x=4.$

Now, let us consider a matrix $M_F$ whose columns are scalar coefficients of each vectors above.

Here, we can see that this is a upper triangular matrix of order 12 × 12 and rank of this matrix is 12 which is equal to the number of vectors in $F$ and hence $F$ is linearly independent.

Therefore, any such $F$ is not necessarily linearly dependent.

So, the correct answer is option-2, $\to$ Any such 𝐸 is necessarily linearly dependent but any such 𝐹 is not necessarily linearly dependent