Finding a polynomial $t(x)$ such that $r(x)$ is a linear combination of $p(x)$, $q(x)$ and $t(x).$

Question

Let $$p(x)=x^2+5x-3$$ $$q(x)=4x^2+18x+4$$

Find a second degree polynomial $t(x)$ that satisfies: For every second degree polynomial $r$, it should be possible to find three constants $a$, $b$ and $c$ such that $$r(x)=a*p(x)+b*q(x)+c*t(x)$$.

My thoughts: If $r(x) = dx^2+ex+f$ (a general second degree polynomial) then $t(x)$ takes the form $t(x) = gx^2+hx+i$. I guess I have to end up with an augmented matrix, but i'm not sure how to proceed with this information and put it into a matrix. I have not yet learned about vector spaces, basis or ranks yet (beginning of the linear algebra course) so I would prefer some hints as to how I can solve this using row reduction of a matrix.

Answer 1

We can simply refer to the result obtained here Is $s(x)$ a linear combination of $a \cdot p(x) $and $b \cdot q(x)$ since $p(x),q(x)$ and $s(x)$ are linearly independent, just consider $t(s)=s(x)$ and we are done.

If we are given only $p(x)$ and $q(x)$, with respect to the standard basis $\{1,x,x^2\}$ we have

• $v_p=(-3,5,1)$
• $v_q=(4,18,4)$

then we can take the cross product

$$v_t=v_p \times v_q=\begin{vmatrix}i&j&k\\-3&5&1\\4&18&4\end{vmatrix}=(2,16,-74)$$

such that $t(x)=-37x^2+8x+1$ is linearly independent.

Note that the method by cross product is valid only in dimension $3$, more in general we can consider the big matrix and perform RREF algorithm

$$\begin{bmatrix}-3&5&1\\4&18&4\\1&0&0\\0&1&0\\0&0&1\end{bmatrix}\to \begin{bmatrix}-3&5&1\\0&74&16\\0&15&3\\0&0&-3\\0&0&0\end{bmatrix}\to \begin{bmatrix}-3&5&1\\0&37&8\\0&5&1\\0&0&1\\0&0&0\end{bmatrix}\to \begin{bmatrix}-3&5&1\\0&37&8\\0&0&-3\\0&0&1\\0&0&0\end{bmatrix}\to \begin{bmatrix}-3&5&1\\0&37&8\\0&0&1\\0&0&0\\0&0&0\end{bmatrix}$$

to conclude that also $t(x)=1$ is a good choice.

Note also that the same result can be obtained in a more straightforward and clever way by

$$\det\begin{bmatrix}-3&5&1\\4&18&4\\0&0&1\end{bmatrix}=\det\begin{bmatrix}-3&5\\4&18\end{bmatrix}=-74 \neq 0$$

Answer 2

.Let us abstract out the problem by writing $(a,b,c)$ to represent $ax^2 + bx +c$.

Then, what we seek is the following question : does there exist a polynomial $t(x) = (d,e,f)$, such that for all polynomials $r(x) = (g,h,i)$, there exist constants $a_r,b_r,c_r$ such that $(g,h,i) = a_r(1,5,-3) + b_r(4,18,4) + c_r(d,e,f)$.

That is, we are essentially saying : can you find $d,e,f$ such that the set $\{(1,5,-3),(4,18,4),(d,e,f)\}$ spans $\mathbb R^3$? So that every $(g,h,i)$ us a linear combination of these polynomials.

It is sufficient that they are linearly independent, and therefore that the following determinant : $$ \begin{vmatrix} 1 & 5 & -3 \\ 4 & 18 & 4 \\ d & e & f\\ \end{vmatrix} $$ is non-zero.

Expand the determinant to obtain some inequation in $d,e,f$, and see that it is very simple to choose $d,e,f$ in a manner satisfying the constraints.

Answer 3

However, I love the @gimusi way for your question and that you are not working on a certain interval i.e. $I=[a,b]$; The following way may be practical and tasty. Look at Gram determinant. Accordingly, the set $\{f_1,f_2,f_3\}$ which all are continuous on interval $I$ is linearly dependent iff the following determinant equals to $0$: $$ \begin{vmatrix}\int_a^bf_1^2&\int_a^bf_1f_2&\int_a^bf_1f_3\\\int_a^bf_2f_1&\int_a^bf_2^2&\int_a^bf_2f_3\\\int_a^bf_3f_1&\int_a^bf_3f_2&\int_a^bf_3^2\end{vmatrix} $$