Science:Math Exam Resources/Courses/MATH307/December 2008/Question 02

MATH307 December 2008

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Question 02
Let $\mathbb {P} _{3}$ be the vector space of polynomials of degree $\leq 3$. The monomials $1,x,x^{2},x^{3}$ form a basis of $\mathbb {P} _{3}$. Consider the linear mapping from $\mathbb {P} _{3}$ into $\mathbb {P} _{3}$ given by $p(x)\mapsto p''(x)$. Find the matrix A of the mapping with respect to the monomial basis and determine its null space.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
What is the dimension of $\mathbb {P} _{3}$? How can you represent a general polynomial of $\mathbb {P} _{3}$ in vector notation, that is, as a linear combination of the given basis?

Hint 2
How does A act on the basis vectors? In other words, calculate the second derivative of the basis vectors and write your results in vector notation.

Hint 3
To find the null space look at the zero columns of A. To check your answer, which of the basis monomials get mapped to the zero polynomial when you take the second derivative?

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The number of basis vectors equals the dimension of the vector space, so $\mathbb {P} _{3}$ has dimension 4. To begin with, we represent the basis monomials as vectors: ${\begin{aligned}1\leftrightarrow e_{1}={\begin{bmatrix}1\\0\\0\\0\end{bmatrix}},\quad x\leftrightarrow e_{2}={\begin{bmatrix}0\\1\\0\\0\end{bmatrix}},\quad x^{2}\leftrightarrow e_{3}={\begin{bmatrix}0\\0\\1\\0\end{bmatrix}},\quad x^{3}\leftrightarrow e_{4}={\begin{bmatrix}0\\0\\0\\1\end{bmatrix}}\end{aligned}}$ Then, a general polynomial $\displaystyle a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}$ can represented as $a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}=a_{3}e_{4}+a_{2}e_{3}+a_{1}e_{2}+a_{0}e_{1}={\begin{bmatrix}a_{0}\\a_{1}\\a_{2}\\a_{3}\end{bmatrix}}$ in the given basis. Next we calculate how A acts on the basis: ${\begin{aligned}Ae_{1}&=1''=0\\Ae_{2}&=x''=0\\Ae_{3}&=(x^{2})''=2=2e_{1}\\Ae_{4}&=(x^{3})''=6x=6e_{2}\end{aligned}}$ This determines the columns of A, i.e. $A={\begin{bmatrix}0&0&2&0\\0&0&0&6\\0&0&0&0\\0&0&0&0\end{bmatrix}}$ To double check this we calculate ${\begin{aligned}(a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0})''&\leftrightarrow A{\begin{bmatrix}a_{0}\\a_{1}\\a_{2}\\a_{3}\end{bmatrix}}\\6a_{3}x+2a_{2}&\leftrightarrow {\begin{bmatrix}2a_{2}\\6a_{3}\\0\\0\end{bmatrix}}\\6a_{3}e_{2}+2a_{2}e_{1}&\leftrightarrow 6a_{3}{\begin{bmatrix}0\\1\\0\\0\end{bmatrix}}+2a_{2}{\begin{bmatrix}1\\0\\0\\0\end{bmatrix}}\quad \checkmark \end{aligned}}$ Finally, since the second derivative of the first two monomials is zero, the two corresponding vectors form a basis of the null space of A: $N(A)={\text{span}}\left(\left\{{\begin{bmatrix}1\\0\\0\\0\end{bmatrix}},{\begin{bmatrix}0\\1\\0\\0\end{bmatrix}}\right\}\right)$

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