# Matrices and Linear Transformation

I am having trouble with a certain question.

We denote $D$ the differentiation operator and $T:V \rightarrow V$ a linear transformation which maps p(x) onto xp'(x)

Let W be the image of V under TD such that TD=T[D(p(x))]. Find the bases for V and for W relative to which the matrix of TD is in diagonal form.

• What is $V$? Doesn't $T$ only multiplies with $x$, and thus $xp'$ is just $TD(p)$ (for polynomial $p$)? – Berci Feb 18 '13 at 23:31
• V is the linear space of all real polynomials of degree $\leq 3$ (sorry about that) – Carpediem Feb 18 '13 at 23:32
• Please edit that into the question --- people shouldn't need to read the comments to understand the question. – Gerry Myerson Feb 18 '13 at 23:33
• No it is explicitly mentioned in the problem – Carpediem Feb 18 '13 at 23:49
• I think we can think of $D$ as $V\to V$ too, only that it kills the fourth coordinate, and we're fine. – Pedro Tamaroff Feb 18 '13 at 23:54

Since the matrix you get isn't square I'm going to assume that "diagonal form" means you want all nonzero entries in the matrix to be on the diagonal that extends from the upper left corner. This is not a diagonal matrix though... so I think this is a strangely worded question.

Hint: $TD$ lowers degree's by $1$ so $W$ is going to be polynomials of degree $\leq 2$. The standard basis for $V$ is $\{1, x, x^2, x^3\}$ and the standard basis for $W$ is $\{1, x, x^2\}$. These bases will work, just not in that order.

Keep the basis for $W$ in the order that it's in and try changing the order of the basis for $V$ to make the matrix diagonal.

• We can work with a row of zeroes as DonAntonio does. – Pedro Tamaroff Feb 19 '13 at 0:01

\begin{align*}TD(1)&=&0&=0\cdot 1+0\cdot x+0\cdot x^2+0\cdot x^3\\ TD(x)&=T(1)=&0&=0\cdot 1+0\cdot x+0\cdot x^2+0\cdot x^3\\ TD(x^2)&=T(2x)=&2x&=0\cdot 1+2\cdot x+0\cdot x^2+0\cdot x^3\\ TD(x^3)&=T(3x^2)=&6x^2&=0\cdot 1+0\cdot x+6\cdot x^2+0\cdot x^3\end{align*}

Thus, wrt the given basis, the matrix corresponding to $\,TD\,$ is

$$A:=\begin{pmatrix}0&0&0&0\\0&0&2&0\\0&0&0&6\\0&0&0&0\end{pmatrix}\Longrightarrow p_A(t):=\det(tI-A)=\begin{vmatrix}t&0&0&0\\0&t&-2&0\\0&0&t&-6\\0&0&0&t\end{vmatrix}=t^4$$

So the matrix is nilpotent but certainly not diagonalizable as the only nilpotent diagonalizable matrix is the zero matrix (why? Check the minimal polynomial....)

• Is there a way to do this without the concept of determinant. We still haven't seen this – Carpediem Feb 19 '13 at 0:00
• I can't think right now from the top of my head of any other method but "guessing" basis that gives a diagonal form, which seems a ridiculous method indeed. How do you "find" then the diagonal form, in case it exists, of a matrix?! – DonAntonio Feb 19 '13 at 0:04
• @DonAntonio I'm trying to help in the chat. One might try and model two bases $B_1,B_2$ with a diagonal matrix and show they can't exist. – Pedro Tamaroff Feb 19 '13 at 0:08
• I really have no the slightest, palest idea how to do that choosing out of $\,\aleph_0\,$ options to choose those bases, @PeterTamaroff – DonAntonio Feb 19 '13 at 0:09
• Translating your idea to simplest words, you can say the following: "Saying the matrix is diagonalizable means there exist two basis such that $|TD|_{B_1B_2}$ is diagonal which means there exists a matrix $U$ (a matrix to change of basis) such that, if $M$ is your canonical matrix, $UMU^{-1}=D$ where $D$ is diagonal, not identically zero. But $M$ is nilpotent, which means $U$ can't exist." – Pedro Tamaroff Feb 19 '13 at 0:28

The matrix of $TD=p\mapsto xp''$ in the standard basis $(1,x,x^2,x^3)$ is: (The image of $i^\text{th}$ basis element is written in the $i^\text{th}$ column -- unless I miscalculated:) $$[TD]=\pmatrix{0&0&0&0\\0&0&2&0\\0&0&0&6\\0&0&0&0}$$ Now find the eigenvectors for this. (It is seen that $e_1$ and $e_2$ are already eigenvectors with eigenvalue $0$.)

• I haven't seen eigenvectors still – Carpediem Feb 19 '13 at 0:03
• I was mislead by P.Tamaroff's comment. I have edited the matrix. Hmm.. haven't seen eigenvectors? .. – Berci Feb 19 '13 at 0:06
• @Berci Sorry. I corrected it later on. Will delte the misleading comment. – Pedro Tamaroff Feb 19 '13 at 0:09
• No problem. You convinced me for a minute.. :) – Berci Feb 19 '13 at 0:13