Linear algebra -Transformations and eigenvectors

Question

Consider the operator $T:\mathbf{P}_2\rightarrow\mathbf{P}_2$ by differentiation, $T(p) = p'$. Is there a basis of eigenvectors of this operator? Can someone help me please?

Thanks

Answer 1

There are many ways to do this.

If $T:V\to V$ is a linear map, then there exists a basis for $V$ consisting of eigenvectors for $T$ if and only if there exists a basis $\beta$ for $V$ such that the matrix $[T]_\beta$ is diagonalizable.

Now, we have the map $T:P_2(\Bbb R)\to P_2(\Bbb R)$ given by $T(p)=p^\prime$. Note that $\beta=\{1,x,x^2\}$ is a basis for $P_2(\Bbb R)$. Moreover, we may write \begin{array}{rcrcrcr} T(1) & = & 0\cdot 1 & + & 0\cdot x & + & 0\cdot x^2 \\ T(x) & = & 1\cdot 1 & + & 0\cdot x & + & 0\cdot x^2 \\ T(x^2) & = & 0\cdot 1 & + & 2\cdot x & + & 0\cdot x^2 \end{array} It follows that $$ [T]_\beta= \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{bmatrix} $$ Note that $[T]_\beta$ is not diagonalizable (prove this!). Since $[T]_\alpha$ is similar to $[T]_\beta$ for any other basis $\alpha$ we have that $[T]_\alpha$ is never diagonalizable. Hence there is no basis for $P_2(\Bbb R)$ consisting of eigenvectors of $T$.

Of course we could have done this more directly by examining the equation $$ p^\prime(x)=\lambda\cdot p(x) $$ and finding that the only solution is $p(x)=a$ for some constant $a$ and $\lambda=0$. This tells us that total eigenspace of $T$ has dimension $1$.