Linear Algebra

Is the Derivative Linear Transformation Diagonalizable?

Diagonalization Problems and Solutions in Linear Algebra

Problem 690

Let $\mathrm{P}_2$ denote the vector space of polynomials of degree $2$ or less, and let $T : \mathrm{P}_2 \rightarrow \mathrm{P}_2$ be the derivative linear transformation, defined by
\[ T( ax^2 + bx + c ) = 2ax + b . \]

Is $T$ diagonalizable? If so, find a diagonal matrix which represents $T$. If not, explain why not.

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Solution.

The standard basis of the vector space $\mathrm{P}_2$ is the set $B = \{ 1 , x , x^2 \}$. The matrix representing $T$ with respect to this basis is
\[ [T]_B = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{bmatrix} . \]

The characteristic polynomial of this matrix is
\[ \det ( [T]_B – \lambda I ) = \begin{vmatrix} -\lambda & 1 & 0 \\ 0 & -\lambda & 2 \\ 0 & 0 & -\lambda \end{vmatrix} = \, – \lambda^3 . \] We see that the only eigenvalue of $T$ is $0$ with algebraic multiplicity $3$.

On the other hand, a polynomial $f(x)$ satisfies $T(f)(x) = 0$ if and only if $f(x) = c$ is a constant. The null space of $T$ is spanned by the single constant polynomial $\mathbb{1}(x) = 1$, and thus is one-dimensional. This means that the geometric multiplicity of the eigenvalue $0$ is only $1$.

Because the geometric multiplicity of $0$ is less than the algebraic multiplicity, the map $T$ is defective, and thus not diagonalizable.

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