# Eigenspace for a linear transformation + diagonalizability

I am new to linear algebra, and was asked the question below. I am just looking for some feedback regarding my proposed answer.

$$T:R_4[x] \to R_4[x]$$ is a linear transformation defined by $$T(p(x))=p(x)-3p''(x)$$.

For each eigenvalue of $$T$$, find its eigenspace. Is $$T$$ diagonalizable?

The standard basis of $$R_4[x]$$ is $$E = ((1x^3,0x^2,0x,0),(0x^3,1x^2,0x,0),(0x^3,0x^2,1x,0),(0x^3,0x^2,0x,1))$$

Applying $$T$$ to the standard basis gives us:

$$T(1x^3,0x^2,0x,0)=(1x^3,0x^2,0x,0)-3(0x^3,0x^2,6x,0)=(1x^3,0x^2,-18x,0)$$

$$T(0x^3,1x^2,0x,0)=(0x^3,1x^2,0x,0)-3(0x^3,0x^2,0x,2)=(0x^3,1x^2,0x,-6)$$

$$T(0x^3,0x^2,1x,0)=(0x^3,0x^2,1x,0)-3(0x^3,0x^2,0x,0)=(0x^3,0x^2,1x,0)$$

$$T(0x^3,0x^2,0x,1)=(0x^3,0x^2,0x,1)-3(0x^3,0x^2,0x,0)=(0x^3,0x^2,0x,1)$$

Therefore, the representative matrix of $$T$$, which we shall name $$B$$ is: $$\begin{bmatrix} 1&0&0&0\\0&1&0&0\\-18&0&1&0\\0&-6&0&1\end{bmatrix}$$

$$|λI-B|=$$ $$\begin{vmatrix} λ-1&0&0&0\\0&λ-1&0&0\\18&0&λ-1&0\\0&6&0&λ-1\end{vmatrix}$$

The characteristic polynomial comes out as (λ-1)(λ-1)(λ-1)(λ-1), so we have one eigenvalue λ=1, with algebraic multiplicity 4. I then entered λ = 1 into $$|λI-B|$$ and converted it into a matrix representing a homogenous system. Applying Gauss-Jordan to that homogenous system gave me

$$x_1=0$$

$$x_2=0$$

$$x_3=x_3$$

$$x_4=x_4$$

and therefore $$x_3(0,0,1,0)+x_4(0,0,0,1)$$

So $$P(1I-A)=Sp\{(0,0,1,0)(0,0,0,1)\}$$

These vectors are linearly independent so this set is a basis for the eigenspace of eigenvalue λ=1.

The geometric multiplicity is 2, whereas the algebraic multiplicity is 4, so $$T$$ is not diagonalizable.

Thank you!

Your conclusion is correct. (Your notation for elements of $$\mathbb R_4[x]$$ is a bit curious, but I'm assuming you're following a convention set by your textbook).
If you just wanted to conclude the operator is not diagonalizable, you could get by a bit quicker by noting that your matrix is similar to $$\begin{bmatrix} 1 & -18 \\ & 1 \\ & & 1 & -6 \\ &&& 1 \end{bmatrix}$$ simply by writing your basis is a different order (namely $$x, x^3, 1, x^2$$). Since this matrix is in Jordan form and not already diagonal, it is not diagonalizable.