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?

My answer is:

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:





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





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.


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