Multiple choice: linear algebra re: diagonalizability

Question

I have this exercise I can't solve properly:

• (i) $T:V→V$ is a linear transformation in $\Bbb R^4$.
• (ii) $X(\lambda) = (\lambda - 2)\cdot q(\lambda)$.
• (iii) $q(\lambda)$ is a grade 3 polynomial.
• (iv) $Im(T-3I)={(x,y,z,t)\in\Bbb R^4:x+y=0,x-z=0,3t+z=0}.$

Affirmations:

1. T is diagonalizable
2. $q(3)=0$
3. $m^G$(3)=2

Answers: (The only one that matters is the correct one, according to the test's solution, which is:)

• only 1 and 2 are correct

So...

From (iv) I can get that 3 is an eigenvalue with $m^G$(3)=1, then clearly $q(3)=0$ because it's a root of the characteristic polynomial.

But I can't get to T being diagonalizable. I guess I need to find information about the existance of two more different eigenvectors. Meaning, $q(\lambda)$ has either:

• one more real root $\lambda_3 ≠ 2 ≠ 3 $ with $m^G(\lambda_3)=2$, or
• two more real roots

Where is that information?

Thanks!

Answer 1

You have made a mistake when computing the geometric multiplicity of $3$; since $Im(T-3I)$ has dimension $1$, the eigenspace for $3$, $Ker(T-3I)$, has dimension $3$. Then $T$ is diagonalisable, since it has another eigenvalue.