$\phi$ is diagonalizable iff $\lambda_1, \lambda_2,\lambda_3$ are distinct

Question

We have the matrix $A=\begin{pmatrix}\lambda_1 & 1 & 0 \\ 0 & \lambda_2 & 1 \\ 0 & 0 & \lambda_3\end{pmatrix}$ and $\phi (x)=Ax$, where $\lambda_1, \lambda_2,\lambda_3\in \mathbb{R}$.

I have shown that $\lambda_1, \lambda_2,\lambda_3$ are the eigenvalues of $\phi$.

I want to show that $\phi$ is diagonalizable iff $\lambda_1, \lambda_2,\lambda_3$ are distinct.

If all eigenvalues are distinct then it follows that $\phi$ is diagonalizable.

Why how can we show the other direction?

Do we have to use the dimension of the eigenspace and the algebraic multiplicity? Or can we show that?

Answer 1

In fact, $\phi$ is diagonalizable iff $\phi$ has three linearly independent eigenvectors.

If $\lambda_i=\lambda_j$ for some $1\le i\ne j\le 3$, then there is only one eigenvector of the eigenvalue $\lambda_i$. Thus, $\phi$ has only two linearly independent eigenvectors, that leads to contradiction.

Answer 2

for converse

use the result for a square matrix is diagonalizable iff arithmetic multiplicity(A.M.)= geometric multiplicity(G.M.) for each eigenvalue.

proof by contradiction

case 1: any two are equal particular $\lambda_1$ = $\lambda_2$ = $\lambda$(say) then A.M. of $\lambda$ is 2 and G.M. of $\lambda$ is 1 so not diagonalisible

case 2: if all eigenvalue is equal $\lambda_1$ = $\lambda_2$ = $\lambda_3$=$\lambda$(say) then A.M. of $\lambda$ is 3 and clearly G.M. of $\lambda$ is 1 so not diagonalisiable

Hence for a given matrix is diagonalizable $iff$ all eigenvalue are distinct.

Answer 3

Due to the position of the $1$s, the matrix $A - \lambda I$ will have nullity at most $1$ for any $\lambda$. (Whether or not it is an eigenvalue.)

If $A$ is diagonalisable, then $\operatorname{nullity}(A - \lambda I)$ has to be exactly equal to the algebraic multiplicity of $\lambda$ for every eigenvalue $\lambda$.
By the first paragraph, it is forced that the algebraic multiplicity is $1$ for all eigenvalues. The claim follows.