Diagonalizable matrix known rank, trace and a vector property

Question

Given a matrix $A:\mathbb{R}^3 \to \mathbb{R}^3$, assume it has rank 2 and trace 6. We also know that there exists a vector such that $A\vec{x}=\vec{x}$. The thing is that I don't know how to prove the matrix is diagonalizable.

Answer 1

Here's how to prove it:

Since $A$ has rank $2$, there is a non-zero vector $\vec y$ such that

$A \vec y = 0; \tag{1}$

this is true because one of the columns must be linearly dependent on the other two; thus if $\vec B$, $\vec C$, $\vec D$ are the columns of $A$, so that

$A = \begin{bmatrix} \vec B & \vec C & \vec D \end{bmatrix}; \tag{2}$

then since there exist real $a$, $b$, $c$ such that

$a \vec B + b \vec C + d \vec D = 0, \tag{3}$

we may write

$A \begin{pmatrix} a \\ b \\ c \end{pmatrix} = 0, \tag{4}$

and thus take

$\vec y = (a, b, c)^T. \tag{5}$

We conclude from (1) that $0$ is an eigenvalue of $A$.

From

$A \vec x = \vec x, \tag{6}$

we further conclude that $1$ is another eigenvalue of $A$.

Since

$Tr(A) = 6, \tag{7}$

we find that the remaining eigenvalue of $A$ must be $5$; $A$ has three distinct eigenvalues; hence, it may be diagonalized.