Find values of $k$ where $A$ is diagonalizable

Question

I have the matrix

A= $\begin{pmatrix}1&1&k\\ 1&1&k\\ 1&1&k\end{pmatrix}$

And need to find the values of $k$ where $A$ is diagonalizable.

I'm not too sure how to go about this. Do I first row reduce it and find the eigenvalues?

Answer 1

It is easily seen that the rank of the matrix $A$ is $1$. So $0$ must be an eigenvalue of $A$ for any value of $k$. Also since sum of each rows of $A$ is $k+2$, so $k+2$ is also an eigenvalue of $A$. Now consider two separated cases:

Case:I [$k+2=0$] Here $0$ is the only eigenvalue of $A$, since $A^2=0$. Thus $A$ is not diagonalizable (minimal polynomial of $A$ is $m_a(t)=t^2$, not a product of distinct factors).

Case:II [$k+2\neq0$] Here $A$ has two different eigenvalue, namely $0$ and $k+2$. Using $dimE_\lambda=n-rank(A-\lambda I)$, for a $n\times n$ matrix $A$; $dimE_0=3-rankA=2$ which is equal to algebraic multiplicity of $0$. Also since algebraic multiplicity of $k+2$ is $1$, so $A$ is diagonalizable.

Thus for any real value $k\neq-2$, $A$ is diagonalizable over $\mathbb{R}.$

Answer 2

0 is clearly and eigenvalue. Can you find the eigenvector(s) associated with this eigenvalue?

Can you find any other eigenvectors? They are pretty easy to guess at. (Which is unusual.)

However for some value of $k$ these eigenvectors will not be linearly independent, in which case there is some missing "generalized eigenvector" and the matrix is not diagonalizable.

Answer 3

The matrix has rank one, so there are two zero eigenvalues, and one non-zero. The eigenvectors for eigenvalue zero are just the vectors orthogonal to $(1, 1, k),$ so if the non-zero eigenvalue is real, you are golden. When is it?

Answer 4

For $k+2\neq 0$ we can easily diagonalise $A$ using $$ S=\frac{1}{k+2}\begin{pmatrix} -(2k+1)(k+2) & -(k+1) & -1\cr k+2 & 1 & -1 \cr 2(k+2) & 1 & -1 \end{pmatrix}. $$ For $k=-2$ the Jordan form is $$ J_A=\begin{pmatrix} 0 & 1 & 0 \cr 0 & 0 & 0 \cr 0 & 0 & 0\end{pmatrix}, $$ so it is not diagonalizable.

Answer 5

$A$ has rank one. Therefore, $A$ is diagonalisable if and only if $A^2\ne0$.