All values of $k$ for which matrix is diagonisable

Question

Find all values of k for which matrix is diagonalizable: $$A= \begin{bmatrix} 1 & 1 & k\\ 1 & 1 & k\\ 1 & 1 & k \end{bmatrix} $$ The question contains multiple matrices but I'm stuck on this one. I'm even doubtful that my technique to solve these kind of questions is right. What I do is start by writing out the characteristic polynomial and then get the Eigen values. After replacing value of my Eigen value in $A-\lambda I$ ($\lambda$ is eigenvalue), I solve $[A-\lambda I | 0]$. In this particular problem, I'm having a hard time figuring out the Eigen value itself.

I'll really appreciate any help.!

Answer 1

Just by inspection note that rank of this matrix is $1$ hence it is non-invertible. So one eigen value of this matrix is $\lambda=0$. Also sum of each row is $k+2$, thus another (possibly different) eigen value is $\lambda=k+2$. The trace (sum of diagonal elements) which also happens to be the sum of eigen values is $k+2$. Thus the third eigen value has to be zero as well. This means the eigen values are $\lambda=0,0,k+2$.

Now consider the system $A-\lambda I$ for each $\lambda$. If you have three linearly independent eigen vectors then $A$ will certainly be diagonalizable.

Let $\lambda =0$. Then $$ A-\lambda I = \begin{bmatrix} 1 & 1 & k\\ 1 & 1 & k\\ 1 & 1 & k \end{bmatrix} \quad \xrightarrow{\text{reduced to }} \quad \begin{bmatrix} 1 & 1 & k\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} $$ Since this matrix has rank $1$. Thus nullity must be $2$. In which case we will get two linearly independent eigen vectors.

Now there are two possibilities: $k+2 =0$ and $k+2 \neq 0$ (i.e. third eigen value is same or different).

If $k+2 \neq 0$, then the eigen value $\lambda=k+2$ will have a third linearly independent eigen vector. Hence $A$ is diagonalizable.

But if $k+2 = 0$, then from above it is clear that we cannot have three linearly independent eigen vector for $\lambda=0$. Hence $A$ is not diagonalizable.

So for $k \neq -2$, the matrix is diagonalizable.

Answer 2

Note that $A$ is not invertible then $0$ will be an eigenvalue. You find indeed that $$\chi A=X^2(2+k-X),$$ and then your minimum polynomial has to be :

• $\mu_A=X$ if $k=-2,$ and we see that it's impossible
• $\mu_A=X(2+k-X)$ if $k\neq-2,$ which implies $$\begin{pmatrix}1&1&k\\1&1&k\\1&1&k\end{pmatrix}\begin{pmatrix}2+k-1&-1&-k\\-1&2+k-1&-k\\-1&-1&2+k-k\end{pmatrix}=0_{\mathcal{M}_3(\mathbb{R})}$$ which is always true for $k\neq 2.$

Finally, $k=-2$ is the only value such that $M$ is not diagonalisable.