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.!


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.

  • $\begingroup$ Hi, thanks for your answer. But the answer is All real values of k. (behind the book) $\endgroup$ – Anirudh Gangwal Dec 6 '15 at 0:16
  • $\begingroup$ You say some of each row is k +2, therefore k +2 is another possible eigenvalue. How do you conclude this? $\endgroup$ – Anirudh Gangwal Dec 9 '15 at 19:56
  • $\begingroup$ @AnirudhGangwal because with $\mathbf{x}=(1,1,1)$ we get $A\mathbf{x}=(k+2)\mathbf{x}$. Thus $(1,1,1)$ is an eigen vector corresponding to $k+2$ as eigen value. $\endgroup$ – Anurag A Dec 10 '15 at 15:47

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.

  • $\begingroup$ @AnuragA : I edited that, thank you ! $\endgroup$ – Balloon Dec 6 '15 at 0:09
  • $\begingroup$ Hi, thanks for your answer. But the answer is All real values of k. $\endgroup$ – Anirudh Gangwal Dec 6 '15 at 0:17
  • $\begingroup$ You can read "$M$ is no diagonalisable iff $k=-2$" as "$M$ is diagonalisable iff $k\neq -2,$" and we checked all values for $k\in\mathbb{R}$ ? $\endgroup$ – Balloon Dec 6 '15 at 0:19
  • $\begingroup$ Yes, I understand that. It would mean that for A to be diagonisable, k ∈ R - {-2}. But the answer is R. $\endgroup$ – Anirudh Gangwal Dec 6 '15 at 0:26
  • $\begingroup$ But AnuragA and I showed that it is $\mathbb{R}-\{-2\}.$ Aren't you convinced ? $\endgroup$ – Balloon Dec 6 '15 at 0:30

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