2
$\begingroup$

I have the matrix

$$A=\begin{bmatrix} 2 && -2 && k \\ -1 && 2 && 0 \\ -2 && 0 && 2 \end{bmatrix}$$

Now I need to calculate the parameter $k$ so the matrix is similar to a diagonal matrix. After I have found $k$ I need to calculate the matrix $P$ so that the matrix $P^{-1}AP$ is a diagonal matrix.

I calculated the characteristic polynomial of the matrix $A$: $(2 - \lambda)(\lambda^2 - 4 \lambda + 2k + 2)$ (correct me if I'm wrong here). From the polynomial I can se that one eigenvalue is $2$. Now I need to find other eigenvalues from the second part of the polynomial. Here I get kind of stuck. I know that the polynomial is equal to $0$ if $2k + 2 = 0$. From here I get $k=-1$. Now how do I get the eigenvalues? Can I return the calculated $k$ to the characteristic polynomial and then calculate $\lambda^2 - 4\lambda = 0$. Then I would get $\lambda=0$ and $\lambda=4$. Hereby, my matrix $A$ would have 3 different eigenvalues and would be similar to the a diagonal matrix and $P$ would be a matrix formed of eigenvectors I guess. I just need someone to correct me if I'm wrong, or propose a different solution. And also, is this the only possible value of $k$? Thanks

$\endgroup$
1
$\begingroup$

So, the problem is to find $k$ such that the matrix is diagonalizable. From the characteristic polynomial , $(\lambda-2)(\lambda^2-4\lambda+2k+2)$ we get 2 as one eigenvalue. So, next to find the value of $k$ for which $\lambda^2-4\lambda+2k+2$ does not give us the same root. $i.e.\quad k\neq1$.

Moreover, we can see that for $k=1$, the characteristic polynomial becomes $(\lambda-2)^3$. $i.e.$ 2 is an eigenvalue with algebraic multiplicity 3.But $Null(A-2I)=1$. Hence $A$ no diagonalizable.

NOTE: Please add if I have missed something.

$\endgroup$
0
$\begingroup$

You did not say which is the field that you are working with. I shall assume that it happens to be $\mathbb R$.

If the polynomial $P(\lambda)=\lambda^2-4\lambda+2k+2$ has two real roots, none of which is $2$, then your matrix is similar to a diagonal matrix. Since $P(2)=2k-2$, you want to have $k\neq1$. Furthermore, $P(\lambda)$ has $2$ real roots distinct if and only if $8-8k>0$, that is, if and only if $k<1$. Therefore, your matrix is similar to a diagonal matrix if $k<1$ and it is not similar to a diagonal matrix if $k>1$.

If $k=1$, then the only eigenvalue is $2$. Since your matrix is not twice the identity matrix, it is not similar to a diagonal matrix.

$\endgroup$
  • $\begingroup$ Yes, it is the $\mathbb R$ field, forgot to write that. Okay, so the matrix will have 3 distinct eigenvalues for any value of $k<1$. But what about the values greater than 1. How are we sure that the matrix won't be diagonalizable if it has less than 3 distinct eigenvalues? It might still have 3 linear independent eigenvectors. And how would I find the matrix $P$ then? If you could just clear that up for me, I am kind of confused. Thanks ;) $\endgroup$ – ivana14 May 28 '17 at 18:34
  • $\begingroup$ If $k>1$, then there are eigenvalues in $\mathbb{C}\setminus\mathbb{R}$ and therefore the matrix will not be similar to a diagonal matrix then. $\endgroup$ – José Carlos Santos May 28 '17 at 18:37
  • $\begingroup$ And how would I find $P$? Do I take any eigenvalues I could get when $k<1$ and calculate eigenvectors which form $P$ or is there a better approach? I guess I $P$ could take many different values.. $\endgroup$ – ivana14 May 28 '17 at 19:06
  • $\begingroup$ Indeed, that is how I would do it. But I am baffled by your exepression "any eigenvalues I could get when $k<1$". I proved that there are exactly $3$ eigenvalues. $\endgroup$ – José Carlos Santos May 28 '17 at 19:28
  • $\begingroup$ sorry, what I meant was would the eigenvalues $A$ would have if $k$ was, for example, $0$, be different than the eigenvalues $A$ would have if $k$ was, for example $-1$. I guess they would be, so I asked if I could use any of these sets to find $P$. $\endgroup$ – ivana14 May 28 '17 at 19:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.