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We're given the folowing matrix. $$A = \begin{bmatrix}0 & 0&2h\\1 & 0&-2-3h\\0&1&3+h \end{bmatrix}$$ and we're asked to find for which $h \in \mathbb{R}$ this matrix is diagonalisable.

Here's my approach. I've been able to find the characteristic polynomial of the matrix, which is: $$-\lambda^3+(3+h)\lambda^2 - (2+3h)\lambda + 2h $$

Since I didn't know what to do from here, I tried to see if the matrix is diagonalisable with $h=1$. This gives me the following polynomial: $$-\lambda^3+4\lambda^2-5\lambda+2 \\ = \ - (\lambda-2)(\lambda-1)^2$$ Therefore, to see if the matrix is diagonalisable for $h=1$, I need to see if $\mathrm{dim}(\mathrm{Ker}(A- 1 \cdot I))$ (where $I$ is the indentity matrix) is equal to $2$.$$ A-1\cdot I= \begin{bmatrix}-1&0&2\\1 & -1&-5\\0&1&3 \end{bmatrix}$$ which in echelon form gives me: $$\begin{bmatrix}1 & 0&-2\\0&1&3\\0&0&0 \end{bmatrix}$$

We can see that $\mathrm{dim}(\mathrm{Ker}(A- 1 \cdot I)) \neq 2$. Thus I would conclude that for $h=1$, the matrix A is not diagonalisable.

But the correction sheet says that A is diagonalisable for all $h\in \mathbb{R}$.

So first of all, I don't understand why it doesn't work $h=1$ (I guess I made a mistake somewhere, but I can't find it). Then, I don't know how I'm supposed to find out that the matrix is diagonalisable for all $h \in \mathbb{R}$.

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  • $\begingroup$ For $h\neq1$ and $h\neq2$ $A$ is diagonalizable since you have three distinct eigenvalues. For $h=1$ or $h=2$ you have to check if you can find a sufficient number of eigenvectors. From your calculation it seems that for $h=1$ you can only find one eigenvector thus the matrix is not diagonalizable. In this case you can only find the Jordan form. $\endgroup$ Commented Jan 3, 2018 at 18:53
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    $\begingroup$ Aside from @gimusi's comment, no-one seems to have addressed what I take to be your question: have you made a mistake for $h = 1$? No, you haven't (unless the mistake is in copying out the matrix). This matrix is not diagonalisable for $h = 1$ (or $h = 2$). If you're taking this course with an instructor, it might be worth showing him or her your work and asking if he or she agrees that the textbook solution has a mistake; that isn't at all uncommon. $\endgroup$ Commented Jan 3, 2018 at 23:39
  • $\begingroup$ @LSpice Yes, that's also what I think. The teacher must have made a mistake. Thank you though. $\endgroup$ Commented Jan 4, 2018 at 14:05
  • $\begingroup$ Oops, sorry; I missed that you said that it was a correction sheet rather than a textbook solution set. Well, as much as I hate to admit that teachers (as opposed to textbook authors) are fallible, indeed we occasionally do make mistakes. :-) $\endgroup$ Commented Jan 4, 2018 at 17:03

2 Answers 2

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Note that:

$$-\lambda^3+(3+h)\lambda^2 - (2+3h)\lambda + 2h=-\lambda^3+3\lambda^2 -2\lambda +h\lambda^2-3h\lambda + 2h=-\lambda(\lambda^2-3\lambda +2)+h(\lambda^2-3\lambda +2)=(h-\lambda)(\lambda^2-3\lambda +2)=(h-\lambda)(\lambda-1)(\lambda-2)$$

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  • $\begingroup$ Although I still do not understand why it doesn't work for $h=1$ $\endgroup$ Commented Jan 3, 2018 at 18:33
  • $\begingroup$ In this case you have two equals eigenvalues and you need to verify that the dimensions of its eigenspaces is equal to 2. $\endgroup$ Commented Jan 3, 2018 at 18:45
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The characteristic polynomial of $A$ is$$-x^3+(h+3)x^2-(3h+2)x+2h,$$whose discriminant is$$h^4-6h^3+13h^2-12h+4=(h^2-3h+2)^2=(h-1)^2(h-2)^2.$$Therefore, if $h\notin\{1,2\}$, then $A$ has $3$ distinct eigenvalues and then it must be diagonalizable. All you have to do is to see what happens whan $h\in\{1,2\}$.

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