# Diagonalizing matrix with fractions [closed]

I'm revising for an exam in linear algebra, and I've found myself stuck on this one specific exercise.

I'm supposed to decide a matrix $$P$$ and a diagonal matrix $$D$$ from my matrix $$H$$ (which I'll post below, so that $$P^{-1}HP = D$$.

Normally, I know how to solve tasks like these, but the fractions are what's giving me the issues when trying to get the eigenvalues out of the matrix. I'm just clueless on how to get them, so if any of you could help me out I'd greatly appreciate it.

$$H= \begin{pmatrix} \frac{3}{2} & - \frac{1}{2} &0\\ - \frac{1}{2} & \frac{3}{2}&0 \\ 0 &0&1\end{pmatrix}$$

## closed as off-topic by José Carlos Santos, Alexander Gruber♦Jan 17 at 23:17

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• If it's just the fractions that are making it hard for you, then just diagonalize $2H$. Now, all entries are integer. – José Carlos Santos Jan 17 at 15:23
• Whenever I try doing that I get different eigenvalues though. The common problem I have with these sort of tasks are to get the eigenvalues. The rest I got down, but I just keep screwing up on the eigenvalues. Do you possibly think you could show how to get the correct eigenvalues? – wznd Jan 17 at 15:31
• Why? Didn't you write that you know how to solve tasks like these and that you were just having problems with the fractions? – José Carlos Santos Jan 17 at 15:33
• Normally I get through these tasks fine, except for the eigenvalues which is a problem for me. I should probably have mentioned it. – wznd Jan 17 at 15:38

Note that: $$|H-\lambda I|= \begin{vmatrix} \frac{3}{2}-\lambda & -\frac{1}{2} & 0 \\ -\frac{1}{2} & \frac{3}{2}-\lambda & 0 \\ 0 & 0 & 1-\lambda \end{vmatrix}=(1-\lambda)\left(\left(\frac{3}{2}-\lambda\right)^2-\frac{1} {4}\right)= (1-\lambda)(2-3\lambda+\lambda^2)=(1-\lambda)(\lambda-1)(\lambda-2)=-(\lambda-1)^2(\lambda-2)$$ (I calculated the determinant using the Laplace expansion of the third column). The eigenvalues are hence $$1$$ and $$2$$. Can you take it from here?
• Using $(a-b)^2=a^2-2ab+b^2$ we get that $\left(\frac{3}{2}-\lambda\right)^2=(3/2)^2-2(3/2)(\lambda)+\lambda^2=9/4-3\lambda+\lambda^2$. Subtracting $\frac{1}{4}$, we have $\frac{9}{4}-\frac{1}{4}=\frac{8}{4}=2$ so $\left(\frac{3}{2}-\lambda\right)^2-\frac{1}{4}=2-3\lambda+\lambda^2$. Now, note that by finding the roots of that quadratic (using the quadratic formula, for example) we get $1, 2$ and so the polynomial is actually just $(\lambda-1)(\lambda-2)$. Multiplying by the $(1-\lambda)$ we get from the determinant we get the characteristic polynomial I said earlier. – Yuval Gat Jan 17 at 16:40
If you’re having trouble computing eigenvalues, try looking for eigenvectors instead. You should be able to tell at a glance that $$(0,0,1)^T$$ is an eigenvector of $$H$$ since that vector gets mapped to itself—the columns of $$H$$ are the images of the basis vectors. The corresponding eigenvalue is, of course, $$1$$.
Now focus on the upper-right $$2\times2$$ submatrix. Both rows have the same elements, so the row sums are equal. Summing the first two rows of $$H$$ is equivalent to right-multiplying by $$(1,1,0)^T$$, so there’s another eigenvector that’s linearly independent of the first one, with eigenvalue $$\frac32-\frac12=1$$.
You can always get the last eigenvalue “for free” since the sum of the eigenvalues, taking multiplicity into account, is equal to the trace. So, the remaining eigenvalue is $$4-1-1=2$$. To find a corresponding eigenvector, recall that a real symmetric matrix can be orthogonally diagonalized, so any eigenvector of $$2$$ has to be orthogonal to both of the eigenvectors of $$1$$ that you’ve already found. You’re working in $$\mathbb R^3$$, so such a vector can be found via a cross product: $$(1,1,0)^T\times(0,0,1)^T=(1,-1,0)^T$$ (which could also have been found by inspection).