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We have the matrix \begin{equation*}A=\begin{pmatrix}a & b \\ c & d\end{pmatrix}\end{equation*}

I want to find for which real numbers $a,b,c,d$ the matrix is diagonalizable in $\mathbb{R}^{2\times 2}$ and for which in $\mathbb{C}^{2\times 2}$.

The charachteristic polynomial is \begin{align*}\det (A-\lambda I)=\lambda^2-(a+d)\lambda +(ad-cb)\end{align*}

So, the eigenvalues are
\begin{align*}\lambda^2&-(a+d)\lambda +(ad-cb)=0 \\ \Rightarrow &\lambda_{1,2}=\frac{(a+d)\pm \sqrt{(a+d)^2-4(ad-cb)}}{2}=\frac{(a+d)\pm \sqrt{a^2+2ad+d^2-4ad+4cb}}{2}\\ & =\frac{(a+d)\pm \sqrt{a^2-2ad+d^2+4cb}}{2} =\frac{(a+d)\pm \sqrt{(a-d)^2+4cb}}{2}\end{align*}

We have the following cases:

As an element of $\mathbb{R}^{\times 2}$ we have the following:

  • Expression under the root $< 0$: no real eigenvalue, so the matrix is not diagonalizable.
  • Expression under the root $> 0$: two different eigenvalues, that means that the matrix is not diagonalizable, or not?
  • Expression under the root $= 0$, we have an eigenvalue of multiplicity $2$. What do we have in this case?

As an element of $\mathbb{C}^{\times 2}$ we have the following:

  • Expression under the root $\neq 0$: two different eigenvalues , that means that the matrix is not diagonalizable, right?
  • Expression under the root $= 0$, we have an eigenvalue of multiplicity $2$. What do we have in this case?
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  • $\begingroup$ Two different eigenvalues implies diagonalizable. If you have both the same value for the eigenvalue, it means that $$\chi_{A}\left(X\right)=\left(X-a\right)^2$$ hence if it was diagonalizable you would have $A=PDP^{-1}$ where $D=aI_2$ then $A=aI_2$ that's the only case where it will be diagonalizable. $\endgroup$ – Atmos Jan 21 '18 at 13:43
  • $\begingroup$ So, can we not get a further condition that has to be satisfied so that we know when the matrix is diagonizable in the case of two same eigenvalues? @Atmos $\endgroup$ – Mary Star Jan 21 '18 at 13:55
  • $\begingroup$ At condition (2) it says that it has to have at least n real roots, if it has n complex eigenvalues, what does then hold? @Moo $\endgroup$ – Mary Star Jan 21 '18 at 13:58
  • $\begingroup$ I haven't really understood why in general, if a matrix has complex eigenvalues, it is not diagonalizable. Coudyou explain it further to me? @Moo $\endgroup$ – Mary Star Jan 21 '18 at 15:35
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A matrix is diagonalizable iff its minimum polynomial is a product of simple linear factors.

Also, the distinct roots of the minimum polynomial are the same as the distinct eigenvalues of the matrix.

So for a 2x2 matrix, the only case when it is NOT diagonalizable is when the matrix has 2 repeated eigenvalues.

The matrix has repeated eigenvalues when:

$$det(A-\lambda I) = \lambda^2-\lambda \cdot (a+d) + (ad-bc)$$ has repeated roots....ie: $b^2-4ac=0$, which should simplify to: $$(a-d)^2+4bc=0$$

So all values of a, b, c and d that satisfy $(a-d)^2+4bc\ne0$ make A diagonalizable.

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    $\begingroup$ If the matrix has repeated eigenvalues then its characteristic polynomial is $(x-\alpha )^2$ where $\alpha$ is the eigenvalue. The matrix can still be diagonalizable in this case, but this requires further analysis. $\endgroup$ – Dave Jan 21 '18 at 14:26
  • $\begingroup$ So my solution doesn't hold only when A is a diagonal matrix and with both its diagonal elements being the same. Am I correct? $\endgroup$ – Dylan Zammit Jan 21 '18 at 14:34
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    $\begingroup$ Yes, because this is the only case when the matrix with repeated eigenvalue has two dimensional eigenspace. Perhaps a minor special case, but nonetheless a necessary addition. $\endgroup$ – Dave Jan 21 '18 at 14:40
  • $\begingroup$ I got stuck right now. When a 2x2 matrix has two repeated eigenvalues, the matrix is diagonalizable only when the the eigenvectors are independent? @Dave $\endgroup$ – Mary Star Jan 21 '18 at 16:00
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    $\begingroup$ In the repeated eigenvalue case: the matrix is diagonalizable if and only if $c=b=0$. $\endgroup$ – Dave Jan 21 '18 at 16:40

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