Finding possible values for determinant of a $2\times2$ matrix

Question

$M$ is a $2\times2$ matrix. $M$ is diagonalizable over $\mathbb{R}$. $M$ has the values $1$ and $2$ in the main diagonal.

What are the possible values of $\det M$?

(The solution is $\det M \leq \frac{9}{4}$.)

Can someone explain me how to solve this?

Answer 1

For a $2\times2$ matrix, its characteristic polynomial is $\chi_M = X^2 - \mathrm{trace}(M)X + \det (M)$. Suppose $M$ is diagonalisable, so it has two real eigenvalues. Here, they are distinct. Otherwise, $M$ would be a diagonalisable matrix with one eigenvalue, that is a multiple of $I_2$. But as it has two distinct diagonal coefficients, it is not. Thus, $\chi_M$ has two distinct real roots! It then has a positive discriminant. Compute it: \begin{align} \Delta = \mathrm{trace}(M)^2 - 4 \det(M) >0 \end{align}

That is $\det(M) < \dfrac{\mathrm{trace}(M)^2}{4}=\dfrac{9}{4}$

Answer 2

$M=\pmatrix{1&a\cr b&2}$ the caracteristic polynomial of $M$ is $(1-X)(2-X)-ab=X^2-3X-ab+2$ we must have $9-4(2-ab)>0$ and $det(M)=2-ab$

Answer 3

Let $M = \begin{pmatrix}1 & a \\ b & 2\end{pmatrix}$.

Note that if $M$ is diagonalisable over $\Bbb R$, then it must have all its eigenvalues real. (This is necessary.)
The eigenvalues of $M$ are precisely the roots of its characteristic equation. In this case, that is: $$x^2 - 3x + \underbrace{(2 - ab)}_{\det M} = 0.$$

For real roots, we need $D \ge 0$. ($D$ being the discriminant.) This gives us $$3^2 - 4\det M \ge 0$$ or $$\det M \le \dfrac{9}{4}.$$

Now, note that if $\det M < 9/4$, then we have distinct roots and $M$ is necessarily diagonalisable.
For $\det M = 9/4$, we would have that both the roots are $3/2$. However, we see that $M - (3/2)I$ is not the zero matrix and thus, if $\det M = 9/4$, then $M$ cannot be diagonalisable.

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To conclude, we see that:

$$M \text{ is diagonalisable } \iff \det M < \dfrac{9}{4}$$