Values of k for which there exist two distinct real eigenvalues

Question

$\begin{pmatrix}2&k\\ \:1&5\end{pmatrix}$

So the question is the title itself.

So I found the characteristic polynomial to be:

$x^2-7x+10-k$

After factoring it became:

$\left(x-2\right)\left(x-5\right)-k$

So if we ignore the k the eigenvalues would be 2 and 5. But I am not sure how I would treat the problem with k present.

Any help?

Answer 1

Write $$ x^2 - 7x - k + 10 = \left( x-\frac{7}{2}\right)^2 - \frac{49}{4}-k+10 $$ Now, we see that $$ x^2 - 7x - k + 10 = 0 \iff x = \pm\sqrt{10 -k-\frac{49}{4}} +\frac{7}{2} $$ So, there exists two distinct real eigenvalues if and only if $$ 10 -k-\frac{49}{4}>0 \iff k<10-\frac{49}{4} = -\frac{9}{4} $$

Answer 2

There is a theorem implying that if are eigenvalues of a matrix are distinct then the eigenvectors are all linearly independent. Here the characteristic equation is $x^2-7x+10-k=0$. If the discriminant is zero therefore $$49=4(10-k)$$ or $$k=-\dfrac{9}{4}$$therefore if $k\ne-\dfrac{9}{4}$ we attain what we want also the discriminant should be non negative therefore the answer is$$k\lt-\dfrac{9}{4}$$