Matrix $A$ has two distinct real eigenvalues iff $k$ is greater than what?

The matrix $A = \begin{bmatrix}3&k\\8&8\end{bmatrix}$ has two distinct real eigenvalues iff $k > ?$

So I found the determinant by doing:

$(3 - \lambda)(8 - \lambda) - 8k = \lambda^2 - 11\lambda + 24 - 8k \implies \lambda = 8, \lambda = 3$ The thing is, I'm not really sure what they are asking me because I have found what the eigenvalues are: $\lambda_1 = 8, \lambda_2 = 3$.

I'm assuming I need to solve for $k$ somehow but it doesn't seem very straightforward to me, what am I missing here?

• How did you get $\lambda=8$ and $\lambda=3$? The characteristic polynomial is in terms of $k$, so if you solve for $\lambda$, the results should also be in terms of $k$. – solstafir Jul 14 '16 at 22:23
• The eigenvalues you found are correct when $k = 0$, but not for different values of $k$. – Omnomnomnom Jul 14 '16 at 22:25

Note that the characteristic equation of this matrix is given by $$\lambda^2 - 11\lambda + (24 - 8k) = 0$$ The question is to find which values of $k$ produce two distinct real roots $\lambda$ to this equation. It is helpful to consider the discriminant of this quadratic equation, namely $$b^2 - 4ac = 121 - 4(24 - 8k)$$ recall that a quadratic equation will have distinct real roots iff its discriminant is positive.

• Oh, I forgot about that. So $121 - 96 +32k = 25 + 32k = 0 \implies 25 = -32k \implies k > -25/32$ Thanks so much @Om ! – Yusha Jul 14 '16 at 22:28
• I will reward you points because to me, this is the most straight foward explanation considering the discriminant of the quadratic equation. This is very useful – Yusha Jul 14 '16 at 22:29
• @Yusha Appreciate it, and glad you found this useful. – Omnomnomnom Jul 14 '16 at 22:30

$\lambda^2 - 11\lambda + 24 - 8k = 0\\ \lambda = \dfrac {11 \pm \sqrt{25 +32k}}{2}$

$A$ has 2 distinct real egeinvalues if $25 +32k>0$

The determinant is $$(3 - \lambda)(8 - \lambda) - 8k = \lambda^2 - 11\lambda + 24 - 8k.$$

In order to find the eigenvalues, you have to solve the equation $$\lambda^2 - 11\lambda + 24 - 8k = 0.$$

The solutions are:

$$\lambda = \frac{11 \pm \sqrt{11^2 - 4(24-8k)}}{2} = \frac{11 \pm \sqrt{25+32k}}{2}$$

which of course depend on $k$.

The two solutions are different and real if the part under square root is positive. That is:

$$25 +32 k > 0 \Rightarrow k > -\frac{25}{32}.$$

You need to solve the equation with $\lambda$ the variable and $k$ a parameter.

The solutions are $$-\frac{11}{2} \pm\sqrt{\frac{11^2}{4} - (24 - 8k) }$$ These are the eigenvalues of the matrix in dependence of $k$.

These two are distinct and real if and only if $$\frac{11^2}{4} - (24 - 8k)> 0.$$ Solve this to get your $k$.