Linear Algebra Question
The trace of a matrix is the sum of the entries along the main diagonal. Prove that if A is a 2 x 2 matrix with det(A) = 1, and trace(A) 2, then A has two real and distinct eigenvalues. (answer please, will be offline until tomorrow)

Question

Linear Algebra Question
The trace of a matrix is the sum of the entries along the main diagonal. Prove that if A is a 2 x  2 matrix with det(A) = 1, and trace(A) > 2, then A has two real and distinct eigenvalues. (answer please, will be offline until tomorrow)

kinggeorge · Answer

Well, if you write out your matrix as$$\left[\begin{matrix} a&b\c&d \end{matrix}\right]$$Then we know that the eigenvalues are given by $$\left| \begin{matrix} (a-\lambda) &b\c&(d-\lambda)  \end{matrix} \right|=0$$so $$ad-\lambda (a+d)+\lambda^2 -bc = 0$$Now, notice that $ad-bc$ is the determinant of A, and $a+d$ is the trace of A. So we can rewrite this as $\lambda^2-\lambda(a+d)+1=0$ Where $a +d>2$. 

If we let $a+d=2$, then we can factor the equation as $(\lambda -1)^2$ which only has 1 distinct root. Moreover, if we graph this, we see that the vertex touches the x-axis at $x=1$. Now, note that if $a+d>2$, then we are subtracting a larger number, so the graph of the function $\lambda^2-\lambda(a+d)+1$ is shifted down. Since it is an upward opening parabola, that means that it must have two distinct roots if shifted down.  

QED