Show that a matrix is diagonalizable for all values of a

Question

I have to show that the matrix A is Diagonalizable for all values of a.

Given the matrix A.

\begin{pmatrix} a+3 & 4 \\ 5 & 5 \end{pmatrix}

I first found the characteristic polynomial for A:

\begin{equation} p=\lambda ^{2}-8\lambda -a\lambda+5a-5 \end{equation}

Then I tried to calculate the discriminant of the above polynomial p by using

\begin{equation} d=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \end{equation}

I have to show that the discriminant (d) is positive for all values a in the real domain. But I can't figure out how to calculate the discriminant when the equation contains the variable a.

Also I have to use the result from the above calculation (d) to prove that the matrix is diagonalizable for all values a in the real domain.

Any help is appreciated :-)

Answer 1

The discriminant of that polynomial is $a^2-4a+84$. Since this number is equal to $(a-2)^2+80$, it is always greater than $0$.