# Show that a matrix is diagonalizable for all values of a

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:

$$$$p=\lambda ^{2}-8\lambda -a\lambda+5a-5$$$$

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

$$$$d=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$$$$

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 :-)

• “Diagonalizable,” not “diagonal.” Your matrtix can never be diagonal. – amd Mar 29 at 23:35
• Thank you, I will edit my typo – Donatello V. Mar 30 at 0:01

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 $$0$$.
• Write the quadratic as $$\lambda^2 + (-8-a)\lambda + (5a-5).$$ The discriminant is thus $$(-8-a)^2 -4\times 1\times (5a-5)=\ldots.$$(Remember, the discriminant of $A\lambda^2 + B\lambda + C$ is $B^2 -4AC$.) – Minus One-Twelfth Mar 30 at 0:19
• Since the discriminant is greater than $0$, then the characteristic polynomial has two distinct real roots. That's enough: every $n\times n$ real matrix whose characteristic polynomial has $n$ distinct real roots is diagonaliable over $\mathbb R$. – José Carlos Santos Mar 30 at 10:25