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I need to prove that if a matrix $A_{2 \times 2}$ is symmetric, i.e., $A^t = A$, then it is diagonalisable.

I know that a matrix $M_{n \times n}$ is diagonalisable, if and only if there is a basis of $K^n$ consisting of its eigenvectors. That is, I must have $n$ linear independent eigenvectors of $M$.

So, in order to prove that the matrix $A$ is diagonalisable, I must prove that it has $n$ linear independent eigenvector.

Matrix A has the following form: $$ A = \left[ {\begin{array}{cc} a & b \\ b & c \end{array} } \right] $$

Therefore, its characteristic equation is: $$ (\lambda -a)(\lambda -c) -b^2 =0 $$

Now I need to solve this equation in order to find its eigenvalues in terms of $a$, $b$ and $c$.

I'm not sure how to follow from here.

João Daniel
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    You can show directly that $0=(\lambda-a)(\lambda-c)-b^2=\lambda^2-(a+c)\lambda+(ac-b^2)$ have solutions. Calculate $\Delta=(a+c)^2-4(ac-b^2)$ and shows that $\Delta>0$. – Yuki May 17 '12 at 02:23
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    @Yuki: Dear Yuki, Unless $A$ is a scalar matrix, in which case $\Delta = 0$; but this case is particularly easy! Regards, – Matt E May 17 '12 at 02:27
  • Indeed, I missed! Thanks Matt E! – Yuki May 17 '12 at 02:40
  • @João Daniel: The result is true even for $n \times n$ symmetric matrices. –  May 17 '12 at 02:43
  • @Marvis, yes, but it's not quite as easy to prove. – Gerry Myerson May 17 '12 at 02:43
  • @GerryMyerson True. I just wanted to make the OP aware of this :). –  May 17 '12 at 02:48
  • What sufficient conditions do you know for the diagonalizability of a matrix? Do you know that an $n\times n$ matrix with $n$ distinct eigenvalues is diagonalizable? – Arturo Magidin May 17 '12 at 02:59
  • @ArturoMagidin Yes, because $n$ distinct eigenvalues will give me $n$ LI eigenvectors, right? – João Daniel May 17 '12 at 03:04
  • Yes. So then proving that this matrix is either diagonal, or has two distinct eigenvalues, will establish the result you want. – Arturo Magidin May 17 '12 at 03:09

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As has been noted, the conclusion holds even for $n\times n$ matrices, $n\geq 2$. In fact, a symmetric $n\times n$ matrix with real coefficients is not just diagonalizable, it is orthogonally diagonalizable: you can find an orthonormal basis of eigenvectors.

But for your purposes it suffices to show that a $2\times 2$ symmetric matrix is either diagonal (when $b=0$), or else has two distinct eigenvalues.

If $b=0$, there is nothing to do, so assume $b\neq 0$. The characteristic polynomial is, as you note, $$(t-a)(t-c)-b^2 = t^2 - (a+c)t + (ac-b^2).$$ The discriminant of this quadratic is $$(a+c)^2 - 4(ac-b^2) = a^2+2ac+c^2 - 4ac + 4b^2 = a^2-2ac+c^2 +4b^2 = (a-c)^2+(2b)^2.$$ Since this is a sum of squares, it is nonnegative. Since $b\neq 0$, it is actually positive. Since the discriminant of the quadratic is positive, the polynomial has two distinct real roots, so the matrix has two distinct eigenvalues, hence is diagonalizable.

In conclusion, a symmetric $2\times 2$ matrix is either diagonal (and hence diagonalizable), or has two distinct eigenvalues (and hence is diagonalizable).

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