Let A be a symmetric $(2 × 2)$ -matrix. Without using the Spetral Theorem, show that there exist eigenvectors of A, which form a basis of $R^2$
So I made a matrix A = $\begin{bmatrix}a\, b \\b\, d\end{bmatrix}$ The question was a longer and had different parts and I had found the eigenvalues etc but for the sake of simplicity lets say that it has the eigenvalues $\lambda_1$ and $\lambda_2$. I thought the basis was just gonna contain the linearly independent eigenvectors I will find with the help of the eigenvalues. But apparently the answer is $\begin{bmatrix}\lambda_1\,0\\0\,\lambda_2 \end{bmatrix}$. I get they did something with orthogonal diagonalization as in $D = P^TAP $ but doesn't the question say the basis should contain the eigenvectors? Do they always mean give the diagonal if asked like this? I would be grateful if someone can just clear everything up for me.
The characteristic equation of your matrix is $$\lambda^2-(a+d)|\lambda + (ad--b^2)=0$$ The discriminant of this equation is $$D=(a+d)^2-4(ad-b^2)=(a-d)^2+4b^2 \ge 0$$ Case (i) $D>0$. Then there are two distinct real eigenvalues and hence 2 linearly independent real eigenvectors $\mathbf v$ and $\mathbf w$. Then $\lbrace \mathbf {v,w}\rbrace$ is a basis of $\mathbb R^2$. Case (ii) D=0. Then $a=d$ and $b=0.$ Then every non-zero vector is an eigenvector. In particular $\lbrace [1,0],[0,1]\rbrace$ is a basis of $\mathbb R^2.$