Question about diagonalization

Question

Problem: https://i.stack.imgur.com/S7lqr.png

Let $T: \Bbb R^2 \to \Bbb R^2$ be the linear transformation with standard matrix $A$ below. Find a basis $\mathcal B$ such that the matrix of $T$ relative to $\mathcal B$ is diagonal.

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$$ A = \pmatrix{2&3\\3&2} $$

\begin{bmatrix} 1+3i &0 \\ 0 & 1-3i \end{bmatrix} Can this be answer to the question? If not, how would I approach this question.

Answer 1

Sketch of solution: Such a basis $\mathcal B$ must consist of two linearly independent eigenvectors. To find these, we first find the eigenvalues.

To find the eigenvalues, find the characteristic polynomial of the matrix $A$, which is the determinant of the matrix $$ A - \lambda I = \pmatrix{2-\lambda & 3\\ 3 & 2- \lambda}. $$ You should find that this determinant is equal to $\lambda^2 - 4\lambda - 5$. The zeros of this polynomial (both real!) are the eigenvalues. From there, find the eigenvectors of $A$ in the usual fashion. Your answer should be a list containing both of these vectors.

Answer 2

Hint : $$\pmatrix{2&3\\3&2}\pmatrix{1\\-1} = \pmatrix{-1\\1}$$

and $$\pmatrix{2&3\\3&2}\pmatrix{1\\1} = \pmatrix{5\\5}$$