My professor gave us a question and I would like some help in it. The problem goes as follows: Let $T:\mathbb{R}^3\to\mathbb{R}^3$ be the linear transformation represented by: $T(x_1,x_2,x_3)=(x_1-x_2-x_3,x_1+3x_2+x_3,-3x_1+x_2-x_3)$
a) Find the standard matrix $A$ for T.
b) Show that $A$ is diagonalizable.
c) Compute $A^3$.
d) Find a basis $B$ for $\mathbb{R}^3$ such that the matrix $A$ relative to $B$ is diagonal.
I'm not sure what part d) is asking of me. Could anyone explain it in simpler terms?
Given a diagonalizable linear transformation $T:\mathbb{R}^n\to\mathbb{R}^n$, you can always find a basis for $\mathbb{R}^n$ of eigenvectors of $T$. Your job here is to find eigenvectors of $T$ and use them as a basis for $\mathbb{R}^3$.
The point of part d) is that $A = BDB^{-1}$, where $D$ is diagonal and $B$ contains eigenvectors corresponding to the eigenvalues on the diagonal of $D$.