How to show $T$ is diagonalizable? Let $T\colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be linear with distinct eigenvalues $\lambda_1, \lambda_2, \lambda_3$.  Show that $T$ is diagonalizable.  
It seems as if this is a very simple question but I'm a bit confused about how to start!
 A: You need to prove that $\mathbb{R}^3$ has a basis of eigenvectors.  The key is that if you have vectors $\vec{v}_1, \vec{v}_2, \vec{v}_3$ that are eigenvectors for distinct eigenvalues, then the vectors must be linearly independent.
To prove this latter fact, assume that 
$$c_1 \vec{v}_1 + c_2 \vec{v}_2 + c_3 \vec{v}_3 = \vec{0}.$$
Apply $(T - \lambda_1 I)(T -\lambda_2 I) = (T -\lambda_2 I)(T - \lambda_1 I)$ to both sides to find that $c_3 = 0$.  (Here $I$ is the identity operator.)  Note that you must use $\lambda_3 \neq \lambda_1, \lambda_2$ here.  Similar operators can be applied to show that $c_1$ and $c_2$ must equal $0$ as well.
In this case, the argument gives you three linearly independent eigenvectors in $\mathbb{R}^3$, which are therefore a basis.
A: You have $3$ distinct  eigenvalues  (as $\lambda_1 , \lambda_2 , \lambda_3$) and eigenvector corresponding to each of them must have $\dim$ $1$ (if you have doubt, see Hint below) then we have $3$ eigenvector spaces s that they are subspace of $\mathbb R^3$ and sum of them can construct $\mathbb R^3$.
So if we choice basis from these sub-spaces (choice one base from each sub spaces as $v_1$ from eigenvector space correspond $\lambda_1$ eigen value and  $v_2$ from eigen vector space correspond $\lambda_2$ eigen value and similar about $v_3$ ) since each of thses basis ARE eigen vector then our matrix (represented T) will be diagonalizable because $$T(v_1)=\lambda_1 v_1,T(v_2)=\lambda_2 v_2,T(v_3)=\lambda_3 v_3,$$ And so:
our matrix with basis$\{v_1,v_2,v_3\}$ will be:
$$\begin{pmatrix}\lambda_1&0&0\\0& \lambda_2&0\\0&0&\lambda_3\end{pmatrix}$$
Hint:
any eigenvalue as $\lambda$ have at least one $\dim$ because :
$$\det(A-\lambda I)=0$$ 
$(A-\lambda I)x=0$ has at least one non trivial answer. And at this problem it can not has more than one $\dim$ because sum of $\dim$ of $3$ sub-spaces must be $3$.
A: Pick eigenvectors corresponding to each of the three distinct eigenvalues. Then they will be linearly independent since the eigenvalues are distinct (by a theorem). So you can make a basis with these eigenvectors. So the matrix of $T$ with respect to this basis is diagonal.
