This is given about a matrix A:
$A \begin{bmatrix} 1\\2\\4 \end{bmatrix} =9 \begin{bmatrix} 1\\1\\1 \end{bmatrix} , A \begin{bmatrix} 1\\−3\\9 \end{bmatrix} =2 \begin{bmatrix} 1\\−3\\9 \end{bmatrix} , A \begin{bmatrix} 1\\1\\1 \end{bmatrix} =3 \begin{bmatrix} 1\\1\\1 \end{bmatrix}$
Find:
- Eigenvalues & Eigenvectors of A.
- Is A invertible?
- Is A diagonalizable?
I'm pretty sure that only $2$ & $3$ are eigenvalues and only $\begin{bmatrix} 1\\−3\\9 \end{bmatrix}$ & $\begin{bmatrix} 1\\1\\1 \end{bmatrix}$ are eigenvectors (at least from the information given). Since the 2nd and 3rd identities given are in the form $Ax = {\lambda}x$.
A will be diagonalizable if and only if the eigenvectors are linearly independant, which they are, so Yes, A is diagonalizable
I have no idea how to find if the matrix is invertible.
Is my thinking correct? And how do I know if A is invertible?
Edit:
As I look at this further I see: $3A\begin{bmatrix} 1\\1\\1 \end{bmatrix} = 3*3\begin{bmatrix} 1\\1\\1 \end{bmatrix} = 9\begin{bmatrix} 1\\1\\1 \end{bmatrix} =A\begin{bmatrix} 1\\2\\4 \end{bmatrix}$
so
$A\begin{bmatrix} 3\\3\\3 \end{bmatrix} = A\begin{bmatrix} 1\\2\\4 \end{bmatrix}$
Is this significant?