Finding eigenvalues and eigenvectors of an unknown matrix 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?
 A: As you found above,
$$ A \left[\begin{array}{c} 3 \\ 3 \\ 3 \end{array}\right] 
= A \left[\begin{array}{c} 1 \\ 2 \\ 4 \end{array}\right] $$
Subtracting and factoring out $A$, we find
$$ A \left[\begin{array}{c} 2 \\ 1 \\ -1 \end{array}\right] = 0$$
But this means there's a nontrivial vector in the nullspace of $A$, implying another eigenvalue...  
A: The vectors you gave which A acts on are linearly independent, and so they form a basis for three dimensional space. If you know how a matrix acts on a basis, you know how it acts on any vector, (in particular the standard basis) and so you can actually recreate the matrix. Then the remaining questions are straightforward.
A: You don't need to recreate the matrix to answer the questions (although, yes, you can reconstruct it).  With a little more work, as Shaun Alt showed, you see that 0 is an eigenvalue corresponding to $\begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix}$ (you are given the other two eigenvalues and eigenvectors as you know).  So, now you know all 3 eigenvalues and all 3 eigenvectors.  Thus, your eigenvalues are 0, 2, 3.  They are distinct.  Any matrix with distinct eigenvalues is diagonalizable and the diagonal matrices that A is similar to have 0, 2, and 3 (the eigenvalues) on the diagonal.  If you have seen the Jordan canonical form, this is easy to see because all blocks can only be 1 by 1 since each eigenvalue is multiplicity 1.  And, by the way, we know each is multiplicity 1 because there can only be 3 total eigenvalues, counting multiplicity, for a 3 by 3 matrix.  One of the 6 diagonal matrices that $A$ is similar to would be
$$S^{-1} A S = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}, \qquad \text{for some invertible 3 by 3 matrix } S$$
is not invertible because it has a column of 0s.  Since A is similar to this, A also can not be invertible.
A: According to the information you have, 0, 2 and 3 are eigenvalues.
You have already given the eigenvectors for 2 and 3.  As pointed out 
by Shaun Ault, $\begin{bmatrix}2\\ 1\\ -1\end{bmatrix}$ is an eigenvector for $0$. 
It is not invertible because
0 is an eigenvalue, i.e. its kernel is not trivial.
Eigenvectors corresponding to distinct eigenvalues
 are linearly independent and there are 3 distinct eigenvalues. So, there are three linearly independent eigenvectors. So, you have a basis
consisting of eigenvectors. Hence, A is diagonalisable.
