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?
Best Answer
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...