About diagonalization "Let A = $\begin{bmatrix}1 & 1 & 4\\0 & 3 & -4\\0&0&-1\end{bmatrix}$. Is the matrix A diagonalizable?  If so find a matrix P that diagonalizes A. Can you write A as a
linear combination of rank 1 matrices formed from its eigenvectors? Determine the
eigendecomposition $A = PΛP^{-1}$."
My solution attempt:
det(λI - A) = $\begin{vmatrix}λ - 1 & -1 & -4\\0 & λ - 3 &4\\0&0&λ + 1\end{vmatrix}$ = (λ -1) (λ -3) (λ + 1) = 0
$λ_{1}$ = 1, $λ_{2}$ = 3, $λ_{3}$ = -1
($λ_{1}I - A)x_{1}$ = 0
$\begin{bmatrix}0 & -1 & -4\\0 & -2 &4\\0&0&2\end{bmatrix}$ $\begin{bmatrix}x_{11}\\x_{12}\\x_{13}\end{bmatrix}$ = $\begin{bmatrix}0\\0 \\0\end{bmatrix}$ 
$\begin{bmatrix}x_{11}\\x_{12}\\x_{13}\end{bmatrix}$ = $x_{11} \begin{bmatrix}1\\0 \\0\end{bmatrix}$
It follows that $\begin{bmatrix}1\\0 \\0\end{bmatrix}$ is a basis for the eigenspace of A corresponding to $λ_{1}$ = 1.
In a similar way, I found other two basis vectors as $\begin{bmatrix}1/2\\1 \\0\end{bmatrix}$ and $\begin{bmatrix}-2/5\\1 \\1\end{bmatrix}$, and I got $P = \begin{bmatrix}1 & 1/2 & -2/5\\0 & 1 & 1\\0&0&1\end{bmatrix}$. Then my $P^{-1} = \begin{bmatrix}1 & -1/2 & 9/10\\0 & 1 & -1\\0&0&1\end{bmatrix}$. And I said A's left-multiplication by $P^{-1}$ and right-multiplication by $P$ $(P^{-1}AP)$ gives the diagonal matrix Λ that is similar to A. I found Λ as $\begin{bmatrix}1 & 0 & 0\\0 & 3 & 0\\0&0&21/5\end{bmatrix}$. Then I said that if we multiply both sides of the equation $P^{-1}AP = Λ$ by $P$ from the left and by $P^{-1}$ from the right we get $PP^{-1}APP^{-1} = PΛP^{-1} = A$ and that this is the eigendecomposition of A. Here are my questions:


*

*Do you think I understood and did correctly what the asker wanted me to do by saying "Determine the eigendecomposition $A = PΛP^{-1}$."?

*What the hell does the sentence "Can you write A as a linear combination of rank 1 matrices formed from its eigenvectors?" mean?

1) The diagonal matrix always consists of the eigenvalues. Then the matrix $P$ has column vectors as the eigenvectors. So if $\Lambda_{11} = \lambda_{1}$, then the first column vector in $P$ is the eigenvector corresponding to $\lambda_{1}$. 
2) The eigenvectors form a basis for $span(A_{1}, A_{2}, ..., A_{n})$. If $A$ is of full rank, the eigenvectors form a basis of the entire vector space. So can you write each column of $A$ as a linear combination of the eigenvectors?
