Finding an invertible matrix given some eigenvalue-eigenvector pairs I have been given the eigenpairs (1, [1,0,1]), (2, [1,1,1]), and (3, [0,2,1]). I'm trying to find an invertible matrix using these pairs, but unsure of the process.
I know from the given pairs that for a matrix A,
$$A\vec{x} =\lambda\vec{x} , \space\space \vec{x} \neq{0} $$ where the x-coordinate for the given pairs can be plugged in for $\lambda$, and the y-coordinate for $\vec{x}$. I end up with 3 separate equations. Not sure how to proceed from here, as it seems to be working backwards.
 A: First note that the diagonal matrix $D$ with diagonal entries 1,2,3 have eigenvalues of $\lambda_1 := 1,\lambda_2 := 2,\lambda_3 := 3$, but eigenvectors $e_1 := [1,0,0], e_2 := [0,1,0],e_3 := [0,0,1]$.
Let $v_1 := [1,0,1]$, $v_2 := [1,1,1]$, and $v_3 := [0,2,1]$.
Now you want a matrix with the same eigenvalues $\lambda_1,\lambda_2,\lambda_3$, but with eigenvectors $v_1,v_2,v_3$.
To do this, first find a matrix $M$ that will map $e_1$ to $v_1$, $e_2$ to $v_2$, and $e_3$ to $v_3$. This is of course very easy. The matrix $M$ is simply the matrix with columns $v_1,v_2,v_3$.
You want to think of $M$ as a "transitioning matrix" or "change of basis matrix". Fundamentally, $D$ is doing all of the work (namely scaling vectors by 1, 2, and 3), but unfortunately it's scaling the wrong vectors. 
Now note that matrices are like functions. They take vectors as input, and spit out vectors as output. The product of matrices is just composition of functions.
To fix the fact that $D$ is scaling the wrong vectors, you want to imagine a process where you can send $v_i\mapsto \lambda_i v_i$, ideally using objects you already recognize.
One such process is - first send $v_i\mapsto e_i$, then apply $D$ to send $e_i\mapsto \lambda_i e_i$, then apply the inverse of the first part to send $\lambda_i e_i$ back to $\lambda_i v_i$.
As a matrix, this translates to the product $MDM^{-1}$.
In symbols, we have
$$MDM^{-1}v_i = MD(M^{-1}v_i) = MDe_i$$
where the last equality is because $M$ sends $e_i$ to $v_i$, so $M^{-1}$ must do the opposite.
$$\cdots = M(\lambda_i e_i) = \lambda_i (Me_i) = \lambda_i v_i$$
where the second equality comes from linearity of matrix multiplication, and the third comes from the construction of $M$.
Thus, the matrix you want is $MDM^{-1}$.
A: Let $A=[a_1\ a_2\ a_3]$ where $a_i$'s are the columns of $A$.
Let $v_1,v_2,v_3$ be the eigenvectors that you gave, respectively.
We know the following:
1) $Av_1=v_1$. This implies that $a_1+a_3=v_1$.
2) $Av_2=2v_2$. This implies that $a_1+a_2+a_3=2v_2$.
3) $Av_3=2a_2+a_3=3v_3$.
Note that the right hand side of 1), 2), 3) are known. The unknowns are the columns of $A$. 
Subtract 1) from 2). $a_2=[1\ 2\ 1]^T$. Substitute this to 3), $a_3=[-2\ 2\ 1]^T$. Finally, from 1), $a_1=[3\ -2\ 0]^T$. 
Thus, $A=\begin{bmatrix}3& 1& -2\\-2&2&2\\0&1&1\end{bmatrix}$.
