Matrix from eigenvalues and eigenvectors

Find a matrix $A \in \mathbb{F}^{3\times 3}$ that has eigenvalues $1, 2, 3$ and eigenvectors the following:

$$\begin{pmatrix}
 1\\
1\\
3
\end{pmatrix}, \;\; \begin{pmatrix}
1\\
1\\
0
\end{pmatrix}, \;\; \begin{pmatrix}
1\\
0\\
0
\end{pmatrix}$$

Solution



First of all we observe that the matrix $A$ we seek is diagonizable over $\mathbb{F}$ since it has $3$ eigenvalues. Hence, it can be written in the form:

$$A= U \Delta U^{-1} $$

where $\Delta$ is a diagonizable matrix. However, the matrix of the eigenvalue is a diagonizable one and the matrix $U$, we know from theory, that is the matrix of the eigenvectors. Hence:

$$A= \begin{pmatrix}
1 &1  & 1\\
 1& 1 & 0\\
 1&  0& 0
\end{pmatrix}\begin{pmatrix}
1 &  0&0 \\
 0& 2 & 0\\
 0&0  &3
\end{pmatrix}\begin{pmatrix}
1 &1  &1 \\
 1& 1 & 0\\
 1&  0& 0
\end{pmatrix}^{-1}$$

Multyplying the three matrices (after evaluating the inverse matrix) we get that $A$ is the matrix:

$$\begin{pmatrix}
3 &-1  &-1 \\
 0& 2 & -1\\
 0& 0 &1
\end{pmatrix}$$