I have to find a matrix$A_{4 \times 4}$ with eigenvalues $2$ and $4$ and that both $E(2, A)$ and $E(4, A)$ are $1$ dimensional I know that matrix with eigenvalues $2$ and $4$ can be made from just putting $2$ and $4$ on the diagonal. I have tried many different matrices that have the eigenvalue of $2$ and $4$ but I can not find one that satisfies the rule that that both $E(2, A)$ and $E(4, A)$ are $1$ dimensional. I have come up with some that satisfy either $E2$ or $E4$ but I can not come up with one that satisfies both of them.
 A: If a matrix has all eigenvalues distinct, it is diagonalisable and all eigenspaces are one-dimensional. Recall too that the eigenvalues of a diagonal matrix are simply the diagonal entries.
Thus a matrix with eigenvalues $2,4$, both with one-dimensional eigenspaces, is the diagonal matrix with non-zero entries $1,2,3,4$.
A: For another class of examples, consider the matrix $D=\begin{bmatrix}2&0\\0&4\end{bmatrix}$ and consider any invertible matrix $P$.
The matrix $PDP^{-1}$ then has eigenvalues $2$ and $4$ as well, each of which occurring with dimension $1$.
A: Check $A=\begin{bmatrix}
2 & 0 & 0 & 0 \\
1 & 2 & 0 & 0 \\
0 & 0 & 4 & 0 \\
0 & 0 & 1 & 4
\end{bmatrix}$. It satisfies the conditions which are you wanted.
P.S.= If you wanted to know more about such constructions, study Jordan canonical forms. We already know that diagonalizable matrices can be written in the form of $MDM^{-1}$. Jordan canonical form say that every square matrix can be written in the form of $MHM^{-1}$ where $H$ is lower triangular matrix whose diagonal entries are eigenvalues and you may see 1s under the some diagonal entries.
