Eigenvalue of all matrix with all duplicate rows I've been asked to determine an eigenvalue given the following matrix without any calculations:
$\begin{bmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 2 & 3 \end{bmatrix}$
My reasoning was that you can reduce this to the triangular matrix
$\begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
And that $1$, $0$ and $0$ are eigenvalues. Unfortunately the eigenvalues turned out to be $6$, $0$ and $0$ instead. What am I doing wrong?
 A: The rank of the matrix is the dimension of the space spaned by the columns of the matrix so in our case the rank of this matrix is obviously equal $1$ and by rank-nullity theorem the dimension of the kernel is $2$, hence $0$ is an eigenvalue with multiplicity $2$. Finaly the trace of this matrix gives the third eigenvalue $6$:
$$\text{the spectrum of the matrix}=\{0,0,6\}$$
A: To solve your problem, call your original matrix $A$. What is $A(1,1,1)^T$?
Note that you cannot use row operations here. Row operations are equaivalent to left multiplication by elementary matrices. So, the result of row operations is $PA$ for some invertible matrix $P$. Although the operation $A\mapsto PAP^{-1}$ preserve eigenvalues, the operation $A\mapsto PA$ does not.
A: It is probably impossible to do this without any calculation (i.e., with your neurons switched off). However if you can answer to following questions you can find the full set of eigenvalues rapidly:


*

*What is the rank of the matrix? (this gives an upper bound to the number of nonzero eigenvalues, since every nonzero eigenvalue adds at least one to the dimension of the sum of the nonzero eigenspaces (which is contained in the image), and therefore to the rank).

*What is the trace of the matrix? (this gives the sum of all complex eigenvalues, counted with multiplicity)
