I've came across this issue, and couldn't manage to explain it.

If given a square matrix, such as $3\times 3$, and it's rank equals to $1$, when searching its eigenvalues, the Trace of the matrix is one of them, as well as $0$.

I would like to get the explanation of this issue and to ask if another eigenvalues could be possible (except the trace and $0$)?

EDIT: i mean that if a matrix is in its diagonal form it's pretty obvious, but if a matrix is not in its diagonal form or even it's unknown if the matrix is diagonalizable.

  • $\begingroup$ The trace of a 3-by-3 matrix is not necessarily one of its eigenvalues. Consider the the 3-by-3 identity matrix. Its only eigenvalue is 1, but its trace is 3. $\endgroup$ – Fly by Night Jan 20 '17 at 20:27
  • $\begingroup$ but it's rank is not 1.. $\endgroup$ – liranK Jan 20 '17 at 20:28
  • $\begingroup$ Duplicate of (math.stackexchange.com/q/55165) $\endgroup$ – Jean Marie Jan 20 '17 at 21:01

Say $u$ is a linear map from $E$ to itself, where $E$ is some finite dimensional space.

Let $n = \dim(E)$.

Recall that, by definition, the rank of $u$ is $r=\dim(u(E))$.

Suppose that $r=1$. Then $\dim(\ker(u))=n-1$.

Since the multiplicity of an eigenvalue as at least the dimension of the corresponding eigenspace, we get that $0$ is an eigenvalue with multiplicity at least $n-1$.

And since the sum of all eigenvalues (counted with multiplicity) is $\mathrm{tr}\,(u)$, the last eigenvalue is $\mathrm{tr}\,(u)$.

You can now, given some square matrix $A$, apply this to the map canonically attached to $A$.


The matrix $A$ define the linear map $f$ on $E$ Take a base of $E$ such of the form $(e_1,..,e_n,e_{n+1})$ where $(e_1,...,e_n)$ is the base of $ker f$ and write the matrix of $f$ in this base. The eigenvalue of this matrix is clearly the $n+1-n+1$ coefficient which is also the trace.


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