relationship between trace and eigenvalues

Question

Let A be a 10-by-10 matrix and rank(A)=1 then we want to show that the trace of A is an eigenvalue of A. I know there is one nonzero row in echelon form of A and 10-1=9 free variable. and also I know that the characteristic polynomial is (X^10)-tr Ax^9....+detA. Thanks.

Answer 1

Hint:

• Using the rank-nullity theorem you have $$\dim\ker A=9$$
• The trace of $A$ is given by $${\rm tr}A=\sum_{i=1}^{10}\lambda_{i}$$

Answer 2

The rank of the matrix is $1$. That means the image is spanned by a single vector in $\mathbb{R}^{10}$ denoted by $v$. So there is only non-zero eigenvalue $ \lambda$ and the corresponding eigenvector is $v$ and $ \lambda$ eigen space is spanned by just one vector (as it is of rank 1 ) which is $v$.So trace of matrix is $\lambda$. Thus by definition we have$ A(v)= \lambda v$ and hence it is an eigen vector.