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.
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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.
Chirantan Chowdhury
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Hint:
- Using the rank-nullity theorem you have $$\dim\ker A=9$$
- The trace of $A$ is given by $${\rm tr}A=\sum_{i=1}^{10}\lambda_{i}$$
eranreches
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Can we say that trace of a n*n matrix always is an eigenvalue of that matrix? – zeinab Dec 13 '17 at 17:00
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@zeinab No. Take for example the $n\times n$ identity matrix for $n>1$. All its eigenvalues are $\lambda=1$, but ${\rm tr}I=n$. – eranreches Dec 13 '17 at 17:02
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Ok. but in this question how you show that the trace(A) is an eigenvalue of A? – zeinab Dec 13 '17 at 17:08
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@zeinab Try to use the hints I wrote. For example, what can you deduce about the eigenvalues from the first hint? – eranreches Dec 13 '17 at 17:11
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@zeinab I prefer not to. Though, I can give you another hint $$\ker A\equiv{v\in\mathbb{R}^{10}:Av=0=0\cdot v}$$ – eranreches Dec 13 '17 at 17:22
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Do you mean that rank A=1 and dim A=10 and dim(ker A)=9 so A has only one eigenvalue so trace A=eigenvalue A? – zeinab Dec 13 '17 at 17:25
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You should write $\dim V$ instead of $\dim A$. Also, with a little correction your answer is right! $A$ doesn't have only one eigenvector. It has only one non-zero eigenvector. – eranreches Dec 13 '17 at 17:28
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