For given a real matrix $A$ of size $n$ by $n$, we assume that there exist $\lambda$ with multiplicity $2$ and its corresopnding eigenvector is denoted by $x$. In this situation, prove that

if the dimension of the null space of $A-\lambda I$ is $1$, then $x$ belongs to the column space of $A-\lambda I$.

Actually, I tried many times, but I failed to prove this. Can someone let me know the proof? Some hints is also very thank you?


The eigenspace corresponding to the eigenvalue of multiplicity 2 cannot be the nullspace which has dimension 1 so $\lambda$ is not= 0 . Thus Ax = $\lambda$x and so x=(1/$\lambda$)Ax which is in the column space .

  • $\begingroup$ I am very sorry that I do not understand your answer. First of all, what is $\lambda |=0$? $\endgroup$ – Danny_Kim Dec 4 '17 at 13:23
  • $\begingroup$ If $A=\left[\begin{array}{cc}\lambda & c\\ 0 &\lambda\end{array}\right]$, since $A$ is an upper triangular matrix, the eigenvalues are easily found as $\lambda$ with multiplicity $2$. From $A-2I=\left[\begin{array}{CC}0&c\\0&0\end{array}\right]$, we know * The rank of $A-2I$ is $1$. * The nullity of $A-2I$ is $2-1=1$. * The nullspace of $A-2I$, denoted by $N(A-2I)$, is the span of $\begin{bmatrix}1\\0\end{bmatrix}$. I think eigenspace is $\left\{\begin{bmatrix}c\\0\end{bmatrix}:c\in\mathbb{R}\right\}$, which is the nullspace, isn't it? $\endgroup$ – Danny_Kim Dec 4 '17 at 13:24
  • $\begingroup$ Your formula's for A and A-2 I are wrong . Multiplicity and eigenvalue are not the same . $\endgroup$ – StuartMN Dec 5 '17 at 10:01

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