Consider a symmetric matrix $A$ and its spectral decomposition is $UDU^T$. Denote by $u_1$ the first column of $U$ (meaning, the eigenvector of the greatest eigenvalue). Let $v$ be any vector.
I want to show that if $\sum _{i=1} ^n (u_1)_iv_i= u_1^Tv\ne0$ then $v\not \in \text {Image}(A-\lambda_{\min }I)$ where $\lambda_{\min}$ is the smallest eigenvalue of $A$.
I showed that $A+\lambda I$ is invertible for all $\lambda >-\lambda_{\min}$. Then I showed that $\vert\vert(A+\lambda I)^{-1}v\vert\vert>0$ since $u_1 ^Tv\ne0$. This means that $v\in \text {Image}(A+\lambda I)$.
Now I tried to use this fact to show that if $v\in \text {Image}(A-\lambda_{\min }I)$ then $A-\lambda_{\min }I$ is invertible or $u_1 ^T v=0$ (contradiction). But I could not show that and I don't even know if it is even a correct way to prove this. Any ideas?
This isn't true. Take $A=\begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}$ with eigenvectors $\lambda_1=1, u_1=(1,0)$ and $\lambda_2=0$, $u_2=(0,1)$. Then take $v=u_1$, so $(A-I\lambda_{min})u_1=Au_1=u_1=v$.
Since $UU^T=1$, $A-\lambda_{min}I =UD'U^T$, where D' is the eigenvalue matrix with $\lambda_{min}$ subtracted from each diagonal entry.
Since $U$ is a rotation (and hence invertible), it follows that the range of $A-\lambda_{min}I$ is spanned by $u_1,\cdots,u_{n-1}$, as the bottom right entry of $D'$ is 0. So if the projection onto $u_n$, the smallest-eigenvalue-eigenvector is nonzero then $v$ cannot be in the range.