When we want to get the eigenvalues of a square matrix $A$ we calculate $$ det(A - \lambda I) = 0$$ What I don't understand is that what we do here in basic terms. Like, for a simple, square matrix M, if determinant is 0 then the vectors are linearly dependent and the equation system described by M does not have a unique solution (it either has . But what about $A - \lambda I$, how can I describe what happens in $det(A - \lambda I) = 0$ in basic terms? I am trying to understand how the eigenvalues relate to this determinant. I know that eigenvectors S and eigenvalues $\lambda$ are related by $AS=\lambda S$. Please don't overwhelm me with proofs and math-heavy reasoning, I have a very basic understanding of linear algebra.
I understand that the eigenvectors of a linear transformation are the vectors that, when transformed, are simply scaled by a factor of the corresponding eigenvalue. Therefore $$ A\vec{x}=\lambda\vec{x}. $$ We can replace the scaling of the vector $\vec{x}$ by $\lambda$ with its equivalent linear transformation, $\lambda I$. Then we can rearrange this equation as follows: $$ A\vec{x}-(\lambda I)\vec{x}=\begin{bmatrix}0\\0\end{bmatrix}\\  (A-\lambda I)\vec{x}=\begin{bmatrix}0\\0\end{bmatrix}\\  $$ For this to be true either $\vec{x}$ must be a zero vector or the transformation $A-\lambda I$ must have $\vec{x}$ in its null-space. The course I'm following says that we can solve for $\lambda$ by finding when the determinant of $A-\lambda I$ is zero, but why does that guarantee that $\vec{x}$ will become the zero vector when transformed by $A-\lambda I$? I understand that for $\vec{x}$ to be an eigenvector this transformation must have a determinant of zero, but I don't see how the inverse can be true; having a determinant of zero doesn't gaurantee that $\vec{x}$ will be transformed into the zero vector. For example, let $$ (A-\lambda I) = \begin{bmatrix}1&amp;2\\0&amp;0\end{bmatrix}, \vec{x}=\begin{bmatrix}1\\1\end{bmatrix} $$ Then, although $\det\begin{bmatrix}1&amp;2\\0&amp;0\end{bmatrix} = 0$, it does not gaurantee $(A-\lambda I)\vec{x}$ to be the zero vector. $$ \begin{bmatrix}1&amp;2\\0&amp;0\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}3\\0\end{bmatrix}  $$ We can see that the transformation gives us $\begin{bmatrix}3\\0\end{bmatrix}$ and not the zero vector. Why does finding when the determinant of this transformation is equal to zero help us find the eigenvectors? Am I missing something?
Can my child attend a private school whilst I'm on USA ESTA? I am British, currently staying in the USA for 6 weeks on an ESTA Visa Waiver. I am living with a local family and whilst here my child (9 years old) has been attending a private (non-state) school with the children of the local family.  Today somebody told me that by having my child attend a private school I may be in violation of the terms of the ESTA Visa Waiver.  I have found nothing about this at   Where can I find official documentation / online advice about this? What would make a question popular with high views? I was wondering what might be making my questions less attractive to users?  I get very low views most of the time. What should be done to increase a question's popularity? If $\alpha_{1},\ldots,\alpha_{m}$ are vectors different from zero vector, then there is a linear functional $f$ such that $f(\alpha_{i})\neq 0$ Switch statement fallthrough in C#? Let $M_n$ denote the vector space of all $n\times n$ real valued matrices and for any matrix $A\in M_n$ let $A^t$ denote its transpose. Define the linear map $T:M_n\rightarrow M_n$ as $$T(A)=A+A^t.$$ What is the minimal polynomial of $T$? What are the eigenvalues and the corresponding eigenspaces of $T$? What is the characteristic polynomial of $T$?  I learned about minimal polynomials of operators from $\textit{Linear Algebra Done Right}$ by Axler. In his description he explains how in order to find the minimal polynomial you must find the smallest positive integer $m$ such that $T^m$ is a linear combination of $I, T, T^2, \ldots, T^{m-1}$. That is, the smallest integer $m$ such that $$T^m = a_0I+a_1T+a_2T^2+ \cdots +a_{m-1}T^{m-1}.$$  As for the eigenvalues...since we are working with real-valued matrices, then $A$ and $A^t$ have the same eigenvalues, so if $\lambda \in \mathbb{R}$ is an eigenvalue of $A$ (and hence of $A^t$) wouldn't $2\lambda$ be an eigenvalue of $A+A^t$? 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He accidentally drops the tool, which plummets unimpeded to the center of the moon.  He realizes that over time, it will consume and destroy the moon, which will be catastrophic for Earth.  He uses the time-travel device to travel to the moon at an earlier time so that he can prevent himself from dropping the tool.  He makes bigger problems for himself, forcing him to travel back further and further in time.  He travels so far back that he encounters the original alien inhabitants of the base.  Now we are at the present time in the story.  This is his final conflict, he has to escape, travel forward in time and resolve all of the problems he created.  Unfortunately, the alien inhabitants capture him, and the story ends.  Can anyone identify the story and author? Is there a way to enable/disable a wireless network interface via the command line (so I can put it in a bat file)?  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