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Is there a proof for n X n matrices with the nth column containing only zero entries such that it will contain the eigenvector of $E_{\lambda=0}$ = a unit vector in $\mathbb{R}^n$, where a unit vector is a vector with a norm of 1 and is orthogonal to all other vectors within the coordinate space of the matrix in question?

For a more intuitive sense of what I'm asking, consider the below n X n matrix $B$

$B=\begin{bmatrix} a_1 & a_2 & \cdots & a_{n-1} & 0\\ b_1 & b_2 & \cdots & b_{n-1} & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ n_1 & n_2 & \cdots & n_{n-1} & 0\\ \end{bmatrix}\\$

(Where $n$ represents an arbitrary number of rows and columns and not an actual number of rows and columns.)

Given that any matrix with an entire column of only zero entries will contain at least one eigenvalue of zero (because the determinant is zero), is there a proof to show that the corresponding eigenvector of this zero eigenvalue will equal

$E_{\lambda=0} = \begin{bmatrix} 0_1\\ 0_2\\ 0_3\\ \vdots\\ 0_{n-1}\\ 1\\ \end{bmatrix}$?

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  • $\begingroup$ There may be other eigenvectors with eigenvalue $0$, but the one you mention can be verified by direct (and quick) computation. $\endgroup$ – André Nicolas May 11 '14 at 21:32
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It is certainly true that the matrix will have an eigenvector of that form corresponding to $\lambda = 0$; all you need to do for a "proof" is write out the expression for $(B - 0I)v = B\,v$ where $v$ is that vector.

Note that $B$ could, of course, have more eigenvectors corresponding to the same eigenvalue.

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Yes, here $A-\lambda I=A$ (since $\lambda=0$), so when we consider:

$(A-\lambda I)v=0\Rightarrow Av=0\Rightarrow v=(0,0,...,1)^T$ (if we wish to take the unit eigenvector).

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