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This is similar to a question I asked, but it's a different level of generality, it's more specific.

Let's say $x$ is the one-vector of $\mathbb{C}^n$, the vector space of complex numbers, so for example, $[1,1]$ would be the one-vector of $\mathbb{C}^2.$

Then, for a general square matrix $A \in \mathbb{C}^{n\times n},$ is it true that if $Ax = 0,$ for the zero-vector $0,$ that the matrix $A$ is the zero matrix?

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  • $\begingroup$ This is true only if your vector space has dimension at most 1. $\endgroup$ Aug 6 at 0:04
  • $\begingroup$ Although this idea of "unit vector" may have some appeal, it is not actually so useful, I think. I can understand some intuition about some sort of "most non-trivial vector", but it doesn't turn out to work that way. A reasonable idea, though! :) $\endgroup$ Aug 6 at 0:08
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    $\begingroup$ From Unit vector on Wikipedia, "Not to be confused with Vector of ones." But maybe your vector space use a different norm so $[1,1]$ has length $1$. $\endgroup$
    – peterwhy
    Aug 6 at 16:54
  • $\begingroup$ That is different than what I had in mind, I corrected the post. $\endgroup$
    – StackQuest
    Aug 6 at 17:11

2 Answers 2

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No, you could have the matrix $$A=\begin{pmatrix}1&-1\\1&-1\end{pmatrix}$$ for example.

More generally, if you have two vectors $x$ and $y$ and $x^Ty=0$ then it just means that the vectors are orthogonal. Here the matrix $A$ contains row vectors that are orthogonal to the unit vector.

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  • $\begingroup$ What if we add the condition that $A$ is invertible? $\endgroup$
    – StackQuest
    Aug 6 at 0:08
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    $\begingroup$ In that case, $A$ will certainly not be the zero matrix! $\endgroup$
    – PC1
    Aug 6 at 0:15
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    $\begingroup$ If $A$ is invertible, then $Ax=0$ cannot hold for the unit vector $x$; if it did, then $x = A^{-1}(0) = 0$ ?? From a false hypothesis, anything at all follows: for example $A=O$. $\endgroup$
    – GEdgar
    Aug 6 at 0:40
  • $\begingroup$ Well under what darn conditions will this work? What if $A$ is diagonal? $\endgroup$
    – StackQuest
    Aug 6 at 2:43
  • $\begingroup$ @StackQuest If $A$ is diagonal, say $\begin{bmatrix}a_{11}&0&\cdots\\0&a_{22}&\cdots\\\vdots&\vdots&\ddots\end{bmatrix}$, I think you can calculate what $Ax$ would be. $\endgroup$
    – peterwhy
    Aug 6 at 17:03
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@PC1 already gave a fine answer, with a nice counterexample. I would just like to add that another way to see that this cannot be true is that your definition of "unit vector" is basis dependent, whereas the claim that $Ax=0$ implies that $A$ in the zero matrix is basis independent. In other words, there's nothing special about the vector $[1,1]$ in $\mathbb{C}^2$, since any vector can be written that way in some basis. Thus, the falsity of your statement follows from the fact that not every $n\!\times\!n$ matrix with zero as an eigenvalue is the zero matrix, for $n>1$.

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