# Does $Ax = 0$ imply $A = 0$ if $x$ is the unit vector? [closed]

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?

• This is true only if your vector space has dimension at most 1. Aug 6 at 0:04
• 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! :) Aug 6 at 0:08
• 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 $1$. Aug 6 at 16:54
• That is different than what I had in mind, I corrected the post. Aug 6 at 17:11

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.

• What if we add the condition that $A$ is invertible? Aug 6 at 0:08
• In that case, $A$ will certainly not be the zero matrix!
– PC1
Aug 6 at 0:15
• 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 $A=O$. Aug 6 at 0:40
• Well under what darn conditions will this work? What if $A$ is diagonal? Aug 6 at 2:43
• @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. Aug 6 at 17:03

@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 $$n>1$$.