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I came across a Linear Algebra problem and I'm having some trouble understanding it. The problem is 2b from the Stanford CS229 Machine Learning Course's first worksheet. I am not attending the class, I'm just taking it by myself on my own time since the course materials and lecture videos are online.

http://cs229.stanford.edu/materials/ps0.pdf

Problem

Let z ∈ R^n be a non-zero n-vector. Let A = zz^T. What is the null-space of A? What is the rank of A?

I figured out that the null space of A is the set of vectors orthogonal to z^T

What I can't figure out is the rank of A. I saw that zz^T would be symmetric and since the the resulting matrix is the product of a column vector and its transpose, I figured all the columns would be linearly dependent of one another and thus rank(A) = 1. I wasn't 100% sure of my reasoning since I had no way to really prove my hypothesis, so I checked the solution.

It turns out I was right on both accounts, but the reasoning used for the rank was different from mine and didn't make sense? The answer said that because null space of A is the set of vectors orthogonal to z^T, then rank(A) = 1. I don't understand why this is true. Can someone explain?

Also, it mentioned that the null space of A would have n-1 dimensions which I also didn't understand. A bit of clarification on that would also be helpful

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  • $\begingroup$ I see it in this way: first $\forall v$, $Av = zz^t v = \langle z, v \rangle z$. Here, $\langle z, v \rangle$ is the dot product of $z$ and $v$. So, for $v$ to be in the nullspace of $A$, we need $v$ to be orthogonal to $z$ or $z$ to be zero. On the other side, any vector $u$ in $Im(A)$ must be of the form $\alpha z$ where $\alpha$ is the scalar $\langle v, z \rangle$ for $v$ such that $u = Av$, this tell us that $Im(A) = span\{ z\}$ and therefore $rank(A) = dim(Im(A)) = dim(span\{z\}) = 1$. $\endgroup$
    – rarwoan
    Commented Oct 22, 2020 at 2:11

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They implicitly used the rank-nullity theorem, which follows by the general first isomorphism theorem and says that the sum of the rank and the dimension of the kernel of a linear map equals to the dimension of the domain of the map.

They also use that the orthogonal complement of a nonzero vector is $n-1$ dimensional.

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