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If $X$ is a matrix, then a Singular Value Decomposition representation is $X = \sum_{i=1}^{k}d_iu_iv_i^{T} = UDV^{T}$. I read inside a book that $u_iv_i^{T}$ is a rank one matrix. I was able to verify this indirectly but am wondering if anyone had a simple explanation or a way to see why.

My method involves recognizing $rank(AB) \leq min(rank(A), rank(B))$. Then, because the rank of either the vector $u_i$ or $v_i^{T}$ is one, then:

$$ rank(u_iv_i^{T}) \leq 1 $$

Now, since only a zero matrix has rank zero, we have:

$$ 0<rank(u_iv_i^{T}) \leq 1 \implies rank(u_iv_i^{T}) = 1 $$

Is there a more intuitive way to to see this?

user321627
  • 2,714

1 Answers1

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If $u=(u_1,\dots,u_n)^T$ and $v=(v_1,\dots,v_n)^T$, then $$ uv^T=\begin{bmatrix}u_1v_1 &\cdots &u_1v_n\\ u_2v_1 & \cdots & u_2v_n\\ \vdots & \cdots & \vdots \\ u_nv_1 & \cdots & u_nv_n \end{bmatrix} $$

In particular, every row is a multiple of $v$, so $uv^T$ has rank one (unless of course one of $u,v$ is the zero vector).

carmichael561
  • 54,793