Rank one matrix

Question

Given that I know that the matrix $\mathbf{W} \in \mathbb{R}^{d \times m}$ satisfies that: $$\mathbf{W}^\top\mathbf{W} = \mathbf{a} \cdot \mathbf{a}^\top$$ Where $\mathbf{a} \in \mathbb{R}^{m \times 1}$ is a column vector.

Does this mean $\mathbf{W}$ is a rank one matrix, meaning it's columns are all identical up to some multiplicative scale?

Since for all vectors $x$ $(W^\prime W)x=(aa^\prime)x=a(a^\prime x)$ is a scalar multiple of $a,$ the rank of $WW^\prime$ cannot exceed $1$ by definition. You can also use the definition of matrix multiplication to (immediately) conclude the columns are all multiples of $a$ and the scale factor for column $j$ of $W^\prime W$ is $a_j.$ — whuber, Jan 19 '21 at 14:14

score 2 · Accepted Answer · answered Jan 19 '21 at 08:47

2

If $a=0$, then rank of $W$ is $0$.

Otherwise, you are right, the rank of $W$ is $1$. $$rank(W^TW)=1,$$

Let $W=U\Sigma V^T$, then $W^TW=V\Sigma^T\Sigma V^T$, and we can see that the rank of $W^TW$ and the rank of $W$ is the same.

answered Jan 19 '21 at 08:47

Siong Thye Goh

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What is the meaning of $\Sigma$ and $V$ in your answer? – Carlos Jan 22 '22 at 11:42
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I wrote it as the Singular Value Decomposition of $W$. – Siong Thye Goh Jan 22 '22 at 15:40

Rank one matrix

1 Answers1