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Suppose a vector $v_1$ is in the row space of some matrix $A$. Can we say anything about the column space? Gilbert Strang on 3:35 talks about this. Is vector $u_1=Av_1$ simply a "definition" for the vector $u_1$ that is related to $v_1$ through transformation by $A$. I can't quite grasp the connection between the column space and row space here, so I think it must be a definition. I know that the row space and the null space are orthogonal and have a zeroth vector in common; the same is the case for the column space and the left nullspace. But how does one link the row space and the column space?

Am I correct in assuming that $u_1=Av_1$ is simply a definition for $u_1$?

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  • $\begingroup$ What's the meaning of SVD? $\endgroup$
    – Sharpie
    Jun 17, 2016 at 12:38
  • $\begingroup$ @Sharpie Why don't you just take the first result off of google? $\endgroup$
    – Ben Grossmann
    Jun 17, 2016 at 12:59
  • $\begingroup$ See the video at 6:20, in fact the working definition he sticks to later is that both $u_1$ and $v_1$ should be unit vectors. $\endgroup$
    – Ben Grossmann
    Jun 17, 2016 at 13:08
  • $\begingroup$ Sorry @Omnomnomnom, I have already known this concept. My first language is not English, so this concept doesn't know as SVD, but thank for your answer. $\endgroup$
    – Sharpie
    Jun 18, 2016 at 0:22

2 Answers 2

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Stating that $u_1 = Av_1$ is part of the definition, but it is not the full definition. What he's really stating is the problem that characterizes SVD.

In particular, we're looking for a set of orthonormal basis $v_1,\dots,v_n$ such that the corresponding vectors $u_k = A v_k$ (for $k = 1,\dots,n$) turn out to be orthogonal to each other.

Note that this is not exactly what SVD is, as he clarifies later in the video (see my comment above).

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Think of the matrix as a function: the column space is the range of the function, and the row space the pre-image. $u_1=Av_1$ simply says that $A$ maps the vector $v_1$ in the domain of $A$ to the vector $u_1$ in its range.

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