# SVD, the connection between the column space and the row space?

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 $u_1$?

• What's the meaning of SVD? – Sharpie Jun 17 '16 at 12:38
• – Ben Grossmann Jun 17 '16 at 12:59
• 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. – Ben Grossmann Jun 17 '16 at 13:08
• 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. – Sharpie Jun 18 '16 at 0:22

## 2 Answers

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).

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.