Why isn't the SVD simplified to USV (no transpose)?

Question

I've looked at a lot of explanations of the SVD, and they all phrase it as $A = USV^T$ (or $V^*$ for complex matrices), and where V is unitary. But the conjugate of a unitary matrix is always unitary, right? Wouldn't a simpler definition be $A = USV$?

Answer 1

$$\mathrm A = \mathrm U \Sigma \mathrm V^{\top} = \sigma_1 \mathrm u_1 \mathrm v_1^{\top} + \sigma_2 \mathrm u_2 \mathrm v_2^{\top} + \cdots + \sigma_r \mathrm u_r \mathrm v_r^{\top}$$

where $\mathrm u_i$ and $\mathrm v_i$ are the $i$-th columns of $\mathrm U$ and $\mathrm V$, respectively.

Let $\mathrm W := \mathrm V^{\top}$ and write $\mathrm A = \mathrm U \Sigma \mathrm W$. How to denote the rows and columns of a matrix so that we can easily distinguish them? We could write

$$\mathrm A = \mathrm U \Sigma \mathrm W = \sigma_1 \mbox{col}_1 (\mathrm U) \, \mbox{row}_1 (\mathrm W) + \sigma_2 \mbox{col}_2 (\mathrm U) \, \mbox{row}_2 (\mathrm W) + \cdots + \sigma_r \mbox{col}_r (\mathrm U) \, \mbox{row}_r (\mathrm W)$$

but this notation clutters too much. Since $\mbox{row}_i (\mathrm W) = \left(\mbox{col}_i (\mathrm V)\right)^{\top}$,

$$\mathrm A = \sigma_1 \mbox{col}_1 (\mathrm U) \, \left(\mbox{col}_1 (\mathrm V)\right)^{\top} + \sigma_2 \mbox{col}_2 (\mathrm U) \, \left(\mbox{col}_2 (\mathrm V)\right)^{\top} + \cdots + \sigma_r \mbox{col}_r (\mathrm U) \, \left(\mbox{col}_r (\mathrm V)\right)^{\top} = \mathrm U \Sigma \mathrm V^{\top}$$

Answer 2

One way is to write $A=USV^*$ as $AV=US$, which says that the image of the basis in the columns of $V$ is the basis in the columns of $U$ properly scaled. This is the exact content of the SVD.

If you write $A=USV$, then the statement would talk about the image of the basis in the rows of $V$ being the columns of $U$, not very nice.