# What are left and right singular vectors in SVD?

Let $$USV^T$$ be a singular value decomposition of matrix $$A$$. In the textbook "Linear Algebra and Its Applications" by D. C. Lay et. al., where SVD is introduced, it says that "the columns of $$U$$ in such a decomposition are called left singular vectors of $$A$$, and the columns of $$V$$ are called right singular vectors of $$A$$." But it does not make any connections with the eigenvectors of $$A^T\!A$$. It also says that "the matrices $$U$$ and $$V$$ are not uniquely determined by $$A$$.

But in this web page it says that "the eigenvectors of $$A^T\!A$$ make up the columns of $$V$$, the eigenvectors of $$AA^T$$ make up the columns of $$U$$."

I'm confused about the relationship between the left and right singular vectors (that is columns of $$U$$ and $$V$$) and the eigenvectors of $$A^T\!A$$ and $$AA^T$$. Any clarification is appreciated.

• One observation: Suppose that $A$ is a real $n \times n$ matrix and $A = U \Sigma V^T$, where $U$ and $V$ are orthogonal matrices and $\Sigma$ is diagonal. (I'm assuming $A$ is square just for simplicity.) Then $A^T A = V \Sigma^T U^T U \Sigma V^T = V \Sigma^2 V^T$, which implies that $A^T A V = V \Sigma^2$. This equation, when read "column by column", tells us that the columns of $V$ are eigenvectors of $A^T A$. Jan 12, 2021 at 10:29

Let $$A=UDV^*$$. Then $$A^*A=VDU^*UDV^*=VD^2V^*\implies A^*AV=VD^2$$ $$AA^*=UDV^*VDU^*=UD^2U^*\implies AA^*U=UD^2$$ When a matrix is right-multiplied by a diagonal matrix, each column is multiplied by the diagonal term: $$[\mathbf{b}_1,\ldots,\mathbf{b}_n]\begin{pmatrix}\sigma^2_1&\\&\ddots\\&&\sigma_n^2\end{pmatrix}=[\sigma^2_1\mathbf{b}_1,\ldots,\sigma^2_n\mathbf{b}_n]$$
Hence the right-hand expressions are exactly the eigenvalue equations, $$A^*A\mathbf{v}_i=\sigma_i^2\mathbf{v}_i,\qquad {AA}^*\mathbf{u}_i=\sigma_i^2\mathbf{u}_i$$
• @SanyoMn $U$ and $V$ need not be unique for three reasons: (i) the singular values and their corresponding singular vectors can be permuted. (ii) each singular vector is unique only up to a sign, (iii) repeated singular values, especially the zero ones, do not have unique singular vectors - they can be rotated around. Jan 12, 2021 at 12:28