Here is the SVD decomposition calculated in Julia:
Why is the 4th column of $V^{T}$ also zero? How is that connected with columns of $A$?
Appreciate any help!
2026-04-01 07:17:20.1775027840
How does a column of zeros in a matrix influence its right singular vectors?
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Note that if $A$ has singular value decomposition $A = U \Sigma V^T$, then the columns of $V$ (the rows of $V^T$) form a basis for the row space of $A$. Because every row of $A$ is orthogonal to $(0,0,0,1,0)$, this must hold for the columns of $V$ as well.
Equivalently, the nullspace of $V^T$ must be equal to the nullspace of $A$, and the nullspace of $A$ contains the vector $(0,0,0,1,0)$.