I want to calculate the singular value decomposition $U\Sigma V^T=A$ with $3 \times 1$ matrix $A=(1, 0, -1)$. I know that, in order to get the $3 \times 1$ matrix $\Sigma$ I have to calculate $B=A^TA$ which has $B=(2)$ as a result, meaning that $\Sigma=(\sqrt2, 0, 0)$. Now I have to calculate the determinant of $B-\lambda$ to get the eigenvalues $\lambda_{1,..,n}$ and the eigenvectors of $B$ in the next step. But here's my problem: I can't figure out how to do this with just a single number instead of a matrix. It would be great if someone could show me how it is done.
Singular value decomposition of a $3\times 1$ matrix
eigenvalues-eigenvectorslinear algebramatricesmatrix decompositionsvd
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Best Answer
From the definition of SVD we look for orthogonal matrix $U$ of dimension $3\times3$, a matrix $\Sigma$ of dimensions $3\times1$which is actually a column vector, already found by the author of the question: $\Sigma=(\sqrt2,0,0)^T$ and a $1\times1$ matrix $V$ which is a scalar that we take to be $1$ (since it should be an "orthogonal" matrix made of unit vectors). Now solving $U\Sigma V^T=A$ for the elements of $U$ we just get information for the 1st column of $U$ which is: $u_{11}= 1/\sqrt2$, $u_{12}=0$, $u_{13}=-1/\sqrt2$. The other 2 columns of $U$ are found by imposing the orthogonality conditions with unit vectors and we easily find: $ U=\begin{bmatrix} 1/\sqrt2 & 1/\sqrt2 & 0 \\ 0 & 0 & 1\\ -1/\sqrt2 & 1/\sqrt2 & 0 \end{bmatrix} $