Eigenvectors as basis for given matrixes

Question

suppose i have given some quadratic matrix,which i have represented as a multiplication of two orthogonal matrices and one diagonal matrix,for example

A =

 3     2     4
 1     2     4
 0    10    21



[E D,V]=svd(A)

E =

   -0.1892    0.9302    0.3145
   -0.1866    0.2803   -0.9416
   -0.9640   -0.2369    0.1206


D =

   24.1153         0         0
         0    3.0743         0
         0         0    0.0540


V =

   -0.0313    0.9989    0.0339
   -0.4309    0.0171   -0.9022
   -0.9018   -0.0428    0.4299

as i know $E$ and $V$ matrices are orthogonal,because it's columns are orthogonal,also as i know orthogonality implies linear independence of these vectors,could i have considered these vectors separately as a basis?or maybe list of vectors from $E$ and $V$ as basis,which means that can i use these vectors together to get original matrix using some combinations?thanks in advance

Answer 1

Suppose we name the singular values as $d_1,d_2,d_3$, the columns of $E$ as $e_1,e_2,e_3$, and the rows of $V$ as $v_1,v_2,v_3$. Then the singular value decomposition tells you precisely that $A$ is the operator given by $$ Av_j=d_je_j. $$

Answer 2

You could consider the column vectors in $E$ as a basis for $\mathbb{R}^3$. Similarly, the vectors in $V$ is also a basis for $\mathbb{R}^3$. But, it is meaningless to say that vectors in $E$ and $V$ can be used as a basis to get the matrix back. If you are looking for a basis for the matrix, then you are looking at an entirely different vector space. You need a set of matrices, who linear combination can then be used to get the original matrix back. Also, it should span the set of all $3\times 3$ real matrices. In that case, the vector space of $3\times 3$ matrices is a 9-dimensional space. SVD is not a decomposition of the matrix into a linear combination of matrices, but rather it gives a product decomposition. Having said that, note that \begin{align} A=\sum_{j=1}^{r}d_j(e_jv_j^T) \end{align} (verify this yourself). Thus SVD indeed gives the matrix $A$ as a linear combination of some rank-one matrices ($e_jv_j^T$) with the weights being the singular values. But this rank-one matrices, will never be a basis for the whole $3\times 3$ matrices in general.