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(a) Given the column vector $a$ as an $n\times 1$ matrix. Write out its singular value decomposition, showing the matrices $U, \sum$, and $V$ explicitly.

(b) Given the row vector $a^T$ as an $1\times n$ matrix. Write out its singular value decomposition, showing the matrices $U, \sum$, and $V$ explicitly.

I currently get stucked on this question, since I couldn't find the eigenvalues of the $n\times n$ matrix $aa^T$ in part (a), and $a^Ta$ in part (b) to find the columns of $U$ and $V$, respectively. Can someone please help me on this problem?

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Let $a\neq 0$, $u:=a/\|a\|_2$, and $\tilde{U}$ be $n\times (n-1)$ matrix with orthonormal columns such that $\tilde{U}^Ta=0$ (the columns of $\tilde{U}$ form an orthonormal basis of the orthogonal complement of $\mathrm{span}\{a\}$). You can easily verify that $[u,\tilde{U}]$ is a square orthogonal matrix.

With $U_{\mathrm{E}}:=u$, $\Sigma_{\mathrm{E}}:=\|a\|_2$, and $V_{\mathrm{E}}:=1$, we get the so-called economy SVD of $a$: $$\tag{1} a=U_{\mathrm{E}}\Sigma_{\mathrm{E}}V_{\mathrm{E}}^T. $$ With $U:=[u,\tilde{U}]$, $\Sigma:=[\|a\|_2,0,\ldots,0]^T$, $V:=1$, we get the full SVD of $a$: $$\tag{2} a=U\Sigma V^T. $$

If $a=0$, you can select $U$ to be an arbitrary orthogonal matrix (and $u$ its first column). Otherwise, the rest is the same.

For $a^T$, just transpose (1) and (2).

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  • $\begingroup$ Thank you for your help. May you elaborate on your proof for the $[u,\tilde{U}]*[u,\tilde{U}]^{T}] = I$? I tried it out, but it doesn't seem true to me, since the first element on the diagonal is $a_1^2+ \sum_{i=1}^{n-1} {u^{2}_{1(n-1)}}$, which can't be $1$ as it has to be? $\endgroup$ – user177196 Mar 9 '15 at 20:59
  • $\begingroup$ the last expression = $1+ a^{2}_1$, due to the property of $\tilde{U}$ $\endgroup$ – user177196 Mar 9 '15 at 21:14
  • $\begingroup$ @user177196 $[u,\tilde{U}]^T[u,\tilde{U}]=\pmatrix{u^Tu&u^T\tilde{U}\\ \tilde{U}^Tu&\tilde{U}^T\tilde{U}}$. Now since $\|u\|_2=1$, $\tilde{U}^Ta=0$, $u$ is a multiple of $a$, and $\tilde{U}$ has orthonormal columns, this matrix is equal to the identity. $\endgroup$ – Algebraic Pavel Mar 9 '15 at 22:02

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