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Let $ A = \left( {\begin{array}{*{20}{c}} {x + (\frac{3}{4} + y)i}&1&1\\ 0&{(x - \frac{5}{4}) + iy}&1\\ 0&0&{(x + \frac{3}{4}) + iy} \end{array}} \right)$, and $x,y\in \mathbb{R}$.

What are singular value of $A$?(In terms of $x,y$)

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1 Answer 1

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Since the matrix is upper triangular, the determinant is just the product of the diagonal elements. Hence

$$D = \left[x + i\left(\frac{3}{4} + y \right)\right] \cdot \left [\left(x- \frac{5}{4} \right) + iy \right ] \cdot \left[ \left(x + \frac{3}{4} \right) + iy \right]$$

Setting $D$ to zero and computing the values of $x$ and $y$ will give you the zeroes.

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    $\begingroup$ I think you are wrong. The question is about singular values and not eigenvalues. Your solution is complex instead of non-negative. $\endgroup$
    – Gil
    May 3, 2016 at 6:51
  • $\begingroup$ Oh, hm, you're right. can you SVD a complex matrix in that case? $\endgroup$ May 3, 2016 at 6:53
  • $\begingroup$ Sure, you can SVD any matrix. $\endgroup$
    – Gil
    May 3, 2016 at 6:55
  • $\begingroup$ @Gil A quadratic matrix is singular iff zero is an eigenvalue of that matrix. $\endgroup$ May 3, 2016 at 10:10
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    $\begingroup$ @MichaelHoppe, I understand the question different. I don't think they are looking for values of $x,y$ such that the matrix is singular (i.e., eigenvalues zero), but looking for the singular values (SVD) of $A$. A straightforward solution would be taking the square root of the eigenvalues of $A^*A$, but it is kinda ugly... $\endgroup$
    – Gil
    May 3, 2016 at 16:08

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