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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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.
Siddharth Bhat
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1I think you are wrong. The question is about singular values and not eigenvalues. Your solution is complex instead of non-negative. – Gil May 03 '16 at 06:51
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Oh, hm, you're right. can you SVD a complex matrix in that case? – Siddharth Bhat May 03 '16 at 06:53
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Sure, you can SVD any matrix. – Gil May 03 '16 at 06:55
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@Gil A quadratic matrix is singular iff zero is an eigenvalue of that matrix. – Michael Hoppe May 03 '16 at 10:10
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@MichaelHoppe - Please more guide – Under sky May 03 '16 at 13:25
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@Undersky just read http://math.stackexchange.com/questions/503585/show-that-a-matrix-a-is-singular-if-and-only-if-0-is-an-eigenvalue/503644#503644 – Michael Hoppe May 03 '16 at 13:52
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@MichaelHoppe - Thanks, but what are singular value of $A$?(in terms of $x,y$) – Under sky May 03 '16 at 15:00
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@Undersky - Obviously $(0,-3/4)$, $(5/4,0)$, and $(-3/4,0)$ due to the first answer. – Michael Hoppe May 03 '16 at 15:05
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@MichaelHoppe - Why? (Singular value is square root of $det(A-\lambda I)=0$). – Under sky May 03 '16 at 15:15
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@Gil - Please more guide. – Under sky May 03 '16 at 15:17
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1@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... – Gil May 03 '16 at 16:08