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I need a way for compute the eigenvalues of these block matrix \begin{equation}Acc=\begin{bmatrix} A & I \\ D & 0 \\ \end{bmatrix}\end{equation} Where:

  • $A$ is a generic $n \times n$ matrix with know eigenvalues
  • $D$ is a diagonal $n \times n$ matrix
  • $I$ is the identity $n \times n$ matrix
  • $0$ is the null $n \times n$ matrix

I would a closed form from the eigenvalues of $Acc$ and the eigenvalues of $A$ and $D$

If can be helpful, we can consider first the case where $A$ is too diagonal

For example let be consider

  • $$A=\begin{bmatrix} 3 & 0 \\ 0 & 5\\ \end{bmatrix}$$ the eigenvalues are 3 and 5
  • $$D=\begin{bmatrix} 7 & 0 \\ 0 & 9\\ \end{bmatrix}$$ the eigenvalues are 7 and 9
  • I is the indent matrix $2\times2$
  • $0$ is the null matrix $2\times2$

The eigenvalues of Acc are $4.5414,-1.5414,6.4051,-1.4051$

There exist a relation, linear or non linear, from 3,5,7,9, or such other parameters of the matrix A and D, and the eigenvalues of $Acc$ (This questions it was resolved from Pierpaolo Vivo Thanks a lot:-) )

I thought that it is simple pass from the case where A is diagonal to the case where A is a generic matrix

Now WE consider this reformulated problem

Again $$Acc=\begin{bmatrix} A & I \\ D & 0\\ \end{bmatrix}$$

  • $$A=\begin{bmatrix} 1 & 2 & 3 \\ 11 & 5 & 6 \\ 7 & 8 & 9 \\ \end{bmatrix}$$ the eigenvalues of A are $$17.0245; -1.0123 + 1.2010i; -1.0123 - 1.2010i$$
  • $$D=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & 0\\ 0 & 0 & 7 \\ \end{bmatrix}$$ the eigenvalues are 1,3 and 7
  • I is the indent matrix $3\times3$
  • $0$ is the null matrix $3\times3$

The eigenvalues of $Acc$ are $$ 17.3027; -2.2586 + 1.0499i; -2.2586 - 1.0499i; 1.7965; 0.5996; -0.1816 $$

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  • $\begingroup$ If $A$ is diagonal too, then the problem decouples into $n$ separate instances of finding eigenvalues of a $(^{*}_{*}\,{}^1_0)$ matrix. $\endgroup$ Commented Jan 8, 2016 at 18:56
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    $\begingroup$ Wouldn't it be enough to write $A_{CC}-z I$ and compute its determinant using the formulas for the determinant of a block matrix, see en.wikipedia.org/wiki/Determinant#Block_matrices ? $\endgroup$ Commented Jan 8, 2016 at 19:03
  • $\begingroup$ Following Pierpaolo Vivo's excellent suggestion: we have $$Acc-zI_{2n}=\begin{bmatrix}A-zI_n & I_n \\D & -zI_n\end{bmatrix}$$ (where we've used subscripts to indicate the size of the identity matrices). Since $D$ and $-zI_n$ commute, one of the identities from the link P.V. provided (third from the bottom in that section) gives $$\det(Acc-zI_{2n}) = \det\big( (A-zI_n)(-zI_n) - I_nD\big) = \det (-zA + z^2I_n - D).$$ It's not clear to me that a general formula follows from this; however, it should be helpful in practice. $\endgroup$ Commented Jan 8, 2016 at 19:41
  • $\begingroup$ Becouse this problem come from an control problem and n is too large is not helpful. I have already found this formula but it don't help me. $\endgroup$ Commented Jan 8, 2016 at 21:34
  • $\begingroup$ I would a closed form from the eigenvalues of Acc and the eigenvalues of A and D $\endgroup$ Commented Jan 8, 2016 at 21:35

1 Answer 1

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Combining my hint and Greg's solution (using your initial notation, not the notation in the example you added), $$ \det(Acc-zI_{2n}) = \det\big( (A-zI_n)(-zI_n) - I_nD\big) = \det (-zA + z^2I_n - D)=\prod_{i=1}^n (-z a_i+z^2-d_i)\ , $$ where $a_i$ are eigenvalues of $A$ and $d_i$ are the eigenvalue of $D$. Equating to $0$ and solving for $z$, we have that the eigenvalues $\lambda_i$ of $A_{CC}$ are $$ \lambda_i=\left(\frac{1}{2}(a_i\pm\sqrt{a_i^2+4 d_i})\right)\,\qquad i=1,\ldots,n\ . $$ Specializing to your example, where $a_i=\{3,5\}$ and $d_i=\{7,9\}$, we obtain $$ \lambda_1=(1/2)(3+\sqrt{9+28})\approx 4.54138... $$ $$ \lambda_2=(1/2)(3-\sqrt{9+28})\approx -1.54138... $$ $$ \lambda_3=(1/2)(5+\sqrt{25+36})\approx 6.40512... $$ $$ \lambda_4=(1/2)(5-\sqrt{25+36})\approx -1.40512... $$

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  • $\begingroup$ Thanks a lot @PierpaoloVivo. But if A is not diagonal Can you generalize this formula?? For me it is very important the case where A is not diagunal $\endgroup$ Commented Jan 11, 2016 at 21:28
  • $\begingroup$ I thought that it is simple pass from the case where A is diagonal to the case where A is a generic matrix $\endgroup$ Commented Jan 11, 2016 at 21:31
  • $\begingroup$ It should work for a generic matrix $A$. Now if the answer was useful, don't forget to accept it :-) $\endgroup$ Commented Jan 11, 2016 at 21:35
  • $\begingroup$ @PierpaoloVivoSorry but it not work when A is a generic matrix. Now I have rewrite the question $\endgroup$ Commented Jan 11, 2016 at 22:13
  • $\begingroup$ Surely if you resolve the question where A is generic I quote you in my master thesis $\endgroup$ Commented Jan 11, 2016 at 22:14

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