# Eigenvalues of triangular block matrix 2

I need a way for compute the eigenvalues of these block matrix $$Acc=\begin{bmatrix} A & I \\ D & 0 \\ \end{bmatrix}$$ 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 $$17.3027; -2.2586 + 1.0499i; -2.2586 - 1.0499i; 1.7965; 0.5996; -0.1816$$

• If $A$ is diagonal too, then the problem decouples into $n$ separate instances of finding eigenvalues of a $(^{*}_{*}\,{}^1_0)$ matrix. Commented Jan 8, 2016 at 18:56
• 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 ? Commented Jan 8, 2016 at 19:03
• 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. Commented Jan 8, 2016 at 19:41
• 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. Commented Jan 8, 2016 at 21:34
• I would a closed form from the eigenvalues of Acc and the eigenvalues of A and D Commented Jan 8, 2016 at 21:35

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 $$\lambda_1=(1/2)(3+\sqrt{9+28})\approx 4.54138...$$ $$\lambda_2=(1/2)(3-\sqrt{9+28})\approx -1.54138...$$ $$\lambda_3=(1/2)(5+\sqrt{25+36})\approx 6.40512...$$ $$\lambda_4=(1/2)(5-\sqrt{25+36})\approx -1.40512...$$
• It should work for a generic matrix $A$. Now if the answer was useful, don't forget to accept it :-) Commented Jan 11, 2016 at 21:35