# Reconstructing a Matrix in $\Bbb{R}^3$ space with $3$ eigenvalues, from matrices in $\Bbb{R}^2$

I have a matrix which represents a closed loop matrix of a control system with delays (Control Systems Theory) in $\Bbb{R}^3$ space that has $3$ eigenvalues. Through some process I have obtained three different matrices in $\Bbb{R}^2$ and sometimes in $\Bbb{C}^2$ which represent a part of the control system and each matrix has just two of the three original eigenvalues.

What I want is to try to reconstruct the original $\Bbb{R}^3$ matrix. I do not know which operation or algebraic manipulation should I do to get the $\Bbb{R}^3$ system from two of the three systems obtained.

To show a numerical example we have the following:

$$A=\begin{pmatrix}05099 & 0.2649 & 0.01\\ -0.99 & -0.485 & 0.01\\ -1 & -1.5 & 0\end{pmatrix}$$

It has the following eigenvalues: $-0.0875+0.2592i$, $-0.0875-0.2592i$ and $0.1998$. Through some process I have obtained three different systems:

$$A_{11}=\begin{pmatrix}0.5825-0.1494i & 0.2796-0.1092i\\ -0.9170-0.1502i & -0.4702-0.1097i\end{pmatrix}$$

Its eigenvalues are: $0.1998$ and $-0.0875-0.2592i$

$$A_{22}=\begin{pmatrix}0.5825+0.1494i & 0.2796+0.1092i\\ -0.9170+0.1502i & -0.4702+0.1097i\end{pmatrix}$$

Its eigenvalues are: $0.1998$ and $-0.0875+0.2592i$

$$A_{33}=\begin{pmatrix}0.3502 & 0.2249\\ -1.1505 & -0.5252\end{pmatrix}$$

Its eigenvalues are: $-0.0875-0.2592i$ and $-0.0875+0.2592i$

The process that I have done is the following, this matrix $A$, named extended in the language that I am using, has $n+1$ eigenvalues ($3$ eigenvalues in the case that i am talking about). Is possible to obtain these three new models through this $A_{11}=SAMP^kS^T(SMP^kS^T)^{-1}$ which is the same $A_{11}=SMEP^kS^T(SMP^kS')^{-1}$ where $S$ is a selection matrix, in this case is the following $$S=\begin{pmatrix}1 & 0 & 0\\0 & 1 & 0\end{pmatrix}$$ and $P$ is a permutation matrix $$P=\begin{pmatrix}0 & 0 & 1\\ 1 & 0 & 0\\0 & 1 & 0\end{pmatrix}$$ for eigenvectors. $E$ is a diagonal Matrix which contains the eigenvalues and $M$ is the eigenvectors matrix. Another way to express that solutions is $A_{11}=(SMP^{k}S')S(MP^k)^{-1}A(MP^k)S'(SMP^kS')^{-1}$. What I have done is like a projection from $\mathbb{R}^3$ over $\mathbb{R}^2$ or $\mathbb{C}^2$ eliminating one eigenvalue in each case to obtain $A_{11}$, $A_{22}$, $A_{33}$.

Gina Torres

• @TonyK Multiplication by an orthogonal matrix can easily change the eigenvalues of the original matrix. You probably meant conjugation by any invertible matrix, not multiplication by an orthogonal one. – Dan Shved Dec 11 '12 at 12:34
• @Dan: You are too kind! That's not what I meant at all, I was simply wrong. I have deleted my comment. – TonyK Dec 11 '12 at 14:49
• Thanks @TonyK could you please explain about the conjugation that you are talking about, maybe with a short example? – Gina Torres 17 mins ago – Gina Torres Dec 12 '12 at 9:43
• Is that a problem maybe refered to an Inverse Eingenvalue problem where it is determined a structured matrix from a given spectral data?. But i would like you to explain me please about the conjugation. Thanks. – Gina Torres Dec 12 '12 at 10:47
• @Gina, the conjugation of matrix $A$ by the (invertible) matrix $B$ is the matrix $BAB^{-1}$. This conjugation has the same eigenvalues as $A$ $-$ can you see why? – TonyK Dec 12 '12 at 23:14

If you have an unknown $3 \times 3$ matrix $A$ that has $3$ distinct eigenvalues, and these eigenvalues and their corresponding eigenvectors are known, then of course you can restore matrix $A$. You don't even have to use those additional matrices $A_{11},\,A_{22}$ and $A_{33}$. This is how you can do it: $$A = M \cdot D \cdot M^{-1},$$ where $M$ is the matrix whose columns are eigenvectors of $A$ and $D$ is the diagonal matrix whose diagonal elements are eigenvalues of $A$.
• @Gina Frankly, I don't understand what you mean by "reversing the process". And what's the point in using data that we can easily do without (I mean matrices $A_{ii}$)? – Dan Shved Jan 10 '13 at 15:02
• @Gina The question will start to make sense to me again if we aren't allowed to use matrix $M$ at all. If we're allowed to use $M$, then I simply refuse to use any additional data that isn't useful. – Dan Shved Jan 10 '13 at 15:06
• Thanks @Dan Shved, Lets say that we are not allowed to use M, the point is to recover A from $A_{ii}$.... What could be the process to do that? – Gina Torres Jan 14 '13 at 13:39