# Eigenvectors for a sum of diagonal and anti-diagonal matrices

Consider the case of invertible matrix which is the sum of diagonal and anti-diagonal matrices.

An example $\begin{bmatrix} \color{red}1 & 0 & 0 & 0 & \color{red}6 \\ 0 & \color{red}2 & 0 & \color{red}7 & 0 \\ 0 & 0 & \color{red}3 & 0 & 0 \\ 0 & \color{red}8 & 0 & \color{red}4 & 0 \\ \color{red}9 & 0 & 0 & 0 & \color{red}5 \\ \end{bmatrix}$,

Such matrices I name shortly $X$-matrices (even shorter $X$ - do they have more official name?) and it's easy to check that the sum of two $X$-matrices is an $X$-matrix.

Also the product of two matrices is an $X$-matrix as

$X_1X_2=(D_1+A_1) (D_2+A_2)=(D_1D_2+A_1A_2)+(D_1A_2+A_1D_2)$

($D,A$ denoted here as diagonal and antidiagonal part of $X$) and product of two diagonal or two anti-diagonal is always diagonal and product of diagonal and anti-diagonal is anti-diagonal.

Further if $X$-matrix is invertible also its inverse is an $X$-matrix because inverse can be presented as a polynomial of $X$ from Cayley-Hamilton theorem.

Making calculations I have found one more property of these matrices:
i.e. also eigenvectors $v_1, v_2, \dots$ for this type of matrix can be grouped to make $X$-matrix.
For instance for the matrix listed above we have eigenvectors as colummns of

$V=\begin{bmatrix} \color{red}{-0.730} & \color{red}{0.529} & 0.000 & 0.000 & 0.000 \\ 0.000 & 0.000 & \color{red}{0.730} & \color{red}{-0.633} & 0.000 \\ 0.000 & 0.000 & 0.000 & 0.000 & \color{red}{1.000} \\ 0.000 & 0.000 & \color{red}{-0.683} & \color{red}{-0.774} & 0.000 \\ \color{red}{0.683} & \color{red}{0.848} & 0.000 & 0.000 & 0.000 \\ \end{bmatrix}$,

and it's possible to permute columns in order to obtain from them an $X$-matrix.

• How this last property can be proved? How can we prove that there is a permutation of eigenvectors of $X$-matrix which is also an $X$-matrix?
• Could we use for proof the equation $X=VDV^{-1}$ where however $V$, if columns are chosen randomly, can be in the form which is not an $X$-matrix? (but its some permutation supposedly is ...)
• I wonder whether only this fact that X-matrices (invertible) have sum, product and inverse in the form of X-matrices it's sufficient to prove the claim about eigenvectors ... I suppose we can regard them as a group but is this fact solely allows us to infer about its eigenvectors? – Widawensen Oct 5 '17 at 9:03
• Your group is essentially a direct product of several copies of $GL(2,\mathbb R)$ (plus one instance of $\mathbb R$, if the size is odd). – Ivan Neretin Oct 5 '17 at 9:12
• @IvanNeretin What are consequences of it? – Widawensen Oct 5 '17 at 9:13
• I guess your property is one of the consequences. – Ivan Neretin Oct 5 '17 at 9:15
• Reorder your basis to $(x_1,x_5,x_2,x_4,x_3)$. Then the matrix will be $$\begin{pmatrix} 1 & 6 & 0 & 0 & 0 \\ 9 & 5 & 0 & 0 & 0 \\ 0 & 0 & 2 & 7 & 0 \\ 0 & 0 & 8 & 4 & 0 \\ 0 & 0 & 0 & 0 & 3 \\ \end{pmatrix}$$ – Ivan Neretin Oct 5 '17 at 12:03

Your X-matrices are actually a pretty well known family in disguise: block diagonal matrices with blocks $2\times2$. Go read that paragraph in Wikipedia, it is pretty short, and whatever you need from it is shorter yet:

The eigenvalues and eigenvectors of ${A}$ are simply those of $A_{1}$ and $A_{2}$ and ... and $A_{n}$ (combined).

Indeed, in an X-matrix, $x_1$ interacts only with $x_n$, $x_2$ with $x_{n-1}$, and so on. Why wouldn't you reorder your basis vectors so as to put the interacting ones next to each other? With your $5\times5$ example, this implies reordering the basis as $(x_1,x_5,x_2,x_4,x_3)$. The matrix which does that is $$P=\begin{pmatrix}1& 0& 0& 0& 0\\ 0& 0& 0& 0& 1\\ 0& 1& 0& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 1& 0& 0\end{pmatrix}$$ Now apply that to your matrix $X$ and get $$PXP^T=\begin{pmatrix}1& 6& 0& 0& 0\\ 9& 5& 0& 0& 0\\ 0& 0& 2& 7& 0\\ 0& 0& 8& 4& 0\\ 0& 0& 0& 0& 3 \end{pmatrix}$$ Q.e.d.

• Thank you Ivan for the most general answer. I suppose we can even generalize the case for more matrices "in disguise", but probably these so-called X-matrices are the most interesting from the structural point of view.. – Widawensen Oct 6 '17 at 9:07
• Additionaly your solution is very compact as we can, I suppose, write using $P: \ \ \\ \$ $PXP^T=(PVP^T)(PDP^T)(PVP^T)$ $\\ \ \ \ \ \ \ \ \ \ \ \ \$ ... I'm very glad you decided to present this solution.. – Widawensen Oct 6 '17 at 9:11

Hint. In your 5x5 matrix, you can easily observe that there ALWAYS exist:

a. Two eigenvectors of the form $u=(x,0,0,0,y)$ - because $Au=(x',0,0,0,y')$

b. Two eigenvectors of the form $v=(0,x,0,y,0)$ - because $Av=(0,x',0,y',0)$ and

c. the eigenvector $(0,0,1,0,0)$.

Put them together as columns and obtain an X-matrix.

• Yes, such observations can be made.. but why is so and why exactly two? Not any other would be appropriate? What is the explanation of this fact? How to extend this for any dimension ? – Widawensen Oct 5 '17 at 11:55
• Maybe I would go in the direction you are proposing, it seems to be promising.. – Widawensen Oct 5 '17 at 12:02