# Find the eigenvalues of $A$?

Let $a_1, a_2,a_3,\ldots,a_n \in \mathbb{R}$ be such that $a_1 + a_2 +\cdots+a_n = 0$ and denote

$$A=\left(\begin{matrix} a_1^2 +1 & a_1a_2+1 & \cdots & a_1a_n +1 \\ a_2a_1+1 & a_2^2+1 & \cdots & a_2a_n +1 \\ \vdots & \vdots & & \vdots \\ a_na_1 + 1 & a_na_2 + 1 & \cdots & a_n^2 +1 \\ \end{matrix}\right)$$

Find the eigenvalues of $A$

My attempts : first I take $n = 2$ then i got $A = \begin{bmatrix} a_1^2 +1 &a_1a_2 + 1 \\ a_2a_1 &a_2^2 + 1 \end{bmatrix}$

now finding the eigenvalue of $A - \lambda I =0$ then$$A - \lambda I= \begin{bmatrix} a_1^2 +1-\lambda &a_1a_2 + 1 \\ a_2a_1 &a_2^2 + 1 - \lambda\end{bmatrix}$$

$$(a_1^2 +1-\lambda)(a_2^2 + 1 - \lambda)-a_1^2a_2^2 -a_1a_2 =0$$

Here I find difficulty in finding the eigenvalues of $A$.

Plase, help me.

Any hints/solution will be appreciated.

Thank you.

• Are you sure the term in left corner should be $a_2a_1$? Since every single term of this kind has the form $a_na_1+1$ it would be logical if this term would be $a_2a_1+1$. – mrtaurho Aug 1 '18 at 20:14
• Which terms below the main diagonal get $+1$ ? – Donald Splutterwit Aug 1 '18 at 20:15
• @stupid Okay. That is odd but an important fact. I just wanted to make it clear. – mrtaurho Aug 1 '18 at 20:17
• Then it is a typo in the book. Because this is indeed a $BB^T$ with $+1$ – percusse Aug 1 '18 at 20:41
• @mrtaurho If $A$ is supposed to be written in the form $BB^\top$, then it has to be symmetric. So, percusse is right. The matrix shown here in the OP's statement is not symmetric. – Batominovski Aug 1 '18 at 20:47

As discussed in the comment section, we believe that the problem comes with a typo and the correct matrix is $$\mathbf{A}=\begin{bmatrix} a_1^2+1&a_1a_2+1&a_1a_3+1&\cdots&a_1a_n+1\\ a_2a_1+1&a_2^2+1&a_2a_3+1&\cdots&a_2a_n+1\\ a_3a_1+1&a_3a_2+1&a_3^2+1&\cdots&a_3a_n+1\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ a_na_1+1&a_na_2+1&a_na_3+1&\cdots&a_n^2+1 \end{bmatrix}=\boldsymbol{1}\,\boldsymbol{1}^\top+\boldsymbol{a}\,\boldsymbol{a}^\top\,,$$ where $\boldsymbol{1}:=(\underbrace{1,1,1,\ldots,1}_{n\text{ ones}})$ and $\boldsymbol{a}:=(a_1,a_2,a_3,\ldots,a_n)$. I shall provide a generalization below. From this generalization, we conclude that the eigenvalues of this matrix are $n,\sum\limits_{k=1}^n\,a_k^2,\underbrace{0,0,\ldots,0}_{n-2\text{ zeros}}$.

If $\boldsymbol{a}$ is nonzero, then respective eigenvectors are $\boldsymbol{1},\boldsymbol{a},\boldsymbol{b}_1,\boldsymbol{b}_2,\ldots,\boldsymbol{b}_{n-2}$, where $\boldsymbol{b}_1,\boldsymbol{b}_2,\ldots,\boldsymbol{b}_{n-2}$ are any $n-2$ linearly independent vectors orthogonal (in the usual sense) to both $\boldsymbol{1}$ and $\boldsymbol{a}$ (they exist since the map $\boldsymbol{\varphi}:\mathbb{R}^n\to\mathbb{R}^2$ sending $\boldsymbol{c}\in\mathbb{R}^n$ to $\big(\langle\boldsymbol{c},\boldsymbol{1}\rangle,\langle\boldsymbol{c},\boldsymbol{a}\rangle\big)\in\mathbb{R}^2$ is a linear map of rank $2$, where $\langle\_,\_\rangle$ is the usual inner product on $\mathbb{R}^n$). If $\boldsymbol{a}$ is the zero vector, then there are $n-1$ linearly independent vectors $\boldsymbol{b}_1,\boldsymbol{b}_2,\ldots,\boldsymbol{b}_{n-1}$ orthogonal to $\boldsymbol{1}$, and respective eigenvectors are then $\boldsymbol{1},\boldsymbol{b}_1,\boldsymbol{b}_2,\ldots,\boldsymbol{b}_{n-1}$.

Lemma. Let $\mathbb{K}$ be a field and $n$ a positive integer. The vector field $V:=\mathbb{K}^n$ is equipped with the usual nondegenerate symmetric bilinear form $\langle\_,\_\rangle$ defined by $$\big\langle (u^1,u^2,\ldots,u^n),(v^1,v^2,\ldots,v^n)\big\rangle:=u_1v_1+u_2v_2+\ldots+u_nv_n$$ for all $u^1,u^2,\ldots,u^n,v^1,v^2,\ldots,v^n\in\mathbb{K}$. For a nonnegative integer $m$, suppose that $\mathbf{x}_1,\mathbf{x}_2,\ldots,\mathbf{x}_m$ are linearly independent vectors in $V$, which are mutually orthogonal, namely, $\langle \mathbf{x}_i,\mathbf{x}_j\rangle =0$ for all $i,j\in\{1,2,\ldots,m\}$ with $i\neq j$. Write $\lambda_k:=\langle \mathbf{x}_k,\mathbf{x}_k\rangle$ for $k=1,2,\ldots,m$. Let $\xi_1,\xi_2,\ldots,\xi_m$ be arbitrary nonzero scalars (elements of $\mathbb{K}$). Then, the matrix $$\mathbf{X}=\sum_{k=1}^m\,\xi_k\,\mathbf{x}_k\,\mathbf{x}_k^\top$$ is of rank $m$ with eigenvalues $\xi_1\lambda_1,\xi_2\lambda_2,\ldots,\xi_m\lambda_m,\underbrace{0,0,\ldots,0}_{n-m\text{ zeros}}$.

First, we see that $\mathbf{X}$ has rank $m$ (in this paragraph, mutual orthogonality of $\mathbf{x}_1,\mathbf{x}_2,\ldots,\mathbf{x}_m$ is not used). To show this, we define $W$ to be the subspace of $V$ spanned by $\mathbf{x}_1,\mathbf{x}_2,\ldots,\mathbf{x}_m$. Thus, $\dim_\mathbb{K}(W)=m$. Take the linear functional $\psi_k:V\to\mathbb{K}$ to be the map sending $\mathbf{v}\mapsto \langle \mathbf{v},\mathbf{x}_k\rangle$ for each $k=1,2,\ldots,m$ and $\mathbf{v}\in V$. These linear functionals are linearly independent element of the dual space $V^*$ of $V$ because $\langle\_,\_\rangle$ is a nondegenerate bilinear form. In particular, there exist $\mathbf{w}_1,\mathbf{w}_2,\ldots,\mathbf{w}_m\in V$ such that $\psi_i(\mathbf{w}_j)=\delta_{i,j}$ for every $i,j=1,2,\ldots,m$, where $\delta$ is the Kronecker delta. Therefore, $\mathbf{X}\,\mathbf{w}_k=\xi_k\,\mathbf{x}_k$ for every $k=1,2,\ldots,m$. This means $\text{im}(\mathbf{X})=W$, whence $\mathbf{X}$ is a matrix of rank $m$ over $\mathbb{K}$.

Without loss of generality, we can rearrange $\mathbf{x}_1,\mathbf{x}_2,\ldots,\mathbf{x}_m$ so that, for some nonnegative integer $r\leq m$, $\lambda_1,\lambda_2,\ldots,\lambda_r$ are nonzero, whereas $\lambda_{r+1},\lambda_{r+2},\ldots,\lambda_{m}$ are zero. Observe that, for each integer $t\geq 2$, $$\mathbf{X}^t=\sum_{k=1}^r\,\xi_k^t\lambda_k^{t-1}\,\mathbf{x}_k\,\mathbf{x}_k^\top\,.$$ Using the same argument as before, we determine that $\mathbf{X}^t$ has rank $r$ for every $t=2,3,\ldots$. That is, $0$ is an eigenvalue of $\mathbf{X}$ with multiplicity exacly $n-r$.

For the remaining part, we just determine $r$ eigenvectors of $\mathbf{X}$ corresponding to the nonzero eigenvalues $\xi_1\lambda_1,\xi_2\lambda_2,\ldots,\xi_r\lambda_r$. This is easy, as $\mathbf{X}\,\mathbf{x}_k=\xi_k\lambda_k\,\mathbf{x}_k$ for all $k=1,2,\ldots,m$, recalling that $\mathbf{x}_1,\mathbf{x}_2,\ldots,\mathbf{x}_m$ are mutually orthogonal. Thus, $\mathbf{x}_1,\mathbf{x}_2,\ldots,\mathbf{x}_r$ are eigenvectors associated to the eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_r$, respectively. The lemma is now verified.

Remark. We can say a bit more about the Jordan canonical form of $\mathbf{X}$. First, note that $\ker(\mathbf{X})$ is $(n-m)$-dimensional. As $\dim_\mathbb{K}\big(\ker(\mathbf{X}^t)\big)=n-r$ for all $t=2,3,\ldots$, we conclude that, for the eigenvalue $0$, $\mathbf{X}$ has $m-r$ Jordan blocks of size $2$ and $n+r-2m$ Jordan blocks of size $1$. We can also see that $\mathbf{x}_{r+1},\mathbf{x}_{r+2},\ldots,\mathbf{x}_m\in\ker(\mathbf{X})\subseteq \ker(\mathbf{X}^2)=\ker(\mathbf{X}^3)=\ker(\mathbf{X}^4)=\ldots$. Indeed, for each $k=r+1,r+2,\ldots,m$, the two vectors $\mathbf{x}_k$ and $\mathbf{w}_k$ are generalized eigenvectors for a Jordan block of $\mathbf{X}$. You can then use the map $\boldsymbol{\varphi}:\mathbb{K}^n\to\mathbb{K}^{2m-r}$ sending $$\mathbf{v}\mapsto \big(\langle \mathbf{v},\mathbf{x}_1\rangle,\langle\mathbf{v},\mathbf{x}_2\rangle,\ldots,\langle\mathbf{v},\mathbf{x}_m\rangle,\langle\mathbf{v},\mathbf{w}_{r+1}\rangle,\langle\mathbf{v},\mathbf{w}_{r+2}\rangle,\ldots,\langle\mathbf{v},\mathbf{w}_m\rangle\big)$$ for all $\mathbb{v}\in V$ and take $n+r-2m$ linearly independent vectors $\mathbf{y}_1,\mathbf{y}_2,\ldots,\mathbf{y}_{n+r-2m}\in \ker(\boldsymbol{\varphi})$. Then, you have found generalized eigenvectors of $\mathbf{X}$: $$\mathbf{x}_1,\mathbf{x}_2,\ldots,\mathbf{x}_r,\mathbf{x}_{r+1},\mathbf{w}_{r+1},\mathbf{x}_{r+2},\mathbf{w}_{r+2},\ldots,\mathbf{x}_m,\mathbf{w}_m,\mathbf{y}_1,\mathbf{y}_2,\ldots,\mathbf{y}_{n+r-2m}\,.$$

In the case where $\mathbb{K}$ is a symmetric subfield of $\mathbb{C}$, that is $\mathbb{K}$ contains the complex conjugates of all its elements, we can use the same argument, except that $\langle\_,\_\rangle$ is replaced by the standard sesquilinear inner product on $\mathbb{C}^n$ and that the transpose operator $\top$ is replaced by the Hermitian operator $\dagger$. In other words, we have the following corollary.

Corollary. Let $\mathbb{K}$ be a symmetric subfield of $\mathbb{C}$ and $n$ a positive integer. For each The vector field $V:=\mathbb{K}^n$ is equipped with the usual Hermitian form $\langle\_,\_\rangle$ defined by $$\big\langle (u^1,u^2,\ldots,u^n),(v^1,v^2,\ldots,v^n)\big\rangle:=u_1\bar{v}_1+u_2\bar{v}_2+\ldots+u_n\bar{v}_n$$ for all $u^1,u^2,\ldots,u^n,v^1,v^2,\ldots,v^n\in\mathbb{K}$. For a nonnegative integer $m$, suppose that $\mathbf{x}_1,\mathbf{x}_2,\ldots,\mathbf{x}_m$ are linearly independent vectors in $V$, which are mutually orthogonal, namely, $\langle \mathbf{x}_i,\mathbf{x}_j\rangle =0$ for all $i,j\in\{1,2,\ldots,m\}$ with $i\neq j$. Write $\lambda_k:=\langle \mathbf{x}_k,\mathbf{x}_k\rangle$ for $k=1,2,\ldots,m$. Let $\xi_1,\xi_2,\ldots,\xi_m$ be arbitrary nonzero scalars (elements of $\mathbb{K}$). Then, the matrix $$\mathbf{X}=\sum_{k=1}^m\,\xi_k\,\mathbf{x}_k\,\mathbf{x}_k^\dagger$$ is of rank $m$ with eigenvalues $\xi_1\lambda_1,\xi_2\lambda_2,\ldots,\xi_m\lambda_m,\underbrace{0,0,\ldots,0}_{n-m\text{ zeros}}$.

Unlike the lemma, all $\lambda_1,\lambda_2,\ldots,\lambda_m\in\mathbb{K}$ are positive real numbers (i.e., they cannot be $0$). Thus, $\mathbf{x}_1,\mathbf{x}_2,\ldots,\mathbf{x}_m$ are all eigenvectors corresponding to the nonzero eigenvalues $\xi_1\lambda_1,\xi_2\lambda_2,\ldots,\xi_m\lambda_m$. Plus, the linear map $\boldsymbol{\varphi}:\mathbb{K}^n\to\mathbb{K}^m$ sending $\mathbf{v}\mapsto\big(\langle\mathbf{v},\mathbf{x}_1\rangle,\langle\mathbf{v},\mathbf{x}_2\rangle,\ldots,\langle\mathbf{v},\mathbf{x}_m\rangle\big)$ for all $\mathbf{v}\in\mathbb{K}^n$ is of rank $m$, so there exist $n-m$ linearly independent elements $\mathbf{y}_1,\mathbf{y}_2,\ldots,\mathbf{y}_{n-m}$ in $\ker(\boldsymbol{\varphi})$. Then, $\mathbf{y}_1,\mathbf{y}_2,\ldots,\mathbf{y}_{n-m}$ are $n-m$ linearly independent eigenvalues of $\mathbf{X}$ associated to the eigenvalue $0$.

P.S. The matrix $\mathbf{B}$ as in the hint of the problem can be taken to be the $n$-by-$m$ matrix $$\mathbf{B}:=\begin{bmatrix} {\rule[-1ex]{0.5pt}{2.5ex}} & {\rule[-1ex]{0.5pt}{2.5ex}} & & {\rule[-1ex]{0.5pt}{2.5ex}}\\ \textbf{x}_1&\textbf{x}_2&\cdots&\textbf{x}_m\\ {\rule[-1ex]{0.5pt}{2.5ex}} & {\rule[-1ex]{0.5pt}{2.5ex}} & & {\rule[-1ex]{0.5pt}{2.5ex}} \end{bmatrix}\,,$$ so that $\mathbf{A}=\mathbf{B}\,\mathbf{D}\,\mathbf{B}^\top$ (or $\mathbf{A}=\mathbf{B}\,\mathbf{D}\,\mathbf{B}^\dagger$ in the last part of my answer), where $\mathbf{D}$ is the $m$-by-$m$ diagonal matrix $$\mathbf{D}:=\text{diag}\left(\xi_1,\xi_2,\ldots,\xi_m\right)\,.$$ You can see that $\mathbf{B}$ has full rank (i.e., $\text{rank}(\mathbf{B})=m$) and $\mathbf{B}^\top\,\mathbf{B}$ (or $\mathbf{B}^\dagger\, \mathbf{B}$ in the last part of my answer) is the $m$-by-$m$ diagonal matrix $$\text{diag}\left(\lambda_1,\lambda_2,\ldots,\lambda_m\right)\,.$$

• thanks u @Batominovski – jasmine Aug 2 '18 at 10:49