3
$\begingroup$

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.

$\endgroup$
  • 6
    $\begingroup$ 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 $a_2a_1+1$. $\endgroup$ – mrtaurho Aug 1 '18 at 20:14
  • 2
    $\begingroup$ Which terms below the main diagonal get $+1$ ? $\endgroup$ – Donald Splutterwit Aug 1 '18 at 20:15
  • 1
    $\begingroup$ @stupid Okay. That is odd but an important fact. I just wanted to make it clear. $\endgroup$ – mrtaurho Aug 1 '18 at 20:17
  • 3
    $\begingroup$ Then it is a typo in the book. Because this is indeed a $BB^T$ with $+1$ $\endgroup$ – percusse Aug 1 '18 at 20:41
  • 2
    $\begingroup$ @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. $\endgroup$ – Batominovski Aug 1 '18 at 20:47
7
$\begingroup$

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 $$\text{diag}\left(\lambda_1,\lambda_2,\ldots,\lambda_m\right)\,.$$

$\endgroup$
  • $\begingroup$ thanks u @Batominovski $\endgroup$ – jasmine Aug 2 '18 at 10:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.