# How to find the characteristic polynomial of the given matrix?

I am stuck at finding the characteristic polynomial of the following matrix. $$A=\begin{bmatrix} \ell a^2 & a & a & a&\dotso & \dotso& a &a \\ a & t &0&0 & \dotso & \dotso &0&0 \\ a & 0 & t &0 &\dotso & \dotso & 0 &0 \\ a & 0& 0& t &\dotso & \dotso &0 &0 \\ \dotso & \dotso& \dotso& \dotso &\dotso & \dotso & \dotso & \dotso \\ \dotso & \dotso& \dotso& \dotso &\dotso & \dotso & \dotso & \dotso \\ \dotso & \dotso& \dotso& \dotso &\dotso & \dotso & \dotso & \dotso \\ a& 0 &0&0 & \dotso & \dotso&t&0 \\ a& 0 &0&0 & \dotso & \dotso&0&t \end{bmatrix}_{(\ell+1) \times (\ell +1)}.$$

Here $$t,l,a$$ are constants.

My try: The characteristic polynomial of the matrix is $\det(xI-A). We have, $$\det(xI-A)=\begin{bmatrix} x-\ell a^2 & -a & -a & -a&\dotso & \dotso& -a &-a \\ -a & x-t &0&0 & \dotso & \dotso &0&0 \\ -a & 0 & x-t &0 &\dotso & \dotso & 0 &0 \\ -a & 0& 0& x-t &\dotso & \dotso &0 &0 \\ \dotso & \dotso& \dotso& \dotso &\dotso & \dotso & \dotso & \dotso \\ \dotso & \dotso& \dotso& \dotso &\dotso & \dotso & \dotso & \dotso \\ \dotso & \dotso& \dotso& \dotso &\dotso & \dotso & \dotso & \dotso \\ -a& 0 &0&0 & \dotso & \dotso&x-t&0 \\ -a& 0 &0&0 & \dotso & \dotso&0&x-t \end{bmatrix}_{(\ell+1) \times (\ell +1)}.$$ I expanded along the last row. I got $$(x-t)\times\det(M)-a \times \det (N)$$ where $$M=\begin{bmatrix} x-\ell a^2 & -a &-a & -a&\dotso & \dotso& -a \\ -a & t &0&0 & \dotso & \dotso &0 \\ -a & 0 &t &0 &\dotso & \dotso & 0 \\ -a & 0& 0& t &\dotso & \dotso &0 \\ \dotso & \dotso& \dotso& \dotso &\dotso & \dotso & \dotso \\ \dotso & \dotso& \dotso& \dotso &\dotso & \dotso & \dotso \\ \dotso & \dotso& \dotso& \dotso &\dotso & \dotso & \dotso \\ -a& 0 &0&0 & \dotso & \dotso&t\end{bmatrix}$$ and $$N=\begin{bmatrix} -a &-a & -a&\dotso & \dotso& -a \\ x-t &0&0 & \dotso & \dotso &0 \\ 0 &x-t &0 &\dotso & \dotso & 0 \\ 0& 0& x-t &\dotso & \dotso &0 \\ \dotso& \dotso& \dotso &\dotso & \dotso & \dotso \\ \dotso & \dotso& \dotso& \dotso &\dotso & \dotso \\ \dotso & \dotso& \dotso& \dotso &\dotso & \dotso \\ 0 &0&0 & \dotso & x-t&0\end{bmatrix}$$. But I am stuck from here. Can someone please help me out? Also, it is my request to everyone that please don't delete my question. This is my first question. I am trying really hard to show to the community what I have tried in the question. I am not so good at writing, I am a first year student studying Bachelors in Mathematics. • Well, do you know$\det N$? If so, the structure of your new matrix$M$is almost the same as the old matrix, except smaller, except for the issue that the value of$\ell$should be smaller but it isn't. That suggests that you replace the$\ell$in the top left corner with another variable and then induct on$\ell$. Oct 2 at 6:00 • Also it looks like you're only computing the determinant. To compute the characteristic polynomial you'd need to introduce yet another variable. Oct 2 at 6:01 • @QiaochuYuan; I have edited the question. But I am unable to understand how to use induction on$\ell$as you suggested. Will it be possible for you to give a detailed answer Oct 2 at 6:10 • Is$\ell even or odd? Oct 2 at 7:40 ## 2 Answers Let $$u=(\ell a^2-t,a,a,\ldots,a)^T$$ and $$v=(0,a,a,\ldots,a)^T$$. Then $$A=tI+\pmatrix{u&e_1&0&\cdots&0}\pmatrix{e_1^T\\ v^T\\ 0\\ \vdots\\ 0}=:tI+XY.$$ Therefore \begin{aligned} \det(xI-A) &=\det((x-t)I-XY)\\ &=\det((x-t)I-YX)\quad\text{(Sylvester’s secular theorem)}\\ &=\det\left[(x-t)I-\pmatrix{\ell a^2-t&1\\ \ell a^2&0\\ &&0_{(\ell-1)\times(\ell-1)}}\right]\\ &=\det\pmatrix{x-\ell a^2&-1\\ -\ell a^2&x-t\\ &&(x-t)I_{\ell-1}}\\ &=\big[x^2-(\ell a^2+t)x+\ell a^2(t-1)\big](x-t)^{\ell-1}.\\ \end{aligned} Remark. If you prefer a more elementary proof, you may use elementary row and column operations. Let $$B=A-tI =\pmatrix{\ell a^2-t&a&\cdots&a\\ a&0&\cdots&0\\ \vdots&\vdots&&\vdots\\ a&0&\cdots&0}.$$ Subtract the second row of $$B$$ from other rows below it. Then add to the third, fourth,... up to the last column to the second column to obtain the matrix $$C=\left(\begin{array}{c|c}\begin{matrix}\ell a^2-t&\ell a\\ a&0\end{matrix}&\ast\\ \hline 0_{(\ell-1)\times 2}&0_{(\ell-1)\times(\ell-1)}\end{array}\right).$$ Note that the row and column operations above amount to a similarity transform $$C=PBP^{-1}$$ for some invertible matrix $$P$$. It follows that \begin{aligned} \det(xI-A) &=\det((x-t)I-B)\\ &=\det((x-t)I-C)\\ &=\det\left(\begin{array}{c|c}\begin{matrix}x-\ell a^2&-\ell a\\ -a&x-t\end{matrix}&\ast\\ \hline 0_{(\ell-1)\times2}&(x-t)I_{\ell-1}\end{array}\right)\\ &=\big[x^2-(\ell a^2+t)x+\ell a^2(t-1)\big](x-t)^{\ell-1}.\\ \end{aligned} • I am sorry but I don't understand the solution. What is "Sylvester's Secular Theorem"? Is it not possible to do it in some easier way? Maybe by using some inductive steps. Also what ise_1$? Oct 2 at 10:02 • Since its my first year, I dont think I have learned Linear Algebra so much, I guess Oct 2 at 10:02 • @Andrea Sylvester’s secular theorem: for any two square matrices$X$and$Y$over a commutative ring$R$,$XY$and$YX$have the same characteristic polynomial. A short algebraic proof can be found in J. H. Williamson, The characteristic polynomials of$AB$and$BA\$, Edinburgh Mathematical Notes , Volume 39 , January 1954 , p. 13. See also an alternative proof here. Oct 2 at 10:41

By induction

$$\det M=\det \left[\begin{array}{ccccc}a&b_1&b_2&\ldots&b_n\\b_1&c_1&0&\ldots&0\\b_2&0&c_2&\ldots&0\\\ldots&\ldots&\ldots&\ldots&0\\b_n&0&0&\ldots&c_n\end{array}\right]=c_1\ldots c_n\left(a-\sum_{k=1}^n\frac{b_k^2}{c_k}\right).$$ Application: if $$b_1=\dots=b_n=b$$ and $$c_1=\ldots=c_n=c$$ then $$\det(M-\lambda I_{n+1})=(c-\lambda)^{n-1}[(a-\lambda)(c-\lambda)-nb^2].$$ Edit: this shows that $$M$$ is positive definite if and only if $$a>0$$ and $$ac>nb^2.$$