# Is there a fast way to get the characteristic polynomial of this symmetric matrix?

I have to find the characteristic polynomial equation of this matrix

$$A= \begin{bmatrix}2 &1 &1&1 \\1&2&1&1\\1&1&2&1\\1&1&1&2 \end{bmatrix}$$

Is there another way than the rather long $\det(A-\lambda I)$ method ?

Maybe the fact that $A$ is symmetric ($A =A^t$) may be helpful ?

• Could I do something like $A-I\lambda= \begin{bmatrix}2-\lambda &1 &1&1 \\1&2-\lambda&1&1\\1&1&2-\lambda&1\\1&1&1&2-\lambda \end{bmatrix}$ then simplify and calculate the $det(...)$ , or would it change the value of$\lambda$? – HowCanIHelpYou Jan 20 '17 at 18:35
• Subtract the identity matrix and then it becomes pretty easy to show that for $A-I$ the eigenvalues will be 4 and zero (with multiplicity 3) – Sergei Golovan Jan 20 '17 at 18:35
• Interestingly, it seems that the characteristic polynomial of the $n\times n$ matrix with $2$ for each diagonal entry and $1$ for all other entries is $(n+1-\lambda)(1-\lambda)^{n-1}$. This suggests an inductive approach to me. – MPW Jan 20 '17 at 19:00
• Clearly, the eigenvalues of $A$ and $A-I$ differ. Exactly by a unity each. – Sergei Golovan Jan 20 '17 at 19:05
• And it seems that the characteristic polynomial of the $n\times n$ matrix with $a$ for each diagonal entry and $b$ for all other entries is $(a+(n-1)b - \lambda)(a-b-\lambda)^{n-1}$. – MPW Jan 20 '17 at 19:15

There's a formula for determinants of block matrices of the form $\begin{bmatrix} A&B\\B&A\end{bmatrix}$, where $A$ and $B$ are square matrices of the same size: $$\det\begin{bmatrix} A&B\\B&A\end{bmatrix}=\det(A-B)\det(A+B).$$ Applying this formula , we obtain \begin{align}\det(A-\lambda I)&=\begin{vmatrix}2-\lambda&1&1&1\\1&2-\lambda&1&1\\1&1&2-\lambda&1\\1&1&1&2-\lambda \end{vmatrix}=\begin{vmatrix}1-\lambda&0\\0&1-\lambda \end{vmatrix}\cdot\begin{vmatrix}3-\lambda&2\\2&3-\lambda \end{vmatrix}\\ &=(1-\lambda)^2\Bigl[(3-\lambda)^2-4\Bigr]=(\lambda-1)^3(\lambda-5).\end{align}

Here is a matrix in which the columns are eigenvectors of your matrix. Indeed, the columns are pairwise orthogonal.

$$\left( \begin{array}{rrrr} 1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & 0 & 2 & -1 \\ 1 & 0 & 0 & 3 \end{array} \right).$$ It is not an orthogonal matrix, as the columns are of different lengths. However, it can be made orthogonal by dividing each column by its Euclidean length, those being $2, \sqrt 2, \sqrt 6, \sqrt {12}$

The best, and very short way:

First step. Subtract to the rows $2$, $3$ and $4$, the first one.
Second step. Add to the first column the sum of the columns $2$, $3$ and $4.$

We get a triangular determinant.

EDIT: $$\begin{vmatrix}2-\lambda &1 &1&1 \\1&2-\lambda&1&1\\1&1&2-\lambda&1\\1&1&1&2-\lambda \end{vmatrix}\underbrace{=}_{\text{First step}}\begin{vmatrix}2-\lambda &1 &1&1 \\-1+\lambda&1-\lambda&0&0\\-1+\lambda&0&1-\lambda&0\\-1+\lambda&0&0&1-\lambda \end{vmatrix}$$ $$\underbrace{=}_{\text{Second step}}\begin{vmatrix}5-\lambda &1 &1&1 \\0&1-\lambda&0&0\\0&0&1-\lambda&0\\0&0&0&1-\lambda \end{vmatrix}=(5-\lambda)(1-\lambda)^3.$$

There are some simple tricks that you can use. The eigenvalues of $A$ are those values of $\lambda$ such that $A-\lambda I$ is singular, i.e., has rank less than four. Since $$A-1\cdot I=\pmatrix{ 1 & 1 & 1 & 1 \cr 1 & 1 & 1 & 1 \cr 1 & 1 & 1 & 1 \cr 1 & 1 & 1 & 1}$$ has rank equal to one, it follows that $\lambda=1$ is an eigenvalue of $A$ of multiplicity $4-1=3$. Hence $(\lambda-1)^3$ is a factor in the characteristic polynomial.

In the matrix $$A-5\cdot I=\pmatrix{-3 & 1 & 1 & 1\cr 1 & -3 & 1 & 1 \cr 1 & 1 & -3 & 1 \cr 1 & 1 & 1 & -3},$$ the sum of the elements in each row is zero. Hence $A-5\cdot I$ is singular, and $\lambda-5$ is another factor in the characteristic polynomial of $A$.

It follows that $\det(A-\lambda I = (\lambda-5)(\lambda-1)^3$.