0
$\begingroup$

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 ?

$\endgroup$
  • $\begingroup$ 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 $? $\endgroup$ – HowCanIHelpYou Jan 20 '17 at 18:35
  • 2
    $\begingroup$ 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) $\endgroup$ – Sergei Golovan Jan 20 '17 at 18:35
  • 1
    $\begingroup$ 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. $\endgroup$ – MPW Jan 20 '17 at 19:00
  • 1
    $\begingroup$ Clearly, the eigenvalues of $A$ and $A-I$ differ. Exactly by a unity each. $\endgroup$ – Sergei Golovan Jan 20 '17 at 19:05
  • 1
    $\begingroup$ 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 $(a+(n-1)b - \lambda)(a-b-\lambda)^{n-1}$. $\endgroup$ – MPW Jan 20 '17 at 19:15
2
$\begingroup$

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}

$\endgroup$
1
$\begingroup$

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 $2, \sqrt 2, \sqrt 6, \sqrt {12}$

$\endgroup$
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

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.