Find the eigenvalues

**Kiefer** · May 9th 2012, 06:38 PM

Find the eigenvalues for the following matrix:
A=
1 2 2 2
2 1 2 2
2 2 1 2
2 2 2 1

det(A-λI)=
1-λ 2 2 2
2 1-λ 2 2
2 2 1-λ 2
2 2 2 1-λ

Is there a shortcut to computing this? Or do I need to take (1-λ)det(3x3 matrix)-(2)det(3x3 matrix)+(2)det(3x3 matrix)-(2)det(3x3 matrix)?

**saravananbs** · May 10th 2012, 04:41 AM

convert it into a diagonal matrix. all the leading diagonal elements form the eigenvalues.

**DeMath** · May 10th 2012, 04:57 AM

Use this

$\det (A + x) = \begin{vmatrix}a_{1,1} + x&a_{1,2} + x& \ldots &a_{1,n} + x \\ a_{2,1} + x&a_{2,2} + x& \ldots &a_{2,n} + x \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,1}+ x&a_{n,2} + x& \ldots &a_{n,n}+ x \end{vmatrix}= \det A + x \cdot \sum_{i = 1}^n \sum_{j = 1}^n A_{i,j}$ , where $A_{i,j}$ - cofactors of the matrix $A$ .

Let $A+x = \begin{bmatrix}1 - \lambda&2&2&2 \\ 2&1 - \lambda&2&2 \\ 2&2&{1 - \lambda }&2 \\ 2&2&2&1 - \lambda\end{bmatrix}$ , where $x=2$ and $A=\begin{bmatrix}- 1 - \lambda&0&0&0 \\ 0&{ - 1 - \lambda }&0&0 \\ 0&0&{ - 1 - \lambda }&0 \\ 0&0&0&{ - 1 - \lambda } \end{bmatrix}$ .

Since $\det A= (-1-\lambda )^4= (\lambda + 1)^4$ and $A_{i,j}= \begin{cases}0,&i \ne j\\ (-1-\lambda)^3,&i = j\end{cases}$ , it follows that

$\det (A + x) = \det A + 2\sum\limits_{i = 1}^4 \sum\limits_{j = 1}^4 A_{i,j}= (\lambda + 1)^4+2 \cdot 4( -1-\lambda)^3=$ $(\lambda + 1)^4-8( \lambda+1)^3=(\lambda-7)(\lambda + 1)^3$

**HallsofIvy** · May 11th 2012, 12:09 PM

Originally Posted by saravananbs

convert it into a diagonal matrix. all the leading diagonal elements form the eigenvalues.

Great- uh, how do you convert to a diagonal matrix without finding the eigenvalues first?

**saravananbs** · May 11th 2012, 07:11 PM

by elementary transformation. we will get the upper triangular matrix, whose diagonal elements form the eigenvalues.

is anythink wrong?

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