linear algebra

Question

The eigen value of the following matrix in

[ 1 1 1

  1 1 1

  1 1 1 ]

(a) 1, 1, 1 (b) 1, 0, 0

(c) 3, 0, 0 (d) 0, 0, 0

Comments

C )

Answer 1

If we put eigen values in characteristic matrix i.e A-lambda.I than we get that matrix determinant 0

Also sum of eigen value is equal to trace of A

Hence C

Answer 2

Given that $A=\begin{bmatrix} 1 &1 &1 \\ 1 &1 &1 \\ 1 & 1 &1 \end{bmatrix}$

Characteristic equation of the given matrix$:|A-\lambda I|=0$

$\Rightarrow \begin{vmatrix} 1-\lambda &1 &1 \\ 1 &1-\lambda &1 \\ 1&1 &1-\lambda \end{vmatrix}=0$

Now we can solve this

$\Rightarrow(1-\lambda)[(1-\lambda)(1-\lambda)-1]-1(1-\lambda-1)+1(1-1+\lambda)=0$

$\Rightarrow(1-\lambda)[(1-\lambda)^{2}-1]-1(-\lambda)+1(\lambda)=0$

$\Rightarrow(1-\lambda)[1+\lambda^{2}-2\lambda-1]-1(-\lambda)+1(\lambda)=0$

$\Rightarrow(1-\lambda)[\lambda^{2}-2\lambda]+2\lambda=0$

$\Rightarrow\lambda^{2}-2\lambda-\lambda^{3}+2\lambda^{2}+2\lambda=0$

$\Rightarrow 3\lambda^{2}-\lambda^{3}=0$

$\Rightarrow \lambda^{3}-3\lambda^{2}=0$

$\Rightarrow\lambda^{2}(\lambda-3)=0$

$\Rightarrow\lambda=0,0,3$

So.Eigen Value are $0,0,3$

                                                 $(OR)$

Important Property regarding finding the EigenValue of any square matrix$:$

$(1)$Sum of leading diagonal element(Trace of the matrix) is equal to the Sum of all  Eigen Value.

  i.e.$\lambda_{1}+\lambda_{2}+\lambda_{3}=3$

$(2)$Product of all Eigenvalue is equal to Determinant(Det(A)) of the matrix.

i.e.$\lambda_{1}*\lambda_{2}*\lambda_{3}=0$