EE,2016

Question

Consider a $3\times 3$ matrix with every element being equal to 1 , its only non-zero eigenvalue is ....?

$3\times 3 \begin{bmatrix} 1 & 1 &1 \\ 1 &1 &1 \\ 1 &1 &1 \end{bmatrix}$

now i solve in simple way ..directly $\left | A-\lambda I \right |$ THEN I got correct ans 0,0,3 but

i have doubt .. if i solve by some elementary operation and reduce to $\begin{bmatrix} 1 & 1 &1 \\ 0& 0 &0 \\ 0 & 0 &0 \end{bmatrix}$

but i got different eigen values  why this happened ... i missed something ??

Comments

i think i can appy elementary transformation , but after that it should be in echlaon form ,(lower triangular form ) then eigen values will be diagonal element rt ?

@joshi_nitesh

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@sid1221

sorry, but i always use determinant method to find eigen values, so i dont know what you said is correct or not

but i will verify and then tell you.

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ok thanks verify , but i think you will  correct

Answer 1

If given matrix is n*n and all the elements are same. then non-zero eigen value will be  Element*Order Of matrix. Here, every element is 1  So, Non zero Eigen value will be 1*3 = 3 (element *Order Of Matrix)  Similarly check for 3*3 Matrix With every element equal to 2 , Its eigen values will be 0,0,6 where non - zero eigen value = 2*Order of matrix = 2*3 =6 .