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Eigen vectors of $\begin{bmatrix} 1 && \cos \theta \\ \cos \theta && 1 \end{bmatrix}$ are

  1. $\begin{bmatrix} a^n && 1 \\ 0 && a^n \end{bmatrix}$
  2. $\begin{bmatrix} a^n && n \\ 0 && a^n \end{bmatrix}$
  3. $\begin{bmatrix} a^n && na^{n-1} \\ 0 && a^n \end{bmatrix}$
  4. $\begin{bmatrix} a^n && na^{n-1} \\ -n && a^n \end{bmatrix}$
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3 Answers

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4 votes

I think d option is correct

The zero vector by convention is not an Eigen vector, much in the same way that 1 is not a prime number. If we let zero be an eigen vector, we would have to repeatedly say "assume v is a nonzero Eigen vector such that..." since we aren't interested in the zero vector. The reason being that v=0 is always a solution to the system Av=λv

By definition of Eigen vector : a nonzero vector that is mapped by a given linear transformation of a vector space onto a vector that is the product of a scalar multiplied by the original vector —called also characteristic vector

So Option A , B , C are incorrect

However, the eigenspace associated to an eigenvalue always contains the zero vector. .

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If we try to find the Eigen values of ths Matrix we will get λ=1+-$\cos(Theta)$ , now we have Ax=λx , and if we put x=0 then it will be a trivial solution to the given problem , so we will consider the option which doest not contain any zero , only option D stands so D , still looking for better explaination

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