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Let eigenvalues of $2 \times 2$ matrix $A$ be $1,-2$ and eigenvectors be $x_1$ & $x_2$ respectively. Then eigenvalues and eigenvectors of $A^2-3A+4I$ would be?

We know that eigenvalues can be calculated by substituting in the equation of new matrix. But what is the relation of eigenvectors with new matrix in such cases?

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    $\begingroup$ Try computing $(A^2-3A+4I)x_1$ and $(A^2-3A+4I)x_2$, and see what that tells you. $\endgroup$ – JimmyK4542 Sep 23 '19 at 3:30
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Hint: if $A v = \lambda v$, and $P$ is a polynomial, then $P(A) v = P(\lambda) v$.

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The eigenvectors stay the same. For convenience take the eigenvalue 1 and eigenvector x1. Then, ((A^2) - 3A + 4I) .x1=((1) ^2)x1 - 3.(1.x1)+ 4 (x1) =2.x1. Similarly the others.

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  • $\begingroup$ Will the eigen vectors always remain same irrespective of the operations done on matrix? Do they depend on eigen values and may change if eigen values are not distinct or real? $\endgroup$ – Rahul Gosavi Sep 23 '19 at 7:19
  • $\begingroup$ In which cases the eigen vectors may be different? $\endgroup$ – Rahul Gosavi Sep 23 '19 at 7:24
  • $\begingroup$ Read Carefully the first comment and the first answer. $\endgroup$ – Math-Learner Sep 24 '19 at 1:17

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