Are the following equations concerning matrices and their eigenvectors equivalent?

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Let $A$ be a matrix, $a,b$ its eigenvectors and $lambda_a, lambda_b$ their corresponding eigenvalues. Define $a^prime = Aa$ and $b^prime = Ab$. Is it true that
beginequation
lambda_a = fracVert a^primeVertVert aVert iff Vert aVert = frac1sqrtlambda_a
endequation
and
beginequation
lambda_b = fracVert b^primeVertVert bVert iff Vert bVert = frac1sqrtlambda_b ?
endequation










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    Let $A$ be a matrix, $a,b$ its eigenvectors and $lambda_a, lambda_b$ their corresponding eigenvalues. Define $a^prime = Aa$ and $b^prime = Ab$. Is it true that
    beginequation
    lambda_a = fracVert a^primeVertVert aVert iff Vert aVert = frac1sqrtlambda_a
    endequation
    and
    beginequation
    lambda_b = fracVert b^primeVertVert bVert iff Vert bVert = frac1sqrtlambda_b ?
    endequation










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      Let $A$ be a matrix, $a,b$ its eigenvectors and $lambda_a, lambda_b$ their corresponding eigenvalues. Define $a^prime = Aa$ and $b^prime = Ab$. Is it true that
      beginequation
      lambda_a = fracVert a^primeVertVert aVert iff Vert aVert = frac1sqrtlambda_a
      endequation
      and
      beginequation
      lambda_b = fracVert b^primeVertVert bVert iff Vert bVert = frac1sqrtlambda_b ?
      endequation










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      Let $A$ be a matrix, $a,b$ its eigenvectors and $lambda_a, lambda_b$ their corresponding eigenvalues. Define $a^prime = Aa$ and $b^prime = Ab$. Is it true that
      beginequation
      lambda_a = fracVert a^primeVertVert aVert iff Vert aVert = frac1sqrtlambda_a
      endequation
      and
      beginequation
      lambda_b = fracVert b^primeVertVert bVert iff Vert bVert = frac1sqrtlambda_b ?
      endequation







      linear-algebra eigenvalues-eigenvectors linear-transformations






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      asked Sep 7 at 11:10









      TheSodesa

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          No. First of all, note that the statements don't even make sense if $lambda_a=0$ or $lambda_b=0$. On the other hand, $a'=lambda_aa$ and therefore$$fraca'=lvertlambda_arvert.$$So, $lambda_a=fraca'ifflambda_ageqslant0$. And, of course, $lambda_ageqslant0$ is not equivalent to $|a|=frac1sqrtlambda_a$.






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            Hint: $$frac=frac=frac}=frac=frac1sqrtlambda_a$.






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            Hint: $$frac=frac=frac=fracimprove this answer





















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              answered Sep 7 at 11:14









              zzuussee

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                  Hint : $$a' = A a = lambda_a a$$






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                    Hint : $$a' = A a = lambda_a a$$






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                      up vote
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                      Hint : $$a' = A a = lambda_a a$$






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                      Hint : $$a' = A a = lambda_a a$$







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                      answered Sep 7 at 11:14









                      P. Quinton

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