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For any $v \in \Bbb{R}^n$,
$$ A = vv^T+\lambda I$$ show A is symmetric and A has eigenvalue $\lambda$ and $||v||^2 + \lambda$

I was also given the hint to consider eigenvectors $v$ and $w$ where $w \in span(v)^\bot$

Not sure where to start, any help is appreciated

JMoney
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  • Hint: 1/ What is the definition of symmetric matrix? 2/ What do you get if you multiply $A$ with $v$? Then what happens if you multiply $A$ with $w$ where $w$ is some vector that is perpendicular to $v$? – mathmathmath Apr 20 '21 at 22:50

1 Answers1

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To show that $A$ is symmetric, take the transpose $A^T$. See the definition of a symmetric matrix on wiki. For the next part, solve $Av$ and $Aw$ (using distributive property) and see what you end up with. Remember that the formal definition of an eigenvector is $Av = \lambda v$, where $\lambda$ is some scalar.

Oliver Levy
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  • Thanks. For the first part, I got to $A^T = v^Tv+\lambda I$, where do I head next? what is the relation between $v^Tv$ and $vv^T$ – JMoney Apr 20 '21 at 23:01
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    Should look at your transpose again, see https://en.wikipedia.org/wiki/Transpose#Properties – Oliver Levy Apr 20 '21 at 23:04
  • Ooops I forgot to switch it when applying the transpose. Last question, when distributing $w$ in $w(vv^T)$ how would the property of perpendicular vectors work this out? I know $wv = 1$ but what would happen to $v^T$? – JMoney Apr 20 '21 at 23:17
  • Dot product is denoted $<v,w>$ or $v^Tw$ (a row vector and a column vector). See https://mathinsight.org/dot_product_matrix_notation for further explanation.

    Because of that definition above, its important to solve $Av$ and not $vA$. The first gives you $Av = (vv^T + \lambda I)v = vv^Tv + \lambda I v$. Can you take it from here?

     – Oliver Levy Apr 20 '21 at 23:26
  • Thank you, but I was referring to when you do $Aw$ – JMoney Apr 21 '21 at 00:33
  • Substitute $w$ in above: $Aw= vv^tw + \lambda I w$. By our notation, $v^tw$, is the dot product between $v$ and $w$. what does that tell you? – Oliver Levy Apr 22 '21 at 21:07