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Suppose we have a matrix V given by

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The columns of V are orthogonal to each other. We have proved that $V^TV = I$. We also have proved that the columns of V form a basis for RN, and that $<v_i, b>$ = $a$ 1 if $b = a1v1+ ... +anVn$.

Now we are given Q, and its eigenvalues such that

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We've proved that the eigenvectors of Q are given by the columns of V.

Now we have to show that if $Q_2 = Q - λ_1 v_1 v_1^T$, $v_1$ is in Nul($Q_2$) and $v_2 ... v_n$ are eigenvectors

My approach so far was substituting $Q_2=Q - λ_1 v_1 v_1^T$ in $Q_2 x =_? 0$ But when I distribute $v_1$ through inner products, I'm not sure if I can say that $<v_1, v_i v_i^T>$ are 0. I mean, I know that $<v_1, v_i>$ because they're orthonormal, but can I make the other claim? Do I have the right approach to this whole proof?

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You are interested in the null space of $Q_2$ and the eigenvectors with non-zero eigenvalues. You are correct that you may obtain the null space by solving the system of equations $Q_2 x = 0$; however, this may be challenging in general. You actually already have a guess for the nullspace, so you should use this guess!

Here's a hint to get you started: to show that $v_1$ is in the null space of $Q_2$, try multiplying $Q_2 v_1$ and see what you get. Next, to show that $v_2, \dots v_n$ are eigenvectors of $Q_2$, try multiplying $Q_2 v_i$ and see what you get for $i=2, 3, \dots n$.

Bonus: If indeed $v_i$ is an eigenvector for $Q_2$ with $i = 2,3, \dots n$, then what is its eigenvalue?

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  • $\begingroup$ thank you! I was overcomplicating this :(( $\endgroup$ – Melanie Dec 6 '19 at 2:11
  • $\begingroup$ Q = VAV^T though? meaning, can we assume V,Vt also has its v1 removed and the first row is removed from A? $\endgroup$ – Melanie Dec 6 '19 at 2:22
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Note that $$Q_2v_1 = (Q- \lambda_1 v_1 v_1^T)v_1 = Qv_1 - (\lambda_1 v_1v_1^T)v_1 = \lambda_1v_1 - \lambda_1v_1 (v_1^T v_1) = \lambda_1v_1 - \lambda_1 v_1 \cdot 1 = 0$$ where the second last equality follows from the fact that $v_1^T \cdot v_1 = 1$ (as the vectors are orthonormal, this follows from $V^TV = I$)

In the same way, you can prove that $Q_2v_i = \lambda_iv_i$ for each $i \in \{2, \ldots, n\}$ since $v_1^Tv_i = 0$ (orthonormal vectors).

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