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Suppose I have a square symmetric matrix $A$, whose largest eigenvalue is $\lambda_1$ and the corresponding eigenvector is $v_1$. Also, suppose the second largest eigenvalue is $\lambda_2$ and the corresponding eigenvector is $v_2$, but the second largest eigenvalue is not known (not yet calculated).

Can I transform the matrix $A$ to get another matrix $A_2$ whose largest eigenvalue is now $\lambda_2$?

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Consider $B=A-\lambda_1 \frac{v_1 v_1^T}{\| v_1 \|^2}$. This will have the same eigenvectors but $v_1$ will have eigenvalue $0$.

Note that we need the symmetry to ensure that this does not alter the other eigenvectors or eigenvalues, because the eigenvectors for different eigenvalues are orthogonal and therefore $\left (A-\lambda_1 \frac{v_1 v_1^T}{\| v_1 \|^2} \right ) v_i = Av_i - 0 v_1 = \lambda_i v_i$ if $\lambda_i \neq \lambda_1$.

Note that if $\lambda_1$ has an eigenspace of dimension higher than $1$ then you need to subtract off the corresponding projector onto that larger eigenspace.

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  • $\begingroup$ why do we have $||v_1||^2$ in the denominator? Can we use any denominator? $\endgroup$
    – uday
    Oct 4 '15 at 20:27
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    $\begingroup$ @uday $v_1 v_1^T$ applied to $v_1$ gives $\| v_1 \|^2 v_1$. You want it to give $\lambda_1 v_1$, so you multiply by the appropriate constant. $\endgroup$
    – Ian
    Oct 4 '15 at 20:28

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