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Let $A_1$ be a $n \times n$ real symmetric matrix whose eigenvalues are $\lambda_1$, $\lambda_2, \lambda_3, \ldots , \lambda_n$.

Let's assume that if $i < j$, then $|\lambda_i| > |\lambda_j|$ (so, $\lambda_1$ is the dominant eigenvalue...)

Also, for any of these eigenvalues $\lambda_i$, define $v_i$ as being the unit associated eigenvector (So, $A_1v_i = \lambda_i v_i$ and $\|v_i\| = 1$)

Is it true that the matrix

$$A_2 = A_1 - A_1v_1v_1^T$$

has $\lambda_2$ as its dominant eigenvalue?

(So, we somehow eliminate the original dominant eigenvalue $\lambda_1$ with this subtraction).

And, in a more general form, is it true that

$$A_k = A_1 - \sum_{i=1}^{k-1}A_1v_iv_i^T$$

has $\lambda_k$ as its dominant eigenvalue?

I read this proposition on a paper about NIPALS and also on this Wikipedia's section, but both places don't present any proof.

I made a computer program to test it, and it seems to be true. Also, I discovered that when I use eigenvectors whose norm is not equal to 1, the test program fails, so, this hypothesis is needed.

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As a consequence of the spectral theorem, any symmetric matrix of rank $r$ can be expressed in the form $$ A = \sum_{i=1}^r \lambda_i\,v_iv_i^T $$ where each $\lambda_i$ is a non-zero eigenvalue, and the vectors $v_i$ are chosen in any way such that each $v_i$ is an eigenvector of length $1$, and any pair $v_i,v_j$ are mutually orthogonal.

Starting from here, I think you'll find it easy to show that your statement is true.

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  • $\begingroup$ The eigenvectors of a symmetric matrix are linear independent, right? But they are also orthogonal? $\endgroup$ – Hilder Vítor Lima Pereira Aug 13 '15 at 17:49
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    $\begingroup$ We can always choose the eigenvectors to be orthogonal for a symmetric matrix. If all of the eigenvalues are distinct, then no matter how you choose the eigenvectors, they will end up orthogonal. $\endgroup$ – Omnomnomnom Aug 13 '15 at 17:50

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