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LSpice
LSpice
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As explained in Answer 3Answer 3, the shift can be explained by the inverse power iteration. $x'=(A-rI)^{-1}x$ has larger component in $v_1$ if $r$ is closer to $\lambda_1$.

However in QR iteration, we don't use the inverse iteration, but the power iteration.

In the power iteration, $x'=(A-rI)x$ has smaller component in $v_1$ if $r$ is closer to $\lambda_1$. It seems the conclusion is opposite. I don't know what my misunderstanding is. I appreciated if anyone can help. Thanks.

As explained in Answer 3, the shift can be explained by the inverse power iteration. $x'=(A-rI)^{-1}x$ has larger component in $v_1$ if $r$ is closer to $\lambda_1$.

However in QR iteration, we don't use the inverse iteration, but the power iteration.

In the power iteration, $x'=(A-rI)x$ has smaller component in $v_1$ if $r$ is closer to $\lambda_1$. It seems the conclusion is opposite. I don't know what my misunderstanding is. I appreciated if anyone can help. Thanks.

As explained in Answer 3, the shift can be explained by the inverse power iteration. $x'=(A-rI)^{-1}x$ has larger component in $v_1$ if $r$ is closer to $\lambda_1$.

However in QR iteration, we don't use the inverse iteration, but the power iteration.

In the power iteration, $x'=(A-rI)x$ has smaller component in $v_1$ if $r$ is closer to $\lambda_1$. It seems the conclusion is opposite. I don't know what my misunderstanding is. I appreciated if anyone can help. Thanks.

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Tongru Huo
Tongru Huo
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As explained in Answer 3, the shift can be explained by the inverse power iteration. $x'=(A-rI)^{-1}x$ has larger component in $v_1$ if $r$ is closer to $\lambda_1$.

However in QR iteration, we don't use the inverse iteration, but the power iteration.

In the power iteration, $x'=(A-rI)x$ has smaller component in $v_1$ if $r$ is closer to $\lambda_1$. It seems the conclusion is opposite. I don't know what my misunderstanding is. I appreciated if anyone can help. Thanks.