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enogjini
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MHB An Overview of the History of Tea
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The discussion introduces a mathematical concept related to eigenvalues and eigenvectors, specifically focusing on the equation $(A-cI)x=(\lambda-c)x$. It emphasizes that $\lambda$ is an eigenvalue of matrix A if a vector v exists such that $Av= \lambda v$. The simplification of the equation demonstrates how the transformation affects the vector v when adjusted by a scalar c. The conversation highlights the importance of showing effort in mathematical discussions, aligning with forum rules. Overall, the thread combines mathematical theory with community guidelines for effective participation.
- #2
Evgeny.Makarov
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Welcome to the forum. Have you tried simplifying $(A-cI)x=(\lambda-c)x$?
Please the forum rules https://mathhelpboards.com/help/forum_rules/, especially the "Show some effort" rule.
Please the forum rules https://mathhelpboards.com/help/forum_rules/, especially the "Show some effort" rule.
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Kansas Boy
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First. $\lambda$ is an eigenvalue of A if and only if there exist a vector, v, such that $Av= \lambda v$. Of course, for any vector v, Iv= v so for any number c, cIv= cv. Then $(A- cI)v= Av- cIv= \lambda v- cv= (\lambda- c)v$
Hello!
In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way:
I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true?
And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...
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