Let $\lambda $ be an eigenvalue of A, such that no eigenvector of A associated with $\lambda $ has a zero entry. Then prove that every list of n−1 columns of $A-\lambda I$ is linearly independent.

Ivan Buckley
2022-09-25
Answered

Let $\lambda $ be an eigenvalue of A, such that no eigenvector of A associated with $\lambda $ has a zero entry. Then prove that every list of n−1 columns of $A-\lambda I$ is linearly independent.

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booyo

Answered 2022-09-26
Author has **6** answers

The null space of ${A}_{n\times n}-\lambda I$ is the eigenspace of $\lambda $. Since no eigenvector corresponding to $\lambda $ contains a 0 entry, the basis of the eigenspace must contain a single eigenvector of A. This means the nullity of $A-\lambda I$ is 1, and by the rank-nullity theorem, its rank n−1.

Now recall that $(A-\lambda I)v$ is a linear combination of the columns of $A-\lambda I$. This means their exists a tuple $({v}_{1},{v}_{2},...,{v}_{n})$ with ${v}_{i}\ne 0$ that satisfies ${v}_{1}{C}_{1}+{v}_{2}{C}_{2}+\cdot \cdot \cdot +{v}_{n}{C}_{n}=0$. Since the rank of these vectors is n−1, there is one linearly dependent column ${C}_{j}$, and since ${v}_{j}\ne 0$ for all j, that ${C}_{j}$ could be any column

Now recall that $(A-\lambda I)v$ is a linear combination of the columns of $A-\lambda I$. This means their exists a tuple $({v}_{1},{v}_{2},...,{v}_{n})$ with ${v}_{i}\ne 0$ that satisfies ${v}_{1}{C}_{1}+{v}_{2}{C}_{2}+\cdot \cdot \cdot +{v}_{n}{C}_{n}=0$. Since the rank of these vectors is n−1, there is one linearly dependent column ${C}_{j}$, and since ${v}_{j}\ne 0$ for all j, that ${C}_{j}$ could be any column

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a)

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, where $\psi ,\phi $ are scalar functions of the coordinates. Then i,ii designate differentiation of first and second order, and $\Sigma ,\Omega $ are the surface and volyme respectively?

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I would like to integrate

$\int {d}^{3}\overrightarrow{k}\int {d}^{3}\overrightarrow{q}\frac{1}{{q}^{2}}$

under the condition that

$|\overrightarrow{k}|\le A\text{}\text{}\text{}\text{and}\text{}\text{}\text{}|\overrightarrow{k}+\overrightarrow{q}|\le A$

where A is some constant.

Is this executable?

$\int {d}^{3}\overrightarrow{k}\int {d}^{3}\overrightarrow{q}\frac{1}{{q}^{2}}$

under the condition that

$|\overrightarrow{k}|\le A\text{}\text{}\text{}\text{and}\text{}\text{}\text{}|\overrightarrow{k}+\overrightarrow{q}|\le A$

where A is some constant.

Is this executable?