Let's say for the eigenvector equation, $ (\lambda I - A)X = 0 $, some eigenvalue of $A$, $ \lambda_1 $, is found and $ \lambda_1 I - A $ is reduced to solve for its respective eigenvector $ X_1 $. If it is reduced to the identity matrix, $ I $, what can you say about the eigenvector $ X_1 $? Does the eigenvector $X_1$ exist? Is $A$ diagonalizable?
Eigenvector of matrix that reduces to identity
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0What does "some $\lambda_1$ is found" mean? Did someone randomly choose a $\lambda_1 \in \mathbb{R}$? – 2012-10-29
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0I meant some eigenvalue of $ A $, $ \lambda_1 $ is found. – 2012-10-29
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0Then this question is self-contradictory. – 2012-10-29
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3This question is quite confusing. If $\lambda_1$ is indeed an eigenvalue then $(\lambda_1I-A)$ cannot possibly be row equivalent to the identity. The definition of an eigenvalue is so that $(\lambda I-A)$ is singular. – 2012-10-29
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0Sorry bear with me I'm a beginner at linear algebra. I didn't know that $\lambda_1 I - A$ cannot be row equivalent to $I$. Why is that? – 2012-10-29
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0@hesson: By definition, $\lambda_1$ being an eigenvalue means that there exists a **nonzero** $v$ such that $Av=\lambda_1 v$. So $(\lambda_1 I-A)v=0$, and $\lambda_1 I - A$ is singular. But if $\lambda_1 I - A$ can be row-reduced to $I$, then it cannot be singular! – 2012-10-29
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0If $\lambda_1I-A$ is row equivalent to the identity, then the only vector $\mathbf{x}$ such that $$(\lambda_1I - A)\mathbf{x} = \mathbf{0}$$ would be $\mathbf{x} = \mathbf{0}$. By definition, we need eigenvectors to be non-zero, so that means there exists no eigenvectors corresponding to $\lambda_1$. – 2012-10-29
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0Ah yes, I completely missed the condition that $(\lambda_1 I - A)$ cannot be invertible. Thank you wj32 and EuYu for clearing this up! – 2012-10-29
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3I can't tell you how many times I've marked an exam paper where a student found an eigenvalue $\lambda$ and then row-reduced $A-\lambda I$ to the identity. Of course that means there was a mistake in the algebra somewhere, but unfortunately the students rarely realize that and instead just make up some eigenvector. – 2012-10-29
1 Answers
By definition, $x$ is an eigenvector of $A$ for the value $\lambda_1$ if $Ax = \lambda_1 x$, or by rearranging, $(\lambda_1 I - A)x=0$. Also by definition, $\lambda_1$ is an eigenvalue if and only if it has a non-zero eigenvector.
So if $\lambda_1 I-A$ is row-reducible to the identity matrix, then the equation $(\lambda_1 I - A)x=0$ has only the trivial solution $x=0$. But then $\lambda_1$ has no eigenvectors except 0, so $\lambda_1$ is not actually an eigenvalue at all.
In other words, $\lambda_1$ is an eigenvalue of $A$ if and only if $(\lambda_1 I - A)x$ is not row-reducible to the identity.
One other note, your choice of wording implies you might think that each eigenvalue has exactly one eigenvector, which is definitely not the case. The set of eigenvectors of an eigenvalue forms a subspace, so if there is one eigenvector then there are an infinity of them.