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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?

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What does "some $\lambda_1$ is found" mean? Did someone randomly choose a $\lambda_1 \in \mathbb{R}$? – wj32 Oct 29 '12 at 0:00
I meant some eigenvalue of $ A $, $ \lambda_1 $ is found. – hesson Oct 29 '12 at 0:00
Then this question is self-contradictory. – wj32 Oct 29 '12 at 0:02
This 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. – EuYu Oct 29 '12 at 0:03
I 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. – Gerry Myerson Oct 29 '12 at 0:18
up vote 3 down vote accepted

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.

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