# Eigenvector of matrix that reduces to identity

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

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.