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The following is from an old exam.

Let $V$ be a vector space, and $T:V\rightarrow V$ a linear transformation that is not diagonalizible.

Show that if $1$ and $2$ are eigenvalues of $T$, then $dimV>2$.

I don't understand the given answer. There is a theorem in our book which states that:

If $\lambda_1,\dots ,\lambda_k$ are distinct eigenvalues of a linear transformation $T$, and if $v_1,\dots ,v_k$ are their corresponding eigenvectors, then $v_1,\dots ,v_k$ are linearly independent.

The answer then refers to the above theorem and concludes that there are at least 2 linearly independent eigenvectors and therefore $dimV\geq 2$.

From where do they conclude that there are 2 linearly independent eigenvectors? As far as I know there is a theorem which states that the geometric multiplicity is always less than or equal to the algebraic multiplicity, but I'm not aware of anything which states that the geometric multiplicity is always at least 1 ( since that's not true ).

The only thing I can think of is that the existence of two distinct non-zero eigenvalues might imply $dimN(A-\lambda I)\geq1$ for each $\lambda_i$.

But of course the point is they don't say anything about that subject and I just don't get how they conclude that there are at least two linearly independent eigenvectors.

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The geometric multiplicity is always at least 1. If the dimension of the eigenspace were zero, then there wouldn't be any eigenvectors! –  Christopher A. Wong Nov 6 '12 at 20:38
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3 Answers

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You are told there are two distinct eigenvalues $1$ and $2$. Every eigenvalue must have a corresponding eigenvector, it's part of the definition of eigenvalue. The theorem says an eigenvector for $1$ and an eigenvector for $2$ are linearly independent, hence dimension is at least $2$.

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Eck, sometimes the most obvious things are the things you miss.... –  Robert S. Barnes Nov 6 '12 at 21:13
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But the geometric multiplicity is at least one. That's what it means to be an eigenvalue! When you find the eigenvalues, you're finding the values such that $A-\lambda I$ is singular, that means by definition each eigenvalue has at least a one dimensional eigenspace.

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Let's apply the theorem to our case. Let's say that $v_1$ is the eigenvector with eigenvalue $1$ and that $v_2$ is the eigenvector with eigenvalue $2$. Since $1 \neq 2$ the theorem tells us that $v_1$ and $v_2$ are linearly independent, i.e. $v_1$ and $v_2$ form a basis for a two dimensional subspace of $V$. If $V$ contains a two dimensional linear subspace then it must be at least two dimensional itself. Hence: $\dim V \ge 2.$

Consider the case $\dim V = 2$. If $\dim V = 2$ then $v_1$ and $v_2$ will form a basis for $V$. Using $v_1$ and $v_2$ as a basis for $V$, the transformation $T$ is such that $T(v_1) = 1 \cdot v_1 + 0 \cdot v_2$ and $T(v_2) = 0 \cdot v_1 + 2 \cdot v_2$, and so the matrix representation of $T$ with respect to the basis $\{v_1,v_2\}$ is a diagonal matrix with $1$ and $2$ along the leading diagonal. But hang on! You said that $T$ was not diagonalisable. We have a contradiction. Thus $\dim V \neq 2$ and so $\dim V > 2.$

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