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(I will phrase the question in terms of $\mathbb{R}^n$)

Is the following statement a standard well-known linear algebra fact that I can quote without proving? (Perhaps more importantly, is it even true?)

Notation that we will use:

$a_M(\lambda) =$ algebraic multiplicity of $\lambda$ for matrix/ linear map $M$.

$g_M(\lambda) =$ same for geometric multiplicity.

Statement:

Let $A \in \mathbb{R}^{n \times n}$. Then

$\mathbb{R}^n = \textrm{ker}A \oplus \textrm{range}A \Leftrightarrow a_A(0) = g_A(0).$

Thanks,

Julian.

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I think you should precise your question. Do you want to know if as abstract vector spaces, $\mathbb{R}^n \simeq \textrm{ker}A \oplus \textrm{range}A$, or if $\mathbb{R}^n$ is the direct sum of its subspaces $\ker A$ and $\mathrm{range} A$? Note that in the second case, there is a canonical morphism $\ker A \oplus\mathrm{range} A \rightarrow \mathbb{R}^n$ and we can wonder wether it is an isomorphism or not. Dylan Moreland answers the first question. The second one is equivalent to $\ker A = \ker A^2$. –  Plop Jun 19 '11 at 12:01
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1 Answer 1

up vote 0 down vote accepted

The statement is indeed correct and essentially follows right from the definition.

The rank-nullity theorem tells us that ${\rm dim Ker} A + {\rm dim Im} A = {\rm dim} V$ so we only need to prove that ${\rm Ker} A \cap {\rm Im} A = 0$. But this is precisely the condition for the geometric multiplicity being equal to algebraic multiplicity.

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