As you mentioned, the geometric multiplicity of an eigenvalue $\lambda$ is the dimension of its eigenspace. The algebraic multiplicity is the dimension of what's called the generalized eigenspace.
If you have a linear transformation, represented by a matrix A, and an eigenvalue $\lambda$, a generalized eigenvector is a vector, $v$ such that for some integer n, $$(A-\lambda I)^nv =0$$
Then, the space formed by taking all such generalized eigenvectors is called the generalized eigenspace and its dimension is the algebraic multiplicity of $\lambda$.
There's a nice discussion of the intuition behind generalized eigenvectors here.
In your first assertion you wrote $\Rightarrow$, which is correct, but in fact the converse $\Leftarrow$ is also (almost) true. There are several equivalent criterion for diagonalizability. I'll list them out for you.
Let $V$ be a finite-dimensional vector space over a field $F$, and let $T:V \to V$ be a linear map. Then the following statements are all equivalent.
- $T$ is diagonalizable.
- There is an ordered basis $\beta$ of $V$ consisting of eigenvectors of $T$.
- We can express the vector space $V$ as a direct sum of eigenspaces of $T$:
\begin{align}
V = \bigoplus \limits_{\lambda \in \sigma(T)} \ker(T-\lambda I)
\end{align}
- The characteristic polynomial of $T$ splits over $F$, and for every eigenvalue $\lambda$ of $T$,
\begin{align}
\dim \ker(T - \lambda I) = \text{algebraic multiplicity of $\lambda$}
\end{align}
(i.e geometric multiplicity = algebraic multiplicity)
There are a few more equivalent statements if you know about Jordan canonical forms and minimal polynomials, however, for now, you should try to prove that the $2^{\text{nd}}$ and $3^{\text{rd}}$ statements are equivalent.
Edit:
After looking at ThorWittich's answer, I modified my answer to include the assumption that the characteristic polynomial splits (I implicitly assumed this fact throughout the discussion as is usually done, but strictly speaking I should have explicitly stated this).
Edit 2:
The assumption that the characteristic polynomial splits over the given field is important. To show this, consider the matrix
\begin{align}
A =
\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix} \in M_{2 \times 2}(\Bbb{R})
\end{align}
So we are working over the field $\Bbb{R}$. It is easy to see that the characteristic polynomial is $\chi_A(t) = t^2 + 1$, which doesn't split over $\Bbb{R}$. Hence, $A$ is NOT diagonalizable over $\Bbb{R}$. However, if we consider $A$ to be an element of $M_{2 \times 2}(\Bbb{C})$, then the characteristic polynomial splits over $\Bbb{C}$, and there are two eigenvalues of $A$, namely $i$ and $-i$, and it's easy to verify (either directly or by using the 4th condition above) that $A$ is diagonalizable over $\Bbb{C}$.
Hence, the statement that the characteristic polynomial splits over the given field is actually important.
Best Answer
Since $f$ is an isometry, $\lambda=\pm1$ for each eigenvalue $\lambda$. In fact, if $v$ is an eigenvector with eigenvalue $\lambda$,$$|\lambda|\|v\|=\|\lambda v\|=\|f(v)\|=\|v\|.$$
Suppose that the geometric multiplicity of $\lambda$ is smaller than the algebraic multiplicity. Then there is an eigenvector $v$ of $f$ with eigenvalue $\lambda$ (and we may assume without loss of generality that $\|v\|=1$) and there is a vector $w$ such that $f(w)=\lambda v+w$ (think about the Jordan normal form). Now, let $w'=w-\langle w,v\rangle v$. Then\begin{align}f\left(w'\right)&=f\bigl(w-\langle w,v\rangle v\bigr)\\&=f(w)-\langle w,v\rangle f(v)\\&=\lambda w+v-\lambda\langle w,v\rangle v\\&=\lambda w'+v.\end{align}But $\left\langle v,w'\right\rangle=0$. Now, note that\begin{align}\left\|f\left(w'\right)\right\|&=\left\|\lambda w'+v\right\|\\&=\sqrt{\lambda^2\|w'\|^2+\|v\|^2}\\&=\sqrt{\|w'\|^2+1}\text{ (since $|\lambda|=\|v\|=1$)}\\&>\|w'\|,\end{align}which is impossible, since $f$ is an isometry.