Why is kernel important? Why are we interested in looking at the kernel and range (image) of a linear transformation on a linear algebra course?
 A: One example to keep in mind is a projection $\pi: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ onto a plane $\Pi$ in $\mathbb{R}^3$.  This type of linear transformation comes up all the time in computer graphics, where you are trying to represent something three-dimensional on a two-dimensional screen.
The image of this linear transformation will be the plane $\Pi$ that you are projecting onto.  The kernel of this linear transformation can be used to answer the following question: when is the projection of two points the same?  (This corersponds to one object 'blocking' the view of the other on the screen.)  The answer is that $\vec{v}$ and $\vec{w}$ are projected to the same point in the plane if and only if $\vec{v} - \vec{w}$ lies in kernel of $\pi$.  In this sense, the kernel allows you to detect when two points are being projected onto the same point.
Linear transformations are more general than just this projection example, but the idea is still the same.  The image of a linear transformation $L : V \rightarrow W$ between two vector spaces just tells you the subspace of vectors in $W$ that are images of vectors in $V$ under the transformation $L$.  In other words, the image tells you where vectors get sent to.  The kernel of a linear transformation helps you detect when two vectors in $V$ are transformed to the same vector in $W$ under $L$.  That is, $L(\vec{v}) = L(\vec{v'})$ if and only if $\vec{v} - \vec{v'} \in \text{ker}(L)$.
A: The Kernel is a measure of injectivity: A linear transformation is injective if and only if the kernel is trivial.
Intuitively, since the kernel consists of the elements sent to $0$, its dimension tells you how much it shrinks the source space into the target space. If it shrinks at all, we squeeze points on top of each other and lose injectivity. Suppose we have a linear transformation $T:V \to V$ where $\dim V=n.$ If the kernel has basis $b_1, b_2, \cdots b_m$ and we extend this to a basis of our original vector space $b_1, b_2, \cdots, b_m, b_{m+1}, \cdots b_n$, then we know that $T(b_{m+1}), \cdots , T(b_n)$ forms a basis of $T(V)\subseteq V.$ Thus having $m$ elements in the kernel caused the image to shrink by $m$ dimensions from the source.
A: Let's say, you want to continue a computer program started by a friend who is not there to help, that uses, for example reflection through a plane $S$ in 3d.
And let's say you find a concrete $3\times 3$ matrix in the code, named 'Refl_S'.
How can you get back $S$?
Let's call now the matrix $M$, and assume it is a reflection through some plane $S$. Then, for each $v$, the midpoint $\displaystyle\frac{v+Mv}2$ is in $S$. So,
$$S=ran \left(\frac{I+M}2\right)$$
and, at the same time
$$S^\perp =ker\left(\frac{I+M}2\right)$$
A: Your question made me think the following theorem which I think is one of the most important theorems in linear algebra:
(Rank-Nullity-Theorem): Let $V,W$ be vector spaces with $\mathrm{dim}V < \infty$ and $T: V \to W$ a linear map. Then we have $$ \mathrm{dim}(\mathrm{im}T) + \mathrm{dim}(\mathrm{ker}T) = \mathrm{dim} V$$
Hope this helps. 
