Can someone help me understand why you would ever want to calculate the kernel of a matrix?

I kept on trying to find applications when $\ker(A) = 0$ is useful but I could not find any.

Can someone please show an example where it is important to calculate $\ker(A)$ and we could obtain some information from calculating the kernel?

  • $\begingroup$ Please refer to math.stackexchange.com/questions/21131/… $\endgroup$ – Anurag A Mar 4 '15 at 8:12
  • $\begingroup$ When the kernel of a matrix is $0$, then it is injective (and so bijective). Would you say this is useful? $\endgroup$ – Moritz Mar 4 '15 at 8:18
  • 2
    $\begingroup$ Finding the kernel of a matrix is equivalent to solving a homogeneous system of linear equations. Surely you have a linear algebra book with applications of solving systems of equations. $\endgroup$ – Jim Mar 4 '15 at 8:25

Often one uses matrices to solve systems of simultaneous linear equations, by forming the matrix of coefficients. If such a matrix has $0$ kernel (a trivial null space), it means a solution to the system is unique, if it exists.

If the kernel (null space) of the matrix is "larger", it provides us with a measure of "how non-unique" our solution is, which is to say how may "free variables" (or parameters) we need to specify to obtain a unique solution.

In many applications, the solution of a system of linear equations is given in the form:

specific solution + homogeneous solution(s).

The kernel of the coefficient matrix tells us about the "homogeneous solutions" part. This is a rough measure of "how much of the domain vector space is shrunk to the zero vector", that is how much "collapsing" or condensation of information takes place. This is often good to know. An extreme example is the $0$-matrix, which annihilates everything, obliterating any useful information the system of equations might tell us (it's "all kernel" and "no range").

A simple example, a system of 3 linear equations:

$x + y = 2$

$x + 2z = 3$

$2x + y + 2z = 5$

Here, the coefficient matrix is:


The kernel of this matrix is $\{t(-2,2,1): t \in \Bbb R\}$ (if we are working in the real numbers). This is a subspace of $\Bbb R^3$ of dimension $1$ (a line through the origin).

It is easy to verify that $(x,y,z) = (1,1,1)$ is a (particular) solution of this system. Thus the "general solution" is of the form:

$(1,1,1) + t(-2,2,1), t \in \Bbb R$.

One needs to determine the kernel to find all the solutions.


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