Examples from Rank-nullity

Kernel and image bases of a matrix

Consider

The third row is the sum of the first two. Replacing $R_3$ by $R_3-R_1-R_2$ therefore gives a zero row, leaving pivots in columns $1$ and $2$. To find the kernel, let $x_3=\lambda_3$ and $x_4=\lambda_4$. The first two equations give

Therefore a basis for the kernel is

The pivot positions are in columns $1$ and $2$, so the corresponding columns of the original matrix give a basis for the image. (Their leading coordinates also make their linear independence immediate.)

Hence $\operatorname{nullity}(A)=2$, $\operatorname{rank}(A)=2$ and

the number of columns.

Kernel basis of a row-echelon matrix

Consider the row-echelon matrix

The leading columns are $1,4,5,7$, so the free variables are $x_2,x_3,x_6,x_8$. Set

Solving $Ax=0$ from the bottom non-zero row upwards gives

The fourth row gives $x_7=-8d$ and substituting this into the third row gives $x_5=-9c+17d$. The second row then gives $x_4$ in the same way.

Finally, the first row gives

Setting each free parameter in turn to $1$ and the others to $0$ gives one kernel basis vector per free variable. Thus a basis for the kernel is

Because the displayed matrix is already row-echelon, no row operation has changed its columns. Its pivot columns therefore give a basis for the image:

There are four pivot and four free columns, so

General solution of an inhomogeneous system

For

take $x_3=\lambda_3$ and $x_4=\lambda_4$. Solving the first two equations gives

Therefore

The two direction vectors are the kernel basis found in the first example and the constant vector is a particular solution: $Ax_p=b$.

Differentiation on polynomial spaces

Over $\mathbb R$ and for $n\ge1$, consider

For $p_1,p_2\in P_n$ and $\lambda\in\mathbb R$,

so $D$ is a linear map. If

then

Its kernel consists of the constant polynomials:

Thus $\operatorname{nullity}(D)=1$. Since $\dim(P_n)=n+1$, rank-nullity gives

The image is $P_{n-1}$: every polynomial of degree at most $n-1$ has a polynomial antiderivative in $P_n$. Hence its dimension is also $n$, agreeing with the rank calculation.