The matrix of a linear map

Can taking a derivative be a matrix? Watch a table of numbers turn a cubic curve into its slope curve.

Linear maps beyond arrows

On A gallery of linear maps a map was linear when it kept adding and scaling. Nothing in that test needs arrows. Take “differentiate”: the derivative of a sum is the sum of the derivatives, and the derivative of $c$ times a function is $c$ times its derivative. So differentiation is a linear map on functions. I’ll write a map with braces, $T\{f\}$, as the gallery page does and as signal processing writes a system. The test is

Other linear maps: integrating, turning, projecting, delaying a signal by one sample. Not linear: adding 1, since $T\{0\}$ would be 1, not 0.

Columns are images

Polynomials of degree at most 3 are a vector space with the basis $1,\ x,\ x^2,\ x^3$, as on Vector spaces and subspaces. Each polynomial’s coordinates in that basis are its four coefficients, constant first: $2-3x+x^2+4x^3$ has the coordinates $(2,-3,1,4)$. To find the matrix of differentiation, do what A matrix moves the plane did with the unit arrows: see where each basis vector goes, and write its coordinates as a column.

• $1\mapsto0$: column $(0,0,0,0)$.
• $x\mapsto1$: column $(1,0,0,0)$.
• $x^2\mapsto2x$: column $(0,2,0,0)$.
• $x^3\mapsto3x^2$: column $(0,0,3,0)$.

For $p(x)=2-3x+x^2+4x^3$: $\mathbf{D}(2,-3,1,4)=(-3,\;2,\;12,\;0)$, which is $p'(x)=-3+2x+12x^2$. The table of numbers does calculus.

Drag the coefficients in the picture and watch the output. Its last entry is always 0, since differentiating lowers the degree, and the constant term never reaches the output at all.

Kernel and range

The null space of a map is also called its kernel, and its column space its range. For $\mathbf{D}$:

• the kernel is the constants, $N(\mathbf{D})=\mathrm{span}\{(1,0,0,0)\}$, since only constants have derivative 0: dimension 1;
• the range is the polynomials of degree at most 2: dimension 3, the rank.

And $3+1=4$, the number of inputs, as on The four fundamental subspaces: rank plus the dimension of the kernel equals the number of columns.

Composing maps multiplies matrices. Differentiating twice is $\mathbf{D}^2$:

and $\mathbf{D}^2(2,-3,1,4)=(2,24,0,0)$, which is $p''(x)=2+24x$. Four derivatives kill every cubic: $\mathbf{D}^4$ is the zero matrix.

Undoing, up to a constant. Integration from degree 2 to degree 3, starting at 0, sends $1\mapsto x$, $x\mapsto x^2/2$, $x^2\mapsto x^3/3$. Its matrix is $4\times3$:

Integrate then differentiate and you are back where you started: $\mathbf{D}\mathbf{A}_\text{int}$, restricted to degree 2, is the $3\times3$ identity. The other order loses the constant: you get $p(x)-p(0)$.

One map, many matrices

A map’s matrix depends on the basis. Use the basis $1,\ x,\ x^2/2,\ x^3/6$ instead, with $\mathbf{V}=\mathrm{diag}(1,1,\tfrac12,\tfrac16)$. By Change of basis, the same map becomes

every entry above the diagonal is 1. Each basis polynomial is sent to the one before it, and the matrix is the Jordan block for the eigenvalue 0, from Similar matrices and the Jordan form. In this basis $p$ has coordinates $(2,-3,2,24)$, and the output $(-3,2,24,0)$ means $-3+2x+24\cdot\tfrac{x^2}2$: the same $p'(x)$.

The picture below makes the same point with arrows.

The projection is one map. On the usual grid its table is full of halves; on a grid along and across the line it is $\mathrm{diag}(1,0)$, keep one coordinate, drop the other. A matrix is a map written in one basis, not the map itself.

Worked example

1. Differentiate. $\mathbf{D}(2,-3,1,4)=(-3,\;2\cdot1,\;3\cdot4,\;0)=(-3,2,12,0)$.

2. Twice. $\mathbf{D}^2(2,-3,1,4)=(2,24,0,0)$, and $\mathbf{D}^2$ applied to $x^3$, coordinates $(0,0,0,1)$, gives $(0,6,0,0)$, that is $6x$.

3. Kernel and range. Kernel: constants, dimension 1. Range: degree at most 2, dimension 3. Total 4.

4. A scaled basis. $\mathbf{V}^{-1}\mathbf{D}\mathbf{V}$ has ones above the diagonal: entry $(1,2)$ is 1, entry $(2,3)$ is $2\cdot\tfrac12=1$, entry $(3,4)$ is $3\cdot\tfrac16\cdot2=1$.

5. A projection in two bases. $\begin{bmatrix}0.5&0.5\\0.5&0.5\end{bmatrix}$ in the usual basis, $\mathrm{diag}(1,0)$ in the basis $(1,1)$, $(1,-1)$. This is the projection matrix of Projections.

Try these before you read the answers.

• Is $f\mapsto f+1$ linear? No: it sends 0 to 1.
• The matrix of “double every polynomial”, in any basis? $2\mathbf{I}$.
• Which matrix undoes $\mathbf{D}$ on polynomials of degree at most 2, up to a constant? The integration matrix.

Where you’ll meet this

Numerical solvers. Programs that solve differential equations with polynomials or smooth series store differentiation as a matrix, then solve the equation as a linear system.

Filters. A filter acting on a block of samples is a matrix, as on A matrix moves the plane, and its different structures are that map in different bases.

Reference card