Loading [MathJax]/extensions/TeX/cancel.js
Skip to main content

Discovery guide 43.1 Discovery guide

Attached to every linear transformation \funcdef{T}{V}{W} is a pair of important subspaces, one in the domain space and one in the codomain space.

kernel of T

the collection of all vectors \uvec{v} in the domain space V for which T(\uvec{v}) = \zerovec_W

\ker T

notation for the kernel of T

image of T

the collection of all image vectors T(\uvec{v}) in the codomain space W

\im T

notation for the image of T

Note that a vector \uvec{w} in W is in \im T precisely when there exists at least one \uvec{v} in V with T(\uvec{v}) = \uvec{w}\text{.}

Discovery 43.1. Kernel of a matrix transformation.

Consider matrix transformation \funcdef{T_A}{\R^n}{\R^m} defined, as usual, by T_A(\uvec{x}) = A \uvec{x} for some m \times n matrix A\text{.}

(a)

Suppose A is a 4 \times 5 matrix whose RREF is

\begin{equation*} \left[\begin{array}{rrrrr} 1 \amp 0 \amp 2 \amp 0 \amp -1 \\ 0 \amp 1 \amp 3 \amp 0 \amp 2 \\ 0 \amp 0 \amp 0 \amp 1 \amp -3 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \end{array}\right]\text{.} \end{equation*}

Determine a basis for \ker T_A as a subspace of \R^5\text{.}

(b)

Connect to previous concepts: \ker T_A is the same as the of A\text{.}

Discovery 43.2. Kernel examples.

In each of the following, describe the kernel vectors in words.

Try to use a more meaningful description than just β€œthe vectors that evaluate to zero in the transformation”.

(a)

\funcdef{T}{\matrixring_n(\R)}{\matrixring_n(\R)} by T(A) = A - \utrans{A}\text{.}

(b)

Evaluation of polynomials at fixed x-value x = a\text{:}

\funcdef{E_a}{\poly(\R)}{\R^1} by E_a(p) = p(a) \text{.}

(c)

Differentiation: let F(a,b) represent the space of functions defined on the interval a \lt x \lt b\text{,} and let D(a,b) represent the subspace of F(a,b) consisting of differentiable functions.

Consider \funcdef{\ddx}{D(a,b)}{F(a,b)} by \ddx(f) = f'\text{.}

(d)

Integration: let C[a,b] represent the space of continuous functions defined on the interval a \le x \le b\text{.}

Consider \funcdef{I_{a,b}}{C[a,b]}{\R^1} by I_{a,b}(f) = \integral{a}{b}{f(x)}{x}\text{.}

Discovery 43.3. Determining kernel basis.

For each of the provided transformations of \matrixring_{2}(\R)\text{,} determine a basis for \ker T by carrying out the following steps.

  1. Starting with an arbitrary matrix
    \begin{equation*} X = \begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix} \end{equation*}
    in \matrixring_{2}(\R)\text{,} determine conditions on parameters a,b,c,d so that T(X) = \zerovec \text{.}
  2. Use those conditions to reduce the number of parameters required to describe an arbitrary matrix in \ker T\text{.}
  3. Determine the basis vector associated to each in the reduced collection of parameters.
(a)

\funcdef{T = \trace}{\matrixring_{2}(\R)}{\R^1}\text{.}

(b)

\funcdef{T}{\matrixring_{2}(\R)}{\matrixring_{2}(\R)} by T(A) = A - \utrans{A}\text{.}

(c)

\funcdef{T = L_B}{\matrixring_{2}(\R)}{\matrixring_{2}(\R)} by L_B(A) = B A\text{,} where

\begin{equation*} B = \begin{bmatrix} 1 \amp 1 \\ 0 \amp 0 \end{bmatrix} \text{.} \end{equation*}
Discovery 43.4. Image of a matrix transformation.

Consider matrix transformation \funcdef{T_A}{\R^n}{\R^m} defined, as usual, by T_A(\uvec{x}) = A \uvec{x} for some m \times n matrix A\text{.}

(a)

Apply the definition: An m-dimensional vector \uvec{b} is in \im T_A precisely when .

(b)

Connect to previous concepts: \im T_A is the same as the of A\text{.}

(c)

If you remembered the relevant previous concept in Task b, then hopefully you also remember how to determine a basis for that special space attached to matrix A\text{.}

Suppose A is a 4 \times 5 matrix whose RREF is

\begin{equation*} \left[\begin{array}{rrrrr} 1 \amp 0 \amp 2 \amp 0 \amp -1 \\ 0 \amp 1 \amp 3 \amp 0 \amp 2 \\ 0 \amp 0 \amp 0 \amp 1 \amp -3 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \end{array}\right]\text{.} \end{equation*}

Describe how to determine a basis for \im T_A as a subspace of \R^4\text{.}

(d)

The matrix in Task c is the same as the one in Discovery 43.1.a. How does \dim (\ker T_A) relate to \dim (\im T_A)\text{?}

Discovery 43.5. Describing images.

Once again, for transformation \funcdef{T}{V}{W}\text{,} a vector \uvec{w} in W is in \im T precisely when there exists a vector \uvec{v} in V so that T(\uvec{v}) = \uvec{w}\text{.}

Suppose the domain space V is finite-dimensional with spanning set S = \{\uvec{v}_1,\uvec{v}_2,\uvec{v}_3\}\text{.}

(a)

Reword the above definition of \im T using these spanning vectors:

a vector \uvec{w} in W is in \im T precisely when there exists a of \uvec{v}_1,\uvec{v}_2,\uvec{v}_3 so that .

(b)

Reword the definition again in terms of the spanning image vectors:

a vector \uvec{w} in W is in \im T precisely when there exists a of T(\uvec{v}_1),T(\uvec{v}_2),T(\uvec{v}_3) so that .

(c)

Summarize: If \{\uvec{v}_1,\uvec{v}_2,\dotsc,\uvec{v}_m\} is a spanning set for V\text{,} then \{T(\uvec{v}_1),T(\uvec{v}_2),\dotsc,T(\uvec{v}_m)\} is for \im T\text{.}

A spanning set is fine, but a basis is better.

Discovery 43.6. Image basis example.

Consider again the transformation \funcdef{T = L_B}{\matrixring_{2}(\R)}{\matrixring_{2}(\R)} from Discovery 43.3.c, defined by L_B(A) = B A for

\begin{equation*} B = \begin{bmatrix} 1 \amp 1 \\ 0 \amp 0 \end{bmatrix} \text{.} \end{equation*}

Also recall the standard basis for \matrixring_{2}(\R)\text{:}

\begin{align*} E_{11} \amp = \begin{bmatrix} 1 \amp 0 \\ 0 \amp 0 \end{bmatrix}, \amp E_{12} \amp = \begin{bmatrix} 0 \amp 1 \\ 0 \amp 0 \end{bmatrix}, \amp E_{21} \amp = \begin{bmatrix} 0 \amp 0 \\ 1 \amp 0 \end{bmatrix}, \amp E_{22} \amp = \begin{bmatrix} 0 \amp 0 \\ 0 \amp 1 \end{bmatrix}\text{.} \end{align*}
(a)

Compute the images of the standard basis vectors:

\begin{equation*} T(E_{11}), \;\; T(E_{12}), \;\; T(E_{21}), \;\; T(E_{22}) \text{.} \end{equation*}

Discovery 43.5.c says these image vectors should form a spanning set for \im L_B\text{.} Do they form a basis for \im L_B\text{?}

(b)

Replace the first two standard basis vectors for V = \matrixring_{2}(\R) with your two basis vectors for \ker L_B that you computed in Discovery 43.3.c.

Write A_1,A_2 for these kernel vectors. Is \{A_1,A_2,E_{21},E_{22}\} still a basis for the domain space V = \matrixring_{2}(\R)\text{?}

If so, then \{T(A_1),T(A_2),T(E_{21}),T(E_{22})\} should again be a spanning set for \im L_B\text{.} Is it a basis for \im L_B\text{?} If not, can it easily be reduced to a basis for \im L_B\text{?}

(c)

Summarize the pattern: To determine a basis for the image of transformation \funcdef{T}{V}{W}\text{,} .

Discovery 43.7.

Create a linear transformation \funcdef{T}{\poly_3(\R)}{\matrixring_2(\R)} that has kernel precisely \Span \{ 1 + x^2, 1 - x^3 \}\text{.}

Hint

A transformation does not have to be specified by a formula; see Procedure 42.3.1.