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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 “the 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.