Discover Linear Algebra

Example 43.4.1.

Using matrix

create the matrix transformation $\funcdef{T_A}{\R^5}{\R^4}$ by defining $T_A(\uvec{x}) = A \uvec{x}\text{,}$ as usual. The RREF of $A$ is

(This is the same RREF as in Discovery 43.1.a and Discovery 43.4.c.)

To determine a basis for $\ker T_A\text{,}$ we solve the homogeneous system $A \uvec{x} = \zerovec\text{.}$ The RREF indicates that we should assign parameters to variables $x_3$ and $x_5\text{.}$ This assignment of parameters leads to general solution

and so

To determine a basis of $\im T_A\text{,}$ we need to determine a basis for the column space of $A\text{,}$ which can be carried out using Procedure 21.3.2. We have already reduced $A$ to RREF above, so identifying leading ones in the first, second, and fourth columns of $\RREF(A)$ leads to

Finally, notice that

the number of columns of $A\text{,}$ as expected.

Example 43.4.2. Symmetric and skew-symmetric matrices.

Consider $\funcdef{T}{\matrixring_2(\R)}{\matrixring_2(\R)}$ by

as considered in Discovery 43.2.a and Discovery 43.3.b. In those discovery activities, we identified the kernel as consisting of symmetric matrices. In the $2 \times 2$ case, an arbitrary symmetric matrix is of the form

so that

In the notation of Procedure 43.3.1, we take

Since $\dim \bigl(\matrixring_2(\R)\bigr) = 4\text{,}$ we only need one more vector to enlarge to a basis for the domain space $\matrixring_2(\R)\text{.}$ The standard basis vector

is not symmetric, and so does not lie in $\ker T\text{.}$ Therefore, taking

the vectors in $\basisfont{K}$ and $\basisfont{K}'$ together form a basis for $\matrixring_2(\R)\text{.}$ To obtain a basis for $\im T\text{,}$ just apply $T$ to the vector in $\basisfont{K}'\text{:}$

so that

Notice that $\im T$ consists of the skew-symmetric $2 \times 2$ matrices, defined by the “skewed” symmetry condition $\utrans{A} = - A\text{.}$

And also notice that

as expected.

Example 43.4.3. Left multiplication of $2 \times 2$ matrices.

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

as in Discovery 43.3.c. An arbitray $2 \times 2$ matrix

is in $\ker L_B$ when

which occurs when $c = -a$ and $d = -b\text{.}$ Inserting these two conditions into the arbitrary matrix above, we have

so that

In the notation of Procedure 43.3.1, we take

Since $\dim \bigl(\matrixring_2(\R)\bigr) = 4\text{,}$ we need two more vectors to enlarge to a basis for the domain space $\matrixring_2(\R)\text{.}$ Neither of the standard basis vectors

is in $\ker T\text{,}$ and the four vectors in $\basisfont{K}$ and

remain independent when taken all together, and so form a basis for the domain space $\matrixring_2(\R)\text{.}$ Putting each of the vectors in $\basisfont{K}'$ through $L_B\text{,}$ we get

so that

This agrees with our earlier calculation of arbitrary input result

where we have replaced the top two entries in the result first result with new arbitrary entries $e,f$ to emphasize that these entries are independently arbitrary through choice of $a,c$ values versus $b,d$ values.

And, once again, we have

as expected.

Example 43.4.4. Differentiation of polynomials.

Consider $\funcdef{\ddx}{\poly_n(\R)}{\poly_n(\R)}$ by $\ddx\bigl(p(x)\bigr) = p'(x)\text{.}$ For simplicity, write $D$ in place of the differential operator $\ddx\text{.}$

Similarly to Discovery 43.2.c, $\ker D$ consists of the constant polynomials, so that

In the notation of Procedure 43.3.1, we take

as a basis for $\ker D\text{.}$ It is straightforward to enlarge this to a basis for the domain space $\poly_n(\R)\text{,}$ as the constant polynomial $1$ is the first vector in the standard basis

So we may take

and differentiate each of these vectors to get

So we have

But since scalar multiples do not affect linear independence, we could instead take

This result should not be surprising, as our knowledge of antidifferentiation leads us to expect that every polynomial in $\poly_{n-1}(\R)$ is the derivative of at least one polynomial in $\poly_n(\R)\text{.}$

And, yet again, we have

as expected.

Example 43.4.5. Kernel and image of a zero transformation.

Consider $\funcdef{\zerovec_{V,W}}{V}{W}$ defined by $\zerovec_{V,W}(\uvec{v}) = \zerovec_W$ for all $\uvec{v}$ in the domain space $V\text{,}$ where $\zerovec_W$ is the zero vector in the codomain space $W\text{.}$ Then clearly

and we have

as expected.

Example 43.4.6. Kernel and image of an identity operator.

Consider $\funcdef{I_V}{V}{V}$ defined by $I_V(\uvec{v}) = \uvec{v}$ for all $\uvec{v}$ in $V\text{.}$ Then clearly

and we have

as expected.

Example 43.4.7. Kernel and image of a coordinate map.

For finite-dimensional space $V$ with basis $\basisfont{B}\text{,}$ consider $\funcdef{\coordmap{B}}{V}{\R^n}$ or $\funcdef{\coordmap{B}}{V}{\C^n}\text{,}$ depending on whether $V$ is a real or complex space, where

for each $\uvec{v}$ in $V\text{.}$ Since coordinate vectors are unique (Theorem 19.5.3), the only vector in $\ker \coordmap{B}$ is the zero vector. And since every column vector of scalars can be traced back to a vector in $V$ by using those scalars as coefficients in a linear combination, as in

(where the $\uvec{v}_j$ are the basis vectors in $\basisfont{B}$), we must have $\im \coordmap{B}$ equal to $\R^n$ or $\C^n\text{,}$ as appropriate.

Example 43.4.8. Kernel and image of pairing with a fixed vector.

Suppose that $V$ is a finite-dimensional inner product space, and $\uvec{u}_0$ is a fixed choice of (nonzero) vector in $V\text{.}$ Let $\funcdef{T_{\uvec{u}_0}}{V}{\R^1}$ represent the linear transformation defined by pairing with $\uvec{u}_0\text{:}$

for all $\uvec{x}$ in $V\text{.}$

Then to be in $\ker T_{\uvec{u}_0}\text{,}$ a vector $\uvec{x}$ must be orthogonal to $\uvec{u}_0\text{.}$ This implies that $\ker T_{\uvec{u}_0} = \orthogcmp{U}$ for $U = \Span \{\uvec{u}_0\}\text{.}$

On the one hand, we know that

since a subspace and its orthogonal complement always form a complete set of independent subspaces of an inner product space (Theorem 37.5.16). On the other hand, the Dimension Theorem says that

From $\ker T_{\uvec{u}_0} = \orthogcmp{U}\text{,}$ we may conclude that

But $\im T_{\uvec{u}_0}$ is a subspace of the codomain space $\R^1\text{,}$ which itself has dimension $1\text{,}$ so we must have $\im T_{\uvec{u}_0} = \R^1\text{.}$