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Section 4.3 Subspaces

The definition of a subspace of a vector space \(V\) is very much in the same spirit as our definition of linear transformations. It is a subset of \(V\) that in some sense respects the vector space structure: in the language of Definition 4.3.1, it is a subset that is closed under addition and closed under scalar multiplication.

In fact the connection between linear transformations and subspaces goes deeper than this. As we will see in Definition 4.3.12, a linear transformation \(T\colon V\rightarrow W\) naturally gives rise to two important subspaces: the null space of \(T\) and the image of \(T\).

Subsection 4.3.1 Definition of subspaces

Definition 4.3.1. Subspace.

Let \(V\) be a vector space. A subset \(W\subseteq V\) is a subspace of \(V\) if the following conditions hold:

  1. \(W\) contains the zero vector.

    We have \(\boldzero\in W\text{.}\)

  2. \(W\) is closed under addition.

    For all \(\boldv_1,\boldv_2\in V\text{,}\) if \(\boldv_1,\boldv_2\in W\text{,}\) then \(\boldv_1+\boldv_2\in W\text{.}\) Using logical notation:

    \begin{equation*} \boldv_1,\boldv_2\in W\implies \boldv_1+\boldv_2\in W\text{.} \end{equation*}
  3. \(W\) is closed under scalar multiplication.

    For all \(c\in \R\) and \(\boldv\in V\text{,}\) if \(\boldv\in W\text{,}\) then \(c\boldv\in W\text{.}\) In logical notation:

    \begin{equation*} \boldv\in W\Rightarrow c\boldv\in W\text{.} \end{equation*}

Let \(V=\R^2\) and let

\begin{equation*} W=\{(t,t)\in\R^2 \colon t\in\R\}\text{.} \end{equation*}

Prove that \(W\) is a subspace.

Solution.

We must show properties (i)-(iii) hold for \(W\text{.}\)

  1. The zero element of \(V\) is \(\boldzero=(0,0)\text{,}\) which is certainly of the form \((t,t)\text{.}\) Thus \(\boldzero\in W\text{.}\)

  2. We must prove the implication \(\boldv_1, \boldv_2\in W\Rightarrow \boldv_1+\boldv_2\in W\text{.}\)

    \begin{align*} \boldv_1,\boldv_2\in W\amp \Rightarrow\amp \boldv_1=(t,t), \boldv_2=(s,s) \text{ for some \(t,s\in\R\) }\\ \amp \Rightarrow\amp \boldv_1+\boldv_2=(t+s,t+s)\\ \amp \Rightarrow\amp \boldv_1+\boldv_2\in W\text{.} \end{align*}

  3. We must prove the implication \(\boldv\in W\Rightarrow c\boldv\in W\text{,}\) for any \(c\in \R\text{.}\) We have

    \begin{align*} \boldv\in W\amp \Rightarrow\amp \boldv=(t,t)\\ \amp \Rightarrow\amp c\boldv=(ct,ct)\\ \amp \Rightarrow\amp c\boldv\in W \end{align*}

Let \(V=\R^n\) and let

\begin{equation*} W=\{(x,y)\in \R^2\colon x, y\geq 0\}\text{.} \end{equation*}

Is \(W\) a vector space? Decide which of the of properties (i)-(iii) in Definition 4.3.1 (if any) are satisfied by \(W\text{.}\)

Solution.

  1. Clearly \(\boldzero=(0,0)\in W\text{.}\)

  2. Suppose \(\boldv_1=(x_1,y_1), \boldv_2=(x_2,y_2)\in W\text{.}\) Then \(x_1, x_2, y_1, y_2\geq 0\text{,}\) in which case \(x_1+x_2, y_1+y_2\geq 0\text{,}\) and hence \(\boldv_1+\boldv_2\in W\text{.}\) Thus \(W\) is closed under addition.

  3. The set \(W\) is not closed under scalar multiplication. Indeed, let \(\boldv=(1,1)\in W\text{.}\) Then \((-2)\boldv=(-2,-2)\notin W\text{.}\)

Remark 4.3.4. Two-step proof for subspaces.

As with proofs regarding linearity of functions, we can merge conditions (ii)-(iii) of Definition 4.3.1 into a single statement about linear combinations, deriving the following two-step method for proving a set \(W\) is a subspace of a vector space \(V\text{:}\)

  1. Show \(\boldzero_V\in W\)

  2. Show that

    \begin{equation*} \boldv_1, \boldv_2\in W\implies c\boldv_1+d\boldv_2\in W\text{,} \end{equation*}

    for all \(c,d\in\R\text{.}\)

Remark 4.3.5. Subspaces are vector spaces.

If \(W\) is a subspace of a vector space \(V\text{,}\) then it inherits a vector space structure from \(V\) by simply restricting the vector operations defined on \(V\) to the subset \(W\text{.}\)

It is important to understand how conditions (ii)-(iii) of Definition 4.3.1 come into play here. Without them we would not be able to say that restricting the vector operations of \(V\) to elements of \(W\) actually gives rise to well-defined operations on \(W\text{.}\) To be well-defined the operations must output elements that lie not just in \(V\text{,}\) but in \(W\) itself. This is precisely what being closed under addition and scalar multiplication guarantees.

Once we know restriction gives rise to well-defined operations on \(W\text{,}\) verifying the axioms of Definition 4.1.1 mostly amounts to observing that if a condition is true for all \(\boldv\) in \(V\text{,}\) it is certainly true for all \(\boldv\) in the subset \(W\text{.}\)

The “existential axioms” (iii) and (iv) of Definition 4.1.1, however, require special consideration. By definition, a subspace \(W\) contains the zero vector of \(V\text{,}\) and clearly this still acts as the zero vector when we restrict the vector operations to \(W\text{.}\) What about vector inverses? We know that for any \(\boldv\in W\) there is a vector inverse \(-\boldv\) lying somewhere in \(V\text{.}\) We must show that in fact \(-\boldv\) lies in \(W\text{:}\) i.e. we need to show that the operation of taking the vector inverse is well-defined on \(W\text{.}\) We prove this as follows:

\begin{align*} \boldv\in W \amp\implies (-1)\boldv\in W \amp (\knowl{./knowl/d_subspace.html}{\text{Definition 4.3.1}}, \text{(iii) } )\\ \amp\implies -\boldv\in W \amp (\knowl{./knowl/th_vectorspace_props.html}{\text{Theorem 4.1.13}}, (iii)) \text{.} \end{align*}

We now know how to determine whether a given subset of a vector space is in fact a subspace. We are also interested in means of constructing subspaces from some given ingredients. The result below tells us that taking the intersection of a given collection of subspaces results in a subspace. In Subsection 4.3.3 we see how a linear transformation automatically gives rise to two subspaces.

Exercise.

Remark 4.3.7. Unions of subspaces.

While the intersection of subspaces is again a subspace, the same is not true for unions of subspaces.

For example, take \(V=\R^2\text{,}\) \(W_1=\{(t,t)\colon t\in\R\}\) and \(W_2=\{(t,-t)\colon t\in\R\}\text{.}\) Then each \(W_i\) is a subspace, but their union \(W_1\cup W_2\) is not.

Indeed, observe that \(\boldw_1=(1,1)\in W_1\subset W_1\cup W_2\) and \(\boldw_2=(1,-1)\in W_2\subset W_1\cup W_2\text{,}\) but \(\boldw_1+\boldw_2=(2,0)\notin W_1\cup W_2\text{.}\) Thus \(W_1\cup W_2\) is not closed under addition. (Interestingly, it is closed under scalar multiplication.)

Subsection 4.3.2 Important subspaces of \(F(X,\R)\)

Let \(X\) be an nondegenerate interval of \(\R\text{:}\) i.e., an interval containing at least two elements. Recall that \(F(X,\R)\) is the set of all functions from \(X\) to \(\R\text{.}\) This is a pretty unwieldy vector space, containing some pathological characters, and when studying functions on an interval we will often restrict our attention to certain more well-behaved subsets: e.g., continuous, differentiable, or infinitely differentiable functions. Not surprisingly, these subsets turn out to be subspaces of \(F(X,\R)\text{.}\)

Definition 4.3.8.

Let \(X\subseteq \R\) be a nondegenerate interval.

  1. We denote by \(C(X)\) the set of all continuous functions on \(X\text{:}\) i.e.,

    \begin{equation*} C(X)=\{f\in F(X,\R)\colon f \text{ is continuous on } X\}\text{.} \end{equation*}

  2. Fix \(n\geq 1\text{.}\) A function \(f\in F(X,\R)\) is \(C^n\) on X if \(f\) is \(n\)-times differentiable on \(X\) and its \(n\)-th derivative \(f^{(n)}(x)\) is continuous. The set of all \(C^n\) functions on \(X\) is denoted \(C^n(X)\text{.}\)

  3. A function \(f\in F(X,\R)\) is \(C^\infty\) on X if \(f\) is infinitely differentiable on \(X\text{.}\) The set of all \(C^\infty\) functions on \(X\) is denoted \(C^\infty(X)\text{.}\)

  4. A polynomial on \(X\) is a function \(f\colon X\rightarrow \R\) of the form \(f(x)=\anpoly\text{,}\) where \(a_i\in \R\) and \(a_n\ne 0\text{.}\) We call \(n\) the degree of f, denoted \(\deg f=n\text{;}\) and we call \(a_n\) the leading term of \(f\).

    The set of polynomials of degree at most \(n\) on \(X\) is denoted \(P_n(X)\text{;}\) the set of all polynomials on \(X\) is denoted \(P(X)\text{.}\) When \(X=\R\text{,}\) we shorten the notation to \(P_n\) and \(P\text{.}\)

The proof amounts to the following observations:

  • The zero function \(0_X\colon X\rightarrow \R\) is an element of all of these sets: i.e. the zero function is continuous, \(C^n\text{,}\) \(C^\infty\text{,}\) a polynomial, etc..

  • If \(f\) and \(g\) both satisfy one of these properties (continuous, \(C^n\text{,}\) \(C^\infty\text{,}\) polynomial, etc.), then so does \(cf+dg\) for any \(c,d\in \R\text{.}\)

The second, “closed under linear combinations” observation is easily seen for \(P(X)\) and \(P_n(X)\) (i.e., the sum of two polynomials of degree at most \(n\) is clearly a polynomial of degree at most \(n\)); for the other spaces, this is a result of calculus properties to the effect that adding and scaling functions preserves continuity and differentiability.

Lastly, that each subset relation holds in the given chain follows from similar observations: polynomials are infinitely differentiable, differentiable functions are continuous, etc..

Remark 4.3.10. Polynomial equality.

An important fact that we will make frequent use of is that for two polynomials \(f(x)=\anpoly\) and \(g(x)=\bmpoly\) of degree \(n\) and \(m\text{,}\)respectively, we have \(f(x)=g(x)\) on an interval \(X\) if and only if (1) \(n=m\text{,}\) and (2) \(a_i=b_i\) for all \(0\leq i\leq n\text{.}\)

In particular, we have \(f(x)=\boldzero\) (the zero function) on \(X\) if and only if \(a_i=0\) for all \(1\leq i\leq n\text{.}\)

With this new wealth of function spaces comes a family of linear transformations, differential operators, and these serve as important tools for studying properties of continuous and differentiable functions.

Remark 4.3.11. Differential operators.

Let \(X\subseteq \R\) be an interval. Define \(T_1\colon C^1(X)\rightarrow C(X)\) as \(T_1(f)=f'\text{:}\) i.e., \(T_1\) takes as input a \(C^1\) function on the interval \(X\text{,}\) and returns its (first) derivative. Note that the definition of \(C^1\) ensures that \(f'\) exists and is continuous on \(X\text{:}\) hence \(f' \in C(X)\text{,}\) as claimed.

The operator \(T_1\) is a linear transformation. Indeed, given \(c,d\in \R\) and \(f,g\in C^1(X)\text{,}\) we have

\begin{align*} T_1(cf+dg) \amp = (cf+dg)' \amp \text{(by def.)} \\ \amp =(cf)'+(dg)' \amp \text{(derivative prop.)} \\ \amp =cf'+dg' \amp \text{(derivative prop.)}\\ \amp = cT_1(f)+dT_1(g)\text{.} \end{align*}

Since taking \(n\)-th derivatives amounts to composing the derivative operator \(T_1\) with itself \(n\) times, it follows from Theorem 4.2.22 that for any \(n\geq 1\) the map

\begin{align*} T_n\colon C^n(X) \amp \rightarrow C(X) \\ f \amp\mapsto f^{(n)} \text{,} \end{align*}

which takes a function \(f\) to its \(n\)-th derivative, is also linear. (Note that we are careful to pick the domain \(C^n(X)\) to guarantee this operation is well-defined!)

Lastly, by Exercise 4.2.6.10, we can add and scale these various operators to obtain more general linear transformations of the form

\begin{equation*} T(f)=c_nf^{(n)}+c_{n-1}f^{(n-1)}+\cdots c_1f'+c_0f\text{.} \end{equation*}

We call such a function a linear differential operator. Understanding the linear algebraic properties of these operators is crucial to the theory of linear differential equations, as Example 4.3.23 illustrates.

Subsection 4.3.3 Null space and image of a linear transformation

Before looking at more examples, we first show how a linear transformation \(T\colon V\rightarrow W\) gives rise to two different subspaces: one insides \(V\text{,}\) and the other inside \(W\text{.}\) These are called the null space and image of \(T\text{,}\) respectively.

Definition 4.3.12. Null space and image.

Let \(T\colon V\rightarrow W\) be a linear transformation.

  1. Null space.

    The null space of \(T\text{,}\) denoted \(\NS T\text{,}\) is defined as

    \begin{equation*} \NS T=\{\boldv\in V\colon T(\boldv)=\boldzero_W\}\text{.} \end{equation*}
  2. Image.

    The image (or range) of \(T\text{,}\) denoted \(\im T\text{,}\) is defined as

    \begin{equation*} \im T=\{\boldw\in W\colon \boldw=T(\boldv) \text{ for some } \boldv\in V \}\text{.} \end{equation*}

Remark 4.3.13.

A few remarks:

  1. Let \(T\colon V\rightarrow W\text{.}\) It is useful to keep in mind where \(\NS T\) and \(\im T\) “live” in this picture: we have \(\NS T\subseteq V\) and \(\im T\subseteq W\text{.}\) In other words, the null space is a subset of the domain, and the image is a subset of the codomain.

  2. Note that the image \(\im T\) of a linear transformation is just its image when considered simply as a function of sets. (See Definition 1.1.18.)

  3. The notion of a null space is analogous to the set of zeros (or roots) of a real-valued function \(f\colon X\rightarrow \R\text{,}\)

    \begin{equation*} \{x\in X\colon f(x)=0\}\text{,} \end{equation*}

    and “the zeros of \(T\)” is a useful English shorthand for \(\NS T\text{.}\) However, there is an important difference between the null space of a linear transformation and the zeros of an arbitrary real-valued function: the null space of a linear transformation comes with the added structure of a vector space (Theorem 4.3.14), whereas the zeros of an arbitrary function in general do not.

    The same observation can be made about the image of a linear transformation (Theorem 4.3.14), in comparison to the image of an arbitrary function.

Null space of \(T\).

We use the two-step technique to prove \(\NS T\) is a subspace.

  1. Since \(T(\boldzero_V)=\boldzero_W\) (Theorem 3.2.11), we see that \(\boldzero_V\in \NS T\text{.}\)

  2. Suppose \(\boldv_1, \boldv_2\in \NS T\text{.}\) Given any \(c,d\in \R\text{,}\) we have

    \begin{align*} T(c\boldv_1+d\boldv_2) \amp=cT(\boldv_1)+dT(\boldv_2) \amp (T \text{ is linear, } \knowl{./knowl/th_trans_props.html}{\text{Theorem 3.2.11}})\\ \amp=c\boldzero_W+d\boldzero_W \amp (\boldv_1, \boldv_2\in \NS T) \\ \amp = \boldzero_W\text{.} \end{align*}

    This shows that \(c\boldv_1+d\boldv_2\in \NS T\text{,}\) completing our proof.

Image of \(T\).

The proof proceeds in a similar manner.

  1. Since \(T(\boldzero_V)=\boldzero_W\) (Theorem 3.2.11), we see that \(\boldzero_W\) is “hit” by \(T\text{,}\) and hence is a member of \(\im T\text{.}\)

Recall that a function \(f\colon X\rightarrow Y\) is surjective if \(\im f=f(X)=Y\text{.}\) (See Definition 1.1.19.) Thus the subspace \(\im T\) of a linear transformation \(T\) is a measure of sorts of the transformation being surjective or not. The next theorem indicates how the null space can be understood as a measure of a transformation being injective (one-to-one) or not. (See Definition 1.1.19.)

Implication: \((1)\implies (2)\).

In general we always have \(\{\boldzero\}\subseteq \NS T\) for a linear transformation \(T\text{.}\) If \(T\) is injective, this inclusion is an equality since

\begin{align*} \boldv\in\NS T \amp \iff T(\boldv)=\boldzero \amp (\text{def. } \NS T)\\ \amp \implies T(\boldv)=T(\boldzero)\\ \amp \implies \boldv=\boldzero \amp (T \text{ injective})\text{.} \end{align*}
Implication: \((2)\implies (1)\).

This implication is a consequence of the equivalence in (4.3.1). Indeed assuming (4.3.1), we have

\begin{align*} T(\boldv)=T(\boldv') \amp\iff \boldv'=\boldv+\boldu \text{ for some } \boldu\in \NS T \\ \amp\implies \boldv'=\boldv+\boldzero \amp (\NS T=\{\boldzero\}) \\ \amp\implies \boldv=\boldv' \text{,} \end{align*}

as desired. Thus it remains only to prove (4.3.1). We do so via a chain of equivalences:

\begin{align*} T(\boldv)=T(\boldv') \amp \iff T(\boldv')-T(\boldv)=\boldzero\\ \amp \iff T(\boldv'-\boldv)=\boldzero \amp (T \text{ is linear})\\ \amp \iff \boldu=\boldv'-\boldv\in\NS T \amp (\text{def. } \NS T)\\ \amp \iff \boldv'=\boldv+\boldu \text{ for some } \boldu\in\NS T \text{.} \end{align*}

Remark 4.3.16.

To determine whether a function of sets \(f\colon X\rightarrow Y\) is injective, we normally have show that for each output \(y\) in the range of \(f\) there is exactly one input \(x\) satisfying \(f(x)=y\text{.}\) Think of this as checking injectivity at every output. Theorem 4.3.15 tells us that in the special case of a linear transformation \(T\colon V\rightarrow W\) it is enough to check injectivity at exactly one ouput: namely, \(\boldzero\in W\text{.}\)

To see that this is a special property of linear transformations, consider the function

\begin{align*} f\colon \R \amp \rightarrow \R\\ x \amp\mapsto f(x)=(x-1)^2 \text{.} \end{align*}

The function is injective at the output \(0=f(1)\text{,}\) but not injective in general.

Theorem 4.3.14 provides us a with a useful indirect way of proving a subset \(W\subseteq V\) is a subspace: namely, find a linear transformation $T$ for which \(W=\NS T\text{.}\) (You could also identify \(W=\im T\) for some linear transformation \(T\text{,}\) but this approach tends to be not as convenient.

Prove that the set \(W\subseteq\R^3\) of all vectors \((x,y,z)\) satisfying \(x+2y+3z=x-y-z=0\) is a subspace of \(\R^3\text{.}\)

Solution.

Let

\begin{equation*} A=\begin{bmatrix}1\amp 2\amp 3\\ 1\amp -1\amp -1 \end{bmatrix}\text{,} \end{equation*}

and let

\begin{align*} T_A\colon \R^3 \amp\rightarrow \R^2 \\ (x,y,z)\amp\mapsto (x+2y+3z, x-y-z) \end{align*}

be its associated linear transformation. Then

\begin{align*} \NS T_A \amp =\{(x,y,z)\in\R^3\colon T_A(x,y,z)=(0,0)\} \\ \amp = \{(x,y,z)\in\R^3\colon (x+2y+3z,x-y-z)=(0,0)\} \\ \amp = \{(x,y,z)\in\R^3\colon x+2y+3z=x-y-z=0\}\\ \amp = W\text{.} \end{align*}

This shows that \(W\) can be identified as the null space of a linear transformation, and hence is a subspace of \(\R^3\text{.}\)

Remark 4.3.18. Lines and planes.

Recall that a line \(\ell\) in \(\R^2\) that passes through the origin can be expressed as the set of solutions \((x_1,x_2)\in\R^2\) to an equation of the form

\begin{equation*} \ell\colon ax_1+bx_2=0\text{.} \end{equation*}

Similarly, a plane \(\mathcal{P}\) in \(\R^3\) that passes through the origin can be expressed as the the set of solutions \((x_1,x_2,x_3)\) to an equation of the form

\begin{equation*} \mathcal{P}\colon ax_1+bx_2+cx_3=0\text{.} \end{equation*}

Using the language of linear algebra, we see that the line \(\ell\) is precisely \(\NS T_A\text{,}\) where \(\underset{1\times 2}{A}=\begin{bmatrix} a \amp b\end{bmatrix}\text{;}\) and the plane \(\mathcal{P}\) is precisely \(\NS T_B\text{,}\) where \(\underset{1\times 3}{B}=\begin{bmatrix}a\amp b\amp c\end{bmatrix}\text{.}\) We conclude from Theorem 4.3.14 that lines in \(\R^2\text{,}\) and planes in \(\R^3\text{,}\) are subspaces, as long as they pass through the origin.

On the other hand, a line or plane that does not pass through the origin is not a subspace, since it does not contain the zero vector.

The question arises: Can we describe all the subspaces of \(\R^2\) or \(\R^3\text{?}\) The answer is yes, as we will see in Section 4.5

Definition 4.3.19. Symmetric and skew-symmetric matrices.

A matrix \(A\in M_{nn}\) is symmetric if \(A^T=A\text{.}\) It is skew-symmetric if \(A^T=-A\text{.}\)

Let \(V=M_{nn}\text{,}\) let \(W_1\) be the set of all symmetric \(n\times n\) matrices, and let \(W_2\) be the set of all skew-symmetric \(n\times n\) matrices. Prove that \(W_1\) and \(W_2\) are subspaces of \(V\text{.}\)

Solution.

Define functions \(T_i\colon M_{nn}\rightarrow M_{nn}\text{,}\) \(i=1,2\text{,}\) as follows:

\begin{align*} T_1(A) \amp = A^T-A \\ T_2(A)\amp = A^T+A\text{.} \end{align*}

Firstly, it is easy to see, using Theorem 3.2.11 that both \(T_1\) and \(T_2\) are both linear transformations. For example, for \(T_1\) we have

\begin{align*} T_1(cA+dB) \amp=(cA+dB)^T-(cA+dB) \\ \amp=cA^T+dB^T-cA-cB \\ \amp = c(A^T-A)+d(B^T-B)\\ \amp = cT_1(A)+dT_1(B)\text{.} \end{align*}

As similar proof applies to \(T_2\text{.}\)

Next, it is easy to see that \(W_1=\NS T_1\) and \(W_2=\NS T_2\text{.}\) Taking \(W_1\) for example, we have

\begin{align*} A \text{ symmetric} \amp\iff A^T=A \\ \amp\iff A^T-A=\boldzero_{n\times n} \\ \amp \iff T_1(A)=\boldzero_{n\times n}\\ \amp \iff A\in \NS T_1\text{.} \end{align*}

Having identified \(W_1\) and \(W_2\) as null spaces of linear transformations, we conclude that they are subspaces of \(M_{nn}\text{.}\)

The next two examples illustrate how to compute the image of a given linear transformation \(T\colon V\rightarrow W\text{.}\) The examples make clear that computing images (i.e., determining the set of elements of \(W\) that are hit by \(T\)) is often more onerous than computing null spaces.

Let \(T=T_A\colon \R^2\rightarrow\R^3\text{,}\) where \(A=\begin{bmatrix}1\amp 1\\ 2\amp 1\\ 3\amp 5 \end{bmatrix}\text{.}\) According to Theorem 4.3.14, \(\im T_A\) is a subspace of \(\R^3\text{.}\) Identify this subspace as a familiar geometric object.

Solution.

By definition \(\im T_A\) is the set

\begin{equation*} \{\boldy\in\R^3\colon \boldy=T_A(\boldx) \text{ for some \(\boldx\in \R^3\) } \}=\left\{\boldy\colon \boldy=A\boldx \text{ for some \(\boldx\in\R^2\) } \right\}\text{.} \end{equation*}

Thus to compute \(\im T_A\) we must determine which choice of \(\boldy=(a,b,c)\) makes the system \(A\boldx=\boldy\) consistent. We answer this using our old friend Gaussian elimination!

\begin{align*} \begin{amatrix}[rr|r] 1\amp 1\amp a\\ 2\amp 1\amp b\\ 3\amp 5\amp c \end{amatrix} \amp \xrightarrow[r_3-3r_1]{r_2-2r_1} \begin{amatrix}[rr|r] 1\amp 1\amp a\\ 0\amp 1\amp 2a-b\\ 0\amp 0\amp -7a+2b+c \end{amatrix}\text{.} \end{align*}

To be consistent we need \(-7a+2b+c=0\text{.}\) We conclude that \(\im T\) is the set of all \((a,b,c)\) satisfying \(-7a+2b+c=0\text{.}\) Geometrically this is the plane passing through \((0,0,0)\) with normal vector \(\boldn=(-7,2,1)\text{.}\)

Consider again the linear transformation \(T_1\colon M_{nn}\rightarrow M_{nn}\text{,}\) \(T_1(A)=A^T-A\text{,}\) from Example 4.3.20. We saw that \(\NS T_1\) was the space of symmetric matrices. Identify \(\im T_1\) as a familiar subspace of \(M_{nn}\text{.}\)

Solution.

Take \(B\in \im T_1\text{.}\) By definition this means \(B=T(A)=A^T-A\) for some \(A\text{.}\) So one, somewhat unsatisfying way of describing \(\im T_1\) is as the set of all matrices of the form \(A^T-A\text{.}\)

Let's investigate further. Notice that if \(B=A^T-A\text{,}\) then \(B^T=(A^T-A)^T=(A^T)^T-A^T=A-A^T=-B\text{.}\) Thus every element \(B\in \im T\) satisfies \(B^T=-B\text{:}\) i.e., all elements of the image are skew-symmetric!

We claim further that in fact \(\im T_1\) is the the set of all skew-symmetric matrices. To prove this, we need to show that given a skew-symmetric matrix \(B\text{,}\) we have \(B\in \im T_1\text{:}\) i.e., that there is a matrix \(A\) such that \(T_1(A)=B\text{.}\)

Suppose \(B\) satisfies \(B^T=-B\text{.}\) Let \(A=-\frac{1}{2}B\text{.}\) Then

\begin{equation*} T_1(A)=T_1(-\frac{1}{2}B)=-\frac{1}{2}(B^T-B)=-\frac{1}{2}(-B-B)=B\text{.} \end{equation*}

This shows that \(B\in \im T_1\text{,}\) and concludes the proof that \(\im T_1\) is the set of all skew-symmetric matrices.

We end with an example that illustrates how the notions of linear transformation, null space, and image can be applied to differential equations.

Fix an interval \(X\subseteq \R\text{.}\) Let \(S\) be the set of functions of \(C^1(X)\) satisfying the differential equation

\begin{equation} f'=f\text{:}\label{eq_diff_eq_ex}\tag{4.3.2} \end{equation}

i.e.,

\begin{equation*} S=\{f\in C^1(\R)\colon f'(x)=f(x) \text{ for all } x\in X\}\text{.} \end{equation*}

Define \(T\colon C^1(X)\rightarrow C(X)\) as the differential operator \(T(f)=f'-f\text{.}\) We have

\begin{align*} f\in S \amp\iff f'=f \\ \amp\iff f'-f=\boldzero \\ \amp\iff T(f)=\boldzero \\ \amp \iff f\in \NS T\text{.} \end{align*}

Thus \(S=\NS T\text{,}\) and we see that the set of solutions to (4.3.2) has the structure of a subspace. That is helpful information for us. For example, since \(S=\NS T\) is closed under vector addition and scalar multiplication, we know that if \(f\) and \(g\) are solutions to (4.3.2), then so is \(cf+dg\) for any \(c,d\in\R\text{.}\)

It is an interesting exercise to try and understand what \(\im T\) means in terms of the given differential equation. By definition, \(\im T\) is the set of \(g\in C(X)\) such that \(g=T(f)=f'-f\text{,}\) for some \(f\in C^1(X)\text{.}\) In the language of differential equations, this is the set of all continuous functions \(g\in C(X)\) for which the differential equation

\begin{equation*} f'-f=g \end{equation*}

can be solved for \(f\text{.}\)

Exercises 4.3.4 Exercises

1.

Let \(V=\R^4\) and define

\begin{equation*} W=\{(x_1,x_2,x_3,x_4)\colon x_1-x_4=x_2+x_3=x_1+x_2+x_3=0\}\text{.} \end{equation*}

  1. Show \(W\) is a subspace by identifying it as \(\NS T\) for an explicit linear transformation \(T\text{.}\)

  2. Give a parametric description of \(\im W\text{.}\)

2.

In each part below a subset \(W\) of \(\R^2\) is described. Sketch \(W\) as a region of \(\R^2\) and determine whether it is a subspace of \(\R^2\text{.}\) Justify your answer either with a proof or explicit counterexample.

  1. Let \(W=\{(x,y)\in \R^2\colon 2x+3y=0\}\text{.}\)

  2. Let \(W=\{(x,y)\in \R^2\colon \val{x}\geq \val{y}\}\text{.}\)

  3. Let \(W=\{(x,y)\in \R^2\colon x^2+2y^2\leq 1\}\text{.}\)

3.

Let \(n\geq 2\text{.}\) In each part below a subset \(W\) of \(M_{nn}\) is described. Determine whether the given \(W\) is a subspace of \(M_{nn}\text{.}\) Justify your answer either with a proof or explicit counterexample.

  1. Let \(W\) be the set of all upper triangular \(n\times n\) matrices.

  2. Let \(W=\{A\in M_{nn}\colon \det A=0\}\text{.}\)

  3. Let \(W=\{A\in M_{nn}\colon \tr A=0\}\text{.}\) (See Exercise 4.2.6.4.)

  4. Fix \(B\in M_{nn}\) and define

    \begin{equation*} W=\{A\in M_{nn} \colon AB=BA\}\text{.} \end{equation*}

4.

Let \(n\geq 1\text{.}\) In each part below a subset \(W\) of \(P_2\) is described. Determine whether the given \(W\) is a subspace of \(P_2\text{.}\) Justify your answer either with a proof or explicit counterexample.

  1. Let \(\ds W=\{f(x)=ax^2+bx+c \colon c=0\}\text{.}\)

  2. Let \(\ds W=\{f(x)=ax^2+bx+c \colon abc=0\}\text{.}\)

  3. Let \(\ds W=\{f(x)\in P_2 \colon xf'(x)=f(x) \}\text{.}\)

5.

In each part below a subset \(W\) of \(C(\R)\) is described. Determine whether the given \(W\) is a subspace of \(C(\R)\text{.}\) Justify your answer either with a proof or explicit counterexample.

  1. Let \(W=\{f\in C(\R)\colon f(4)=0\} \text{.}\)

  2. Let \(W=\{f\in C(\R)\colon f(0)=4\} \text{.}\)

  3. Let \(W=\{f\in C(\R)\colon f(x)=f(-x)\} \text{.}\)

  4. Let \(W=\{f\in C(\R)\colon f(x+\pi)=f(x)\} \text{.}\)

  5. Let \(W=\{f\in C(\R)\colon f\in P \text{ and } \deg f=5\}. \)

6.

Each function \(T\) below is linear. Give parametric descriptions of \(\NS T\) and \(\im T\text{.}\) To do so, you will want to relate each computation to a system of linear equations. (See Example 4.3.21 for an example of computing an image.)

  1. \begin{align*} T\colon \R^4 \amp \rightarrow \R^3 \\ (x,y,z,w)\amp\mapsto (x+z+w, x-y-z,-2x+y-w) \text{.} \end{align*}
  2. \begin{align*} T\colon M_{22}\amp \rightarrow M_{22} \\ A \amp\mapsto \begin{amatrix}[rr]1\amp 1\\ 1\amp 1 \end{amatrix}A \text{.} \end{align*}
  3. \begin{align*} T\colon P_2\amp \rightarrow P_2 \\ f \amp\mapsto f(x)+f'(x)\text{.} \end{align*}

7.

In each part below a linear transformation \(T\) is given, and a claim is made about \(\im T\text{.}\) Prove the claim.

  1. \begin{align*} T\colon \R^\infty \amp \rightarrow \R^\infty \\ (a_1,a_2,\dots)\amp\mapsto (a_2,a_3,\dots) \end{align*}

    . Claim: \(\im T=\R^\infty\)

  2. \begin{align*} T\colon C(\R) \amp \rightarrow C(\R) \\ f(x)\amp\mapsto g(x)=f(x)+f(-x)\text{.} \end{align*}

    Claim: \(\im T\) is the set of all continuous symmetric polynomials. In other words,

    \begin{equation*} \im T=\{f\in C(x)\colon f(-x)=f(x)\}. \end{equation*}