Linear Algebra/Self-Composition/Solutions

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Solutions[edit]

Problem 1

Give the chains of rangespaces and nullspaces for the zero and identity transformations.

Answer

For the zero transformation, no matter what the space, the chain of rangespaces is  V\supset\{\vec{0}\}=\{\vec{0}\}=\cdots\, and the chain of nullspaces is  \{\vec{0}\}\subset V=V=\cdots\, . For the identity transformation the chains are  V=V=V=\cdots and  \{\vec{0}\}=\{\vec{0}\}=\cdots\, .

Problem 2

For each map, give the chain of rangespaces and the chain of nullspaces, and the generalized rangespace and the generalized nullspace.

  1. t_0:\mathcal{P}_2\to \mathcal{P}_2, a+bx+cx^2\mapsto b+cx^2
  2. t_1:\mathbb{R}^2\to \mathbb{R}^2,
    
\begin{pmatrix} a \\ b \end{pmatrix}\mapsto\begin{pmatrix} 0 \\ a \end{pmatrix}
  3. t_2:\mathcal{P}_2\to \mathcal{P}_2, a+bx+cx^2\mapsto b+cx+ax^2
  4. t_3:\mathbb{R}^3\to \mathbb{R}^3,
    
\begin{pmatrix} a \\ b \\ c \end{pmatrix}\mapsto\begin{pmatrix} a \\ a \\ b \end{pmatrix}
Answer
  1. Iterating t_0 twice a+bx+cx^2\mapsto b+cx^2\mapsto cx^2 gives
    
a+bx+cx^2\stackrel{t_0^2}{\longmapsto}cx^2
    and any higher power is the same map. Thus, while \mathcal{R}(t_0) is the space of quadratic polynomials with no linear term \{p+rx^2\,\big|\, p,r\in \mathbb{C}\}, and \mathcal{R}(t_0^2) is the space of purely-quadratic polynomials \{rx^2\,\big|\, r\in \mathbb{C}\}, this is where the chain stabilizes \mathcal{R}_\infty(t_0)=\{rx^2\,\big|\, n\in \mathbb{C}\}. As for nullspaces, \mathcal{N}(t_0) is the space of purely-linear quadratic polynomials \{qx\,\big|\, q\in \mathbb{C}\}, and \mathcal{N}(t_0^2) is the space of quadratic polynomials with no x^2 term \{p+qx\,\big|\, p,q\in \mathbb{C}\}, and this is the end \mathcal{N}_\infty(t_0)=\mathcal{N}(t_0^2).
  2. The second power
    
\begin{pmatrix} a \\ b \end{pmatrix}
\stackrel{t_1}{\longmapsto}\begin{pmatrix} 0 \\ a \end{pmatrix}
\stackrel{t_1}{\longmapsto}\begin{pmatrix} 0 \\ 0 \end{pmatrix}
    is the zero map. Consequently, the chain of rangespaces
    
\mathbb{R}^2
\supset\{\begin{pmatrix} 0 \\ p \end{pmatrix}\,\big|\, p\in\mathbb{C}\}
\supset\{\vec{0}\,\}
=\cdots
    and the chain of nullspaces
    
\{\vec{0}\,\}
\subset\{\begin{pmatrix} q \\ 0 \end{pmatrix}\,\big|\, q\in\mathbb{C}\}
\subset\mathbb{R}^2
=\cdots
    each has length two. The generalized rangespace is the trivial subspace and the generalized nullspace is the entire space.
  3. Iterates of this map cycle around
    
a+bx+cx^2
\stackrel{t_2}{\longmapsto} b+cx+ax^2
\stackrel{t_2}{\longmapsto} c+ax+bx^2
\stackrel{t_2}{\longmapsto} a+bx+cx^2
\;\cdots
    and the chains of rangespaces and nullspaces are trivial.
    
\mathcal{P}_2=\mathcal{P}_2=\cdots
\qquad
\{\vec{0}\,\}=\{\vec{0}\,\}=\cdots
    Thus, obviously, generalized spaces are \mathcal{R}_\infty(t_2)=\mathcal{P}_2 and \mathcal{N}_\infty(t_2)=\{\vec{0}\,\}.
  4. We have
    
\begin{pmatrix} a \\ b \\ c \end{pmatrix}
\mapsto\begin{pmatrix} a \\ a \\ b \end{pmatrix}
\mapsto\begin{pmatrix} a \\ a \\ a \end{pmatrix}
\mapsto\begin{pmatrix} a \\ a \\ a \end{pmatrix}
\mapsto\cdots
    and so the chain of rangespaces
    
\mathbb{R}^3
\supset\{\begin{pmatrix} p \\ p \\ r \end{pmatrix}\,\big|\, p,r\in\mathbb{C}\}
\supset\{\begin{pmatrix} p \\ p \\ p \end{pmatrix}\,\big|\, p\in\mathbb{C}\}
=\cdots
    and the chain of nullspaces
    
\{\vec{0}\,\}
\subset\{\begin{pmatrix} 0 \\ 0 \\ r \end{pmatrix}\,\big|\, r\in \mathbb{C}\}
\subset\{\begin{pmatrix} 0 \\ q \\ r \end{pmatrix}\,\big|\, q,r\in \mathbb{C}\}
=\cdots
    each has length two. The generalized spaces are the final ones shown above in each chain.
Problem 3

Prove that function composition is associative  (t\circ t)\circ t=t\circ (t\circ t) and so we can write t^3 without specifying a grouping.

Answer

Each maps  x\mapsto t(t(t(x))) .

Problem 4

Check that a subspace must be of dimension less than or equal to the dimension of its superspace. Check that if the subspace is proper (the subspace does not equal the superspace) then the dimension is strictly less. (This is used in the proof of Lemma 1.3.)

Answer

Recall that if W is a subspace of V then any basis B_W for W can be enlarged to make a basis B_V for V. From this the first sentence is immediate. The second sentence is also not hard: W is the span of B_W and if W is a proper subspace then V is not the span of B_W, and so B_V must have at least one vector more than does B_W.

Problem 5

Prove that the generalized rangespace \mathcal{R}_\infty(t) is the entire space, and the generalized nullspace \mathcal{N}_\infty(t) is trivial, if the transformation t is nonsingular. Is this "only if" also?

Answer

It is both "if" and "only if". We have seen earlier that a linear map is nonsingular if and only if it preserves dimension, that is, if the dimension of its range equals the dimension of its domain. With a transformation t:V\to V that means that the map is nonsingular if and only if it is onto: \mathcal{R}(t)=V (and thus \mathcal{R}(t^2)=V, etc).

Problem 6

Verify the nullspace half of Lemma 1.3.

Answer

The nullspaces form chains because because if \vec{v}\in\mathcal{N}(t^j) then t^j(\vec{v})=\vec{0} and t^{j+1}(\vec{v})=t(\,t^j(\vec{v})\,)=t(\vec{0})=\vec{0} and so \vec{v}\in\mathcal{N}(t^{j+1}).

Now, the "further" property for nullspaces follows from that fact that it holds for rangespaces, along with the prior exercise. Because the dimension of \mathcal{R}(t^j) plus the dimension of \mathcal{N}(t^j) equals the dimension n of the starting space V, when the dimensions of the rangespaces stop decreasing, so do the dimensions of the nullspaces. The prior exercise shows that from this point k on, the containments in the chain are not proper— the nullspaces are equal.

Problem 7

Give an example of a transformation on a three dimensional space whose range has dimension two. What is its nullspace? Iterate your example until the rangespace and nullspace stabilize.

Answer

(Of course, many examples are correct, but here is one.) An example is the shift operator on triples of reals  (x,y,z)\mapsto (0,x,y) . The nullspace is all triples that start with two zeros. The map stabilizes after three iterations.

Problem 8

Show that the rangespace and nullspace of a linear transformation need not be disjoint. Are they ever disjoint?

Answer

The differentiation operator  d/dx:\mathcal{P}_1\to \mathcal{P}_1 has the same rangespace as nullspace. For an example of where they are disjoint— except for the zero vector— consider an identity map (or any nonsingular map).