# Linear Algebra/Self-Composition/Solutions

## 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 and the chain of nullspaces is . For the identity transformation the chains are and .

- Problem 2

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

- ,
- ,
- ,
**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): {\displaystyle a+bx+cx^2\mapsto b+cx+ax^2}** **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): {\displaystyle t_3:\mathbb{R}^3\to \mathbb{R}^3}**,

- Answer

- Iterating twice gives
**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): {\displaystyle \mathcal{R}(t_0^2)}**is the space of purely-quadratic polynomials**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): {\displaystyle \{rx^2\,\big|\, r\in \mathbb{C}\}}**, this is where the chain stabilizes . As for nullspaces, is the space of purely-linear quadratic polynomials , and is the space of quadratic polynomials with no term , and this is the end . - The second power
**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): {\displaystyle \mathbb{R}^2 \supset\{\begin{pmatrix} 0 \\ p \end{pmatrix}\,\big|\, p\in\mathbb{C}\} \supset\{\vec{0}\,\} =\cdots }**

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): {\displaystyle \{\vec{0}\,\} \subset\{\begin{pmatrix} q \\ 0 \end{pmatrix}\,\big|\, q\in\mathbb{C}\} \subset\mathbb{R}^2 =\cdots }**

- Iterates of this map cycle around
- We have

- Problem 3

Prove that function composition is associative and so we can write **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): {\displaystyle t^3}**
without specifying a grouping.

- Answer

Each maps **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): {\displaystyle 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 is a subspace of **Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): V**
then any basis for can be enlarged to make a basis for . From this the first sentence is immediate. The second sentence is also not hard: is the span of and if **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): W**
is a proper subspace then **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): V**
is not the span of , and so must have at least one vector more than does .

- Problem 5

Prove that the generalized rangespace is the entire space, and the generalized nullspace is trivial, if the transformation 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 that means that the map is nonsingular if and only if it is onto: **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): {\displaystyle \mathcal{R}(t)=V}**
(and thus **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): {\displaystyle \mathcal{R}(t^2)=V}**
, etc).

- Problem 6

Verify the nullspace half of Lemma 1.3.

- Answer

The nullspaces form chains because because if then and and so .

Now, the "further" property for nullspaces follows from that fact that it holds for rangespaces, along with the prior exercise. Because the dimension of plus the dimension of equals the dimension of the starting space , when the dimensions of the rangespaces stop decreasing, so do the dimensions of the nullspaces. The prior exercise shows that from this point **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): 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 **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): {\displaystyle (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 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).