HW 4 - Inverse Maps

3.D: Inverse Maps

#1

Theorem

Suppose $T \in L (U, V)$ and $S \in L (V, W)$ are both invertible linear maps. Then $S T \in L (U, W)$ is invertible and $(S T)^{- 1} = T^{- 1} S^{- 1}$ .

Proof
We'll show that $S T$ is bijective (injective and surjective) first to show that it's invertible.

Suppose $u_{1}, u_{2} \in U$ are arbitrary and that $S T u_{1} = S T u_{2}$ . Since $S$ is invertible then:

\begin{aligned} S T u_{1} & = S T u_{2} \\ S^{- 1} S T u_{1} & = S^{- 1} S u_{2} \\ (S^{- 1} \circ S) T u_{1} & = (S^{- 1} \circ S) T u_{2} \\ I_{V} T u_{1} & = I_{V} T u_{2} \\ T u_{1} & = T u_{2} \end{aligned}

and similarly since $T$ is also invertible:

\begin{aligned} T u_{1} & = T u_{2} \\ T^{- 1} T u_{1} & = T^{- 1} T u_{2} \\ I_{U} u_{1} & = I_{U} u_{2} \\ u_{1} & = u_{2} \end{aligned}

showing injectivity. Now for surjectivity, let $w \in W$ be arbitrary. I claim that the choice of $T^{- 1} S^{- 1} w$ is a vector in $U$ such that:

S T (T^{- 1} S^{- 1} w) = S (T \circ T^{- 1}) S^{- 1} w = S (I_{V}) S^{- 1} w = (S \circ S^{- 1}) w = I_{W} w = w

Thus, showing that $S T$ is surjective. Therefore, $S T$ is bijective and thus invertible.

Notice that $(S T)^{- 1} = T^{- 1} S^{- 1}$ in this case since:

(S T)^{- 1} S T = (T^{- 1} S^{- 1}) S T = T^{- 1} (S^{- 1} \circ S) T = T^{- 1} \circ I_{V} \circ T = T^{- 1} \circ T = I_{U}

and since we get the identity, then this is the actual inverse of the given map. The reverse case showing that $S T (S T)^{- 1} = I_{W}$ is also trivial and follows without loss of generality.
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#2

Theorem

Suppose $V$ is finite-dimensional and $dim (V) > 1$ . Then the set of noninvertible operators on $V$ is not a subspace of $L (V)$ .

Proof
We can prove this by showing that $S = {N \in L (V) : ∄ N^{- 1}}$ is not a subspace.

Since $dim (V) > 1$ say $V$ has dimension $n > 1$ , so then $v_{1}, v_{2}, . . ., v_{n}$ is a basis for $V$ where at least $v_{1}, v_{2}$ exist.

Notice that the zero transformation $O \in S : V \to V$ is noninvertible since clearly $O (v_{1}) = O (v_{2}) = {\vec{0}}_{V}$ but we have $v_{1} \neq v_{2}$ (so it's not injective and therefore not bijective), so we need some different approach as the zero vector is in our "subspace".

We will show that $T_{1} + T_{2} \notin S$ for a chosen $T_{1}, T_{2} \in S$ . Choose $T_{1}$ where $T_{1} v_{1} = \vec{0}$ and $T_{1} v_{i} = v_{i}$ elsewhere. Then define $T_{2}$ where $T_{2} v_{1} = v_{1}$ while $T_{2} v_{i} = \vec{0}$ elsewhere.

Notice that $T_{1} (1 v_{1}) = T_{2} (2 v_{1}) = \vec{0}$ but $1 v_{1} \neq 2 v_{1}$ showing non-injectivity (thus $T_{1}$ isn't invertible). A similar argument shows that $T_{2}$ isn't invertible using instead $v_{2}$ , which we know at least exists.

Notice that $(T_{1} + T_{2})$ is expressed as, for some basis vector $v_{i}$ :

(T_{1} + T_{2}) v_{i} = T_{1} v_{i} + T_{2} v_{i}

Notice that if $i = 1$ then we get $(T_{1} + T_{2}) v_{i} = \vec{0} + v_{1} = v_{1}$ and if $i \neq 1$ then $(T_{1} + T_{2}) v_{i} = v_{i} + \vec{0} = v_{i}$ . Therefore, we get that $(T_{1} + T_{2})$ is just the identity map, which is an invertible operator. Thus $T_{1} + T_{2} \notin S$ showing that $S$ is not closed under vector addition. Hence $S$ is not a subspace.
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#3

Theorem

Suppose $V$ is finite-dimensional, $U$ is a subspace of $V$ and $S \in L (U, V)$ . Then there is some invertible operator $T \in L (V)$ such that $T u = S u$ for all $u \in U$ iff $S$ is injective.

Proof
Consider $\to$ , so suppose that there is some invertible operator $T \in L (V)$ such that $T u = S u$ for all $u \in U$ . We'll prove that $S \in L (U, V)$ is injective.

Let $S u_{1} = S u_{2}$ where $u_{1}, u_{2} \in U$ . Via our supposition, then $T u_{1} = T u_{2} = S u_{1} = S u_{2}$ . Since $T$ is invertible then notice that:

\begin{aligned} T u_{1} & = T u_{2} \\ T^{- 1} T u_{1} & = T^{- 1} T u_{2} \\ I_{U} u_{1} & = I_{U} u_{2} \\ u_{1} & = u_{2} \end{aligned}

Thus showing that $S$ is injective.

Now consider $\leftarrow$ . Suppose that $S$ is injective. We'll need to prove that there is some operator $T \in L (V)$ such that $T u = S u$ for all $u \in U$ . Define $T$ where:

T v = {\begin{cases} S v & v \in U \\ v & v \notin U \end{cases}

First we'll show that $T$ is linear over $V$ . Let $v_{1}, v_{2} \in V$ be arbitrary. Then we have 2 cases:

If $v_{1} + v_{2} \in U$ then:

T (v_{1} + v_{2}) = S (v_{1} + v_{2}) = S v_{1} + S v_{2} = T v_{1} + T v_{2}

since $S$ is a linear operator.
2. If $v_{1} + v_{2} \notin U$ then:

T (v_{1} + v_{2}) = v_{1} + v_{2}

note that then we have either $v_{1}, v_{2} \notin U$ in which case we get that:

T (v_{1} + v_{2}) = v_{1} + v_{2} = T v_{1} + T v_{2}

which is good for showing additivity. WLOG if instead $v_{1} \in U$ and $v_{2} \notin U$ then:

T (v_{1} + v_{2}) = v_{1} + v_{2} = S v_{1} + T v_{2} = T v_{1} + T v_{2}

as by the definition of $T$ . As such, no matter what we get that we have additivity of $T$ .

For homogeneity, let $λ \in F$ be arbitrary.

If $λ v \in U$ then notice that:

T (λ v) = S (λ v) = λ S v = λ T v

since $U$ is closed under scalar multiplication and $S$ is linear.
2. If $λ v \notin U$ then:

T (λ v) = λ v = λ (T v)

since clearly $v \notin U$ as $U$ is closed under scalar multiplication. Thus homogeneity is here.

Hence, $T$ is a linear map, and by the definition of $T$ we have $T u = S u$ for all $u \in U$ .
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#4

Theorem

Suppose $W$ is finite-dimensional and $T_{1}, T_{2} \in L (V, W)$ . Then $null (T_{1}) = null (T_{2})$ iff there is an invertible operator $S \in L (W)$ such that $T_{1} = S T_{2}$ .

Proof
Since $W$ is finite-dimensional (say $n$ dimensional) then there's some basis $w_{1}, . . ., w_{n}$ for $W$ .

Consider $\to$ , so suppose $null (T_{1}) = null (T_{2})$ and have dimension $m$ . By the FTOLM, then $dim (range (T_{1})) = dim (range (T_{2})) = n - m$ . Thus, they both each have some basis $v_{1}, . . ., v_{n - m}$ and $u_{1}, . . ., u_{n - m}$ respectively, where all $v_{i}, u_{i} \in W$ .

Choose an invertible operator $S \in L (W)$ such that $S u_{i} = v_{i}$ for all $i$ .

Notice that this is invertible since surjectivity is implied in the definition (any $v_{i}$ has some $u_{i}$ that it's mapped from, and any $v \in range (T_{1})$ is from our basis $v_{1}, . . ., v_{n - m}$ as defined before, where $S$ is linear since any linear combination from the $v_{i}^{'} s$ can be equated to their corresponding $u_{i}$ 's), and injectivity comes from where we allow $v_{1}, v_{2} \in range (T_{2})$ to be arbitrary, so then if $S v_{1} = S v_{2}$ then $S v_{1} = v_{1}$ so then correspondingly $S v_{2} = v_{2}$ so then $v_{1} = v_{2}$ showing injectivity.

Thus $S$ is a valid invertible operator. Notice also that $S$ is a linear map due to the linearity of the bases $v_{1}, . . ., v_{n - m}$ and $u_{1}, . . ., u_{n - m}$ . But notice that for some arbitrary $v \in V$ that:

\begin{aligned} T_{1} v \in range (T_{1}) \\ \Rightarrow T_{1} v = a_{1} v_{1} + \dots + a_{n - m} v_{n - m} \\ = a_{1} S u_{1} + \dots + a_{n - m} S u_{n - m} \\ = S (a_{1} u_{1} + \dots + a_{n - m} u_{n - m}) \\ = S T_{2} v \end{aligned}

Thus $T_{1} = S T_{2}$ as required.

Now suppose $\leftarrow$ , so $S$ is an invertible operator such that $T_{1} = S T_{2}$ . We'll show $\subseteq$ and $\supseteq$ . Let $w \in null (T_{1}$ ). Then $T_{1} w = \vec{0}$ but notice then that $S T_{2} w = \vec{0}$ consequently. Then $T_{2} w = S^{- 1} \vec{0}$ but since $S$ has an empty nullspace, then $S^{- 1} \vec{0} = \vec{0}$ so then $T_{2} w = \vec{0}$ . Thus $w \in null (T_{2})$ . For the other direction, let $w \in null (T_{2})$ . Then $T_{2} w = \vec{0}$ . Thus $S T_{2} w = S \vec{0} = \vec{0}$ , but then that means that $T_{1} w = \vec{0}$ so then $w \in null (T_{1})$ . Therefore, $null (T_{1}) = null (T_{2})$ .

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#8

Theorem

Suppose $V$ is finite-dimensional and $T : V \to W$ is a surjective linear map of $V$ onto $W$ . Then there is a subspace $U$ of $V$ such that $T |_{U}$ is an isomorphism of $U$ onto $W$ .

Here $T |_{U}$ means that the function $T$ restricted to $U$ , or the function whose domain is $U$ with $T |_{U}$ defined by $T |_{U} (u) = T u$ for all $u \in U$ .

Proof
Let $V$ be $n$ -dimensional and $T : V \to W$ be surjective. Then both:

There's some basis $v_{1}, . . ., v_{n}$ for $V$
For any $w \in W$ there is some $v \in V$ where $T v = w$ .
$T$ is a linear map, so the rules of additivity and homogeneity apply here.

First notice by the FTOLM:

dim (V) = n = dim (null (T)) + dim (range (T))

where since $T$ is surjective then:

n = dim (null (T)) + dim (W) \Rightarrow n - dim (null (T)) = dim (W) \leq n

as such then $W$ must be finite dimensional, so denote the basis of $W$ as $w_{1}, . . ., w_{m}$ where $m \leq n$ .

Consider the following cases. If $m = n$ then we can just choose $U = V$ is a valid subspace (since we have all the conditions as $V$ is a vector space) and $T |_{U}$ is a valid isomorphism since if $m = n$ then $dim (null (T)) = 0$ so then we must have injectivity, so then $T$ is bijective and thus invertible. Hence, we can just have $T |_{U} = T$ be a valid isomorphism.

But now then consider when $m \neq n$ , so then $m < n$ . We can try $U = span (v_{1}, . . ., v_{n - m})$ , where each $v_{i}$ is a basis vector from our basis above. Notice that $\vec{0} \in U$ since $\vec{0}$ is in the span of any part of a basis of $V$ (set all the constants to 0). Further, we have closure since if $u_{1}, u_{2} \in U$ then they are comprised of linear combinations that themselves must be within their own span:

u_{1} + u_{2} = a_{1} v_{1} + \dots + a_{n - m} v_{n - m} + b_{1} v_{1} + \dots + a_{n - m} v_{n - m}

= (a_{1} + b_{1}) v_{1} + \dots + (a_{n - m} + b_{n - m}) v_{n - m} \in span (v_{1}, . . ., v_{n - m}) = U

Showing closure under vector addition. The same argument shows closure under scalar multiplication:

α u = α (a_{1} v_{1} + \dots + a_{n - m} v_{n - m}) = α a_{1} v_{1} + \dots + α a_{n - m} v_{n - m} \in span (v_{1}, . . ., v_{n - m}) = U

But notice still that $T |_{U}$ is a valid isomorphism. If $dim (W) = m - n$ then we have a basis for a subspace for $U$ with the same number of vectors, so then the transformation is isomorphic!

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#9

Theorem

Suppose $V$ is finite-dimensional and $S, T \in L (V)$ . Then $S T$ is invertible iff $S, T$ each are both invertible.

Proof
First, consider $\leftarrow$ , so both $S, T$ are invertible. Then:

T^{- 1} S^{- 1} (S T) = T^{- 1} (S^{- 1} \circ S) T = T^{- 1} \circ I_{V} \circ T = T^{- 1} \circ T = I

and:

S T (T^{- 1} S^{- 1}) = S (T \circ T^{- 1}) S^{- 1} = S \circ I_{V} \circ S^{- 1} = S \circ S^{- 1} = I

so then we've found $T^{- 1} S^{- 1}$ as a valid inverse for $S T$ and thus $S T$ is invertible.

Next, consider $\to$ . Suppose $S T$ is invertible. Assume for contradiction that $T$ (WLOG) is not invertible. That means that $T$ is not bijective, so then it either isn't injective or isn't surjective.

If $T$ isn't injective then there are two vectors $v_{1}, v_{2} \in V$ such that $v_{1} \neq v_{2}$ and therefore $T v_{1} \neq T v_{2}$ . No matter if $S$ is injective or not (from being invertible), it's clear that $S T v_{1} \neq S T v_{2}$ which contradicts the injectivity of $S T$ (from it being invertible).

So clearly $T$ just isn't surjective right? Well then that means that there is some vector $v \in V$ such that for all vectors $v^{'} \in V$ have it that $T v^{'} \neq v$ . But since $S T$ is invertible, it must be surjective. Hence, for any vector $u \in V$ there's some other vector $u^{'} \in V$ such that $S T u^{'} = u$ . We can use $v$ in this statement, so then there's some other vector $v^{″}$ where $S T v^{″} = v$ . Since $T v^{″} \in V$ then we can relabel it as $v^{'} \in V$ . Thus, $T v^{'} = v$ which is a contradiction.

Therefore, $S$ must've been bijective. A similar argument shows $T$ must be bijective as well (since here we didn't assume anything about $T$ whatsoever), so then both $S, T$ are invertible maps.

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#10

Theorem

Suppose $V$ is finite-dimensional and $S, T \in L (V)$ . Then $S T = I$ iff $T S = I$ .

Proof
This proof essentially says that if we find one method of showing that $T S = I$ then $S$ is the inverse as a given. As such, we only consider $\to$ as the other direction is done in the same manner.

Thus, suppose that $S T = I$ . Notice that:

(S T) T = T, T (S T) = T \Rightarrow (S T) T = T (S T) \Rightarrow (S T) T = (T S) T

therefore then $S T = T S$ . But $S T = I$ . Therefore, $T S = I$ .
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#18

Theorem

$V$ and $L (F, V)$ are isomorphic vector spaces.

Proof
We can just show that these two spaces have the same dimension:

dim (L (F, V)) = dim (F) dim (V) = 1 \cdot dim (V) = dim (V)

which comes straight from our Lemma 3.61 from our book. It's already assumed that $V$ is a vector space, and we've proved previously that $L (F, V)$ is a vector space.
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