Lecture 20 - Invertibility (cont.)

From Lecture 18 - Finishing Matrices, Starting Invertibility#^a50ea0, any $T \in L (V, W)$ is invertible implies that there is some transformation $T^{- 1} : W \to V$ that exists where $T \circ T^{- 1} I_{W}, T^{- 1} \circ T = I_{V}$ , and that $T$ is invertible iff $T$ is bijective.

We'll define the idea of isomorphism now.

isomorphism, isomorphic

We say that $V$ and $W$ are isomorphic iff there exists an invertible linear map $T \in L (V, W)$ . Such a map $T$ is called an isomorphism.

A little bit about this word, as it'll come around in abstract algebra a lot. The idea here is that the vectors in each are of the same structure. They just have different labels. For instance, notice that:

a_{0} + a_{1} x + a_{2} x^{2}

If we define our map where:

T (a_{0} + a_{1} x + a_{2} x^{2}) = [\begin{matrix} a_{0} \\ a_{1} \\ a_{2} \end{matrix}]

Notice that $T$ is linear, injective, and surjective, so $T$ is invertible and thus $P_{2} (R)$ and $R^{3}$ is an isomorphism. So these sets are algebraically the same. The vectors only differ in the way that they look.

But notice in our example that the dimensions of $V$ and $W$ are of the same dimension. It turns out that if their dimensions are the same, then they are isomorphic.

Dimension dictates isomorphism

If $V, W$ are finite-dimensional vector spaces. Then $V, W$ are isomorphic iff $dim (V) = dim (W)$ .

The proof is how you expect. See Chapter 3 - Linear Maps#^3e5a28 for more information on the proof.

An Example

Let $R^{2}$ be our vector space. Let $T_{θ}$ be a rotation CCW by some $θ$ radians. Let $B$ be the standard basis for $R^{2}$ . Let's find $M (T_{θ}, B)$ .

Here:

T (\vec{e_{1}}) = \cos (θ) \vec{e_{1}} + \sin (θ) \vec{e_{2}}

and:

T (\vec{e_{2}}) = - \sin (θ) \vec{e_{1}} + \cos (θ) \vec{e_{2}}

thus:

M (T_{θ}, B) = [\begin{matrix} \cos (θ) & - \sin (θ) \\ \sin (θ) & \cos (θ) \end{matrix}]

notice that for any $α \in R$ :

T_{θ} \circ T_{α} = T_{α} \circ T_{θ} = T_{θ + α}

but we can check this with the matrices themselves:

[\begin{matrix} \cos (θ) & - \sin (θ) \\ \sin (θ) & \cos (θ) \end{matrix}] [\begin{matrix} \cos (α) & - \sin (α) \\ \sin (α) & \cos (α) \end{matrix}] = [\begin{matrix} \cos (θ) \cos (α) - \sin (θ) \sin (α) & . . . \\ \sin (θ) \cos (α) - \cos (θ) \sin (α) \end{matrix}]

and the RHS is:

[\begin{matrix} \cos (θ + α) & - \sin (θ + α) \\ \sin (θ + α) & \cos (θ + α) \end{matrix}]

so then equating the matrices then we get various trig identities we've used before!

An Aside

Think about $M (T_{θ})$ . It takes as input some $T$ and outputs a matrix in $F^{m, n}$ .

Theorem

Suppose $V, W$ are finite-dimensional vector spaces. Then: $L (V, W)$ and $F^{dim (V), dim (W)}$ are isomorphic.

This is easy to prove. We just need a ibjective map that takes in a $T$ and gives us some $F^{m, n}$ . That's $M$ !

Proof
Fix some basis for $V$ , say $v_{1}, . . ., v_{n}$ and a basis $w_{1}, . . ., w_{m}$ for $W$ . Call the former $B_{V}$ and the latter $B_{W}$ . We'll show that $M$ defined by:

M (T) = M (T, B_{V}, B_{W})

is an isomorphism. We can first check this is true via injective. If the matrix sends anything to $\vec{0}$ then $T$ must have a zero-vector nullspace, showing injectivity. Surjectivity can be shown by showing that the matrix map always has $W = range (T)$ .

Due to the isomorphism and knowing the dimension for our matrix, we can say they are isomorphic.
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Therefore, then:

dim (L (V, W)) = dim (F^{dim (V), dim (W)}) = (dim (V)) (dim (W))

Notice that since the set of linear transformations is finite-dimensional, then there's some basis of all linear transformations between $V$ and $W$ . Our rotation matrices come from just setting $\cos (θ)$ or similar in one entry, and 0's everywhere else.