Identity matrix and Kronecker delta

$$ \begin{align*} I_n &= \begin{pmatrix} 1 & 0 & \dotsb & 0 \\ 0 & 1 & \dotsb & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 1 \end{pmatrix} \in M_{n \times n}. \end{align*} $$

$I_n$ is the $n \times n$ identity matrix. To construct the matrix, we have the following rule.

$$ \begin{equation*} (I)_{ij} = \delta_{ij} = \begin{cases} 1 \quad \text{if } i = j \\ 0 \quad \text{if } i \neq j \end{cases} \end{equation*} $$

$\delta_{ij}$ is the Kronecker delta. For any matrix $A \in M_{m \times n}$, $AI_n = A$ and $I_mA = A$.
Exercise: Suppose we have a finite basis $\beta$ for $V$, then the identity map is

$$ \begin{align*} I: \ &V \rightarrow V \\ &x \rightarrow x \end{align*} $$

If we compute its matrix representative, we will see that $[I_V]_{\beta}^{\beta} = I_n$.

Inverse Linear Transformation

Next we define the inverse of a linear transformation

Definition

An inverse of $T: V \rightarrow W$ is a map $S: W \rightarrow V$ such that $$ \begin{align*} S \circ T = I_V \text{ and } T \circ S = I_W \end{align*} $$

Example 1

$$ \begin{align*} T : \ &\mathbf{R}^2 \rightarrow \mathbf{R}^2 \\ &(x, y) \rightarrow (y, -x) \end{align*} $$

has an inverse

$$ \begin{align*} S : \ &\mathbf{R}^2 \rightarrow \mathbf{R}^2 \\ &(x, y) \rightarrow (-y, x) \end{align*} $$

We can check that this is true by composing these transformations and check that we get the identity map.

$$ \begin{align*} S \circ T (x,y) &= S(T(x, y)) = S(y, -x) = (x, y) \\ T \circ S (x,y) &= T(S(x, y)) = T(-y, x) = (x, y) \end{align*} $$

Example 2

Suppose we have the following vector spaces:

$$ \begin{align*} V &= P \\ W &= \hat{P} = \{a_1x + a_2x^2 ... + a_kx^k\} \end{align*} $$

$\hat{P}$ is the set of all polynomials without the constant term $a_0$. We claim that $\hat{P}$ is a vector space. The easiest way to verify this to show that $\hat{P}$ is a subspace of $P$. Now consider the following linear map between $P$ to $\hat{P}$ where we multiply the polynomial by $x$.

$$ \begin{align*} T : \ &P \rightarrow \hat{P} \\ &a_0 + a_1x + ... + a_kx^k \rightarrow a_0x + a_1x^2 + ... + a_kx^{k+1} \\ &f \rightarrow xf \end{align*} $$

This map has the following inverse linear transformation:

$$ \begin{align*} S &: \hat{P} \rightarrow P \\ &a_1x + a_2x^2 + ... + a_kx^{k} \rightarrow a_1 + a_2x + ... + a_kx^{k-1} \\ &f \rightarrow \frac{1}{x}f \end{align*} $$

Example 3

The following linear transformation

$$ \begin{align*} T &: P \rightarrow P \\ &f \rightarrow f' \end{align*} $$

has no inverse! (it is onto but not 1-1). Not one-to-one since the derivative of any constant function is 0.

This leads us to the following theorem:

Theorem

Let $T: V \rightarrow W$ be linear.

$T$ has an inverse if and only if $T$ is 1-1 ($N(T)=\{\bar{0}_V\}$) and onto ($R(T) = W$).
If $T$ is invertible, then its inverse is unique $(T^{-1})$.
$T^{-1}$ is linear. (Theorem 2.17 in the book)

The third property is not obvious and requires a proof.

Proof: Let $T: V \rightarrow W$ be a linear map with inverse $T^{-1}: W \rightarrow V$. We want to show that $T^{-1}$ is linear. To do this, we need to show that

$$ \begin{align*} T^{-1}(w_1 + cw_2) = T^{-1}(w_1) + cT^{-1}(w_2). \end{align*} $$

Because $T$ has an inverse then $T$ is onto. This means that $w_1 = T(v_1)$ and $w_2 = T(v_2)$ for some $v_1, v_2 \in V$. Moreover, since $T$ is 1-1, then $v_1 \neq v_2$. Now,

$$ \begin{align*} T^{-1}(w_1 + cw_2) &= T^{-1}(T(v_1) + cT(v_2)) \\ &= T^{-1}(T(v_1 + cv_2)) \text{ (because $T$ is linear!)} \\ &= I(v_1 + cv_2) \text{ (because $T$ has an inverse!)} \\ &= v_1 + cv_2 \\ & = T^{-1}(w_1) + c T^{-1}(w_2) \end{align*} $$

Therefore, $T^{-1}$ is linear. $\blacksquare$

Theorem

Suppose $T: V \rightarrow W$ be linear and invertible. If $\beta$ is a basis for $V$, then $T(\beta)$ is a basis for $W$.

Proof (in the case where $V$ and $W$ are finite dimensional spaces):
Let $T: V \rightarrow W$ be a linear and invertible map and suppose that $\dim(V)=n$. Choose a basis $\beta = \{v_1, ..., v_n\}$ for $V$. Consider the set of images,

$$ \begin{align*} T(\beta) = \{T(v_1),...,T(v_n)\}. \end{align*} $$

We need to show that this set is a basis which means that we need to show that it spans $W$ and so $Span(T(\beta)) = W$ and that it is a linearly independent set. To show that it spans $W$, we need for any $w \in W$, scalars $a_1,...,a_n$ such that

$$ \begin{align*} w = a_1T(v_1) + ... + a_nT(v_n). \end{align*} $$

But we know that $w = T(v)$ for some $v \in V$ since $T$ is onto. We also know that $\beta$ is a basis for $V$ and so we can write $v$ as a linear combination of the vectors in $\beta$ for some scalars $a_1,...,a_n$.

$$ \begin{align*} v = a_1v_1 + ... + a_nv_n. \end{align*} $$

And now because $T$ is linear, we can do the following

$$ \begin{align*} w &= T(v) \\ &= T(a_1v_1 + ... + a_nv_n) \\ & = a_1T(v_1) + ... + a_nT(v_n). \end{align*} $$

Which is what we wanted to show. To show that the vectors in $T(\beta)$ are linearly independent, we need to show that the solution to

$$ \begin{align*} a_1T(v_1) + ... + a_nT(v_n) = \bar{0}_W. \end{align*} $$

is the trivial solution which means that $a_1=0,..,a_n=0$. We can use the linearity of $T$ to show that

$$ \begin{align*} \bar{0}_W &= a_1T(v_1) + ... + a_nT(v_n) \\ &= T(a_1v_1 + ... + a_nv_n). \end{align*} $$

But $T$ is 1-1 and so this implies that $a_1v_1 + ... + a_nv_n$ must be $\bar{0}_V$. Therefore, we must have $a_1=0,...,a_n=0$ as required.

Corollary

Let $T: V \rightarrow W$ be linear. If $\dim(V) = n$ and $T$ is invertible, then $$ \begin{align*} \dim W = \dim V = n \end{align*} $$

(Study Notes:) This is really important. $T$ is invertible and $V$ having dimension $n$ means that $W$ has dimension $n$. Also $T$ is one-to-one and onto. Later we’ll learn if both $V$ and $W$ have the same dimension and $T$ is one-to-one, then this is sufficient to conclude that $T$ is invertible. (next lecture).

I also went through the proof in the book for this corollary. See This

Isomorphism

Definition

$V$ and $W$ are isomorphic if there is an invertible linear map $T: V \rightarrow W$. Such a map $T$ is called isomorphism.

Example 4

$\mathbf{R}^3$ and $P_2$ are isomorphic. To see this, we need an invertible map from one to the other. The maps

$$ \begin{align*} T : \ &\mathbf{R}^3 \rightarrow P_2 \\ &(a_1,a_2,a_3) \rightarrow a_1 + a_2x + a_3x^2 \end{align*} $$

and

$$ \begin{align*} U : \ &\mathbf{R}^3 \rightarrow P_2 \\ &(a_1,a_2,a_3) \rightarrow a_3 + (a_1 + a_2)x + (a_1 + a_2)x^2 \end{align*} $$

are both isomorphisms.

Example 5

$$ \begin{align*} U : \ &P \rightarrow \hat{P} \\ &f \rightarrow xf \end{align*} $$

is an isomorphism and so $P$ and $\hat{P}$ are isomorphic.

Criterion for Isomorphic Finite Dimensional Vector Spaces

Theorem

If $V$ is finite dimensional, then $W$ is isomorphic to $V$ if and only if $\dim W = \dim V$.

Proof:
$\Rightarrow$: If $W$ is isomorphic to $T$, then there exists an isomorphism map $T$. $T$ is onto and one-to-one because it is invertible. Therefore, $\dim(W) = \dim(V)$ by the corollary we stated earlier.

$\Leftarrow$: Suppose that $\dim V = \dim W$. We want to show that they are isomorphic. This means that there exists some invertible map from one to the other. So let $\beta = \{v_1,...,v_n\}$ be a basis for $V$ and $\alpha = \{w_1, ...,w_n\}$ be a basis for $W$.

Define the map $T: W \rightarrow V$ by $[T]_{\alpha}^{\beta} = I_n$. This $T$ works. Why? This $T$ satisfies:

$$ \begin{align*} T(w_i) = v_i. \end{align*} $$

More generally,

$$ \begin{align*} T(w) &= T(a_1w_1 + ... + a_nw_n) \\ &= a_1v_1 + ... + a_nv_n. \end{align*} $$

References

Math416 by Ely Kerman