Raising the bar

Is the inverse of a linear mapping linear?

Orthonormal basis

Exercise 1. I) Let $latex e_j$ denote unit vectors. They possess properties

(1) $latex e_i\cdot e_j=0$ for all $latex i\neq j,$ $latex \left\Vert e_{i}\right\Vert =1$ for all $latex i.$

II) For any $latex x\in R^n$ we have the representation

(2) $latex x=\sum_jx_je_j.$

III) In (2) the coefficients can be found as $latex x_{j}=x\cdot e_{j}:$

(3) $latex x=\sum_j(x\cdot e_j)e_j.$

Proof. I) (1) is proved by direct calculation. II) To prove (2) we write

$latex x=(x_1,...,x_n)=x_1(1,0,...,0)+...+x_n(0,...,0,1)=\sum_jx_je_j.$

III) If we have (2), then it's easy to see that by (1) $latex x\cdot e_i=\sum_jx_j(e_j\cdot e_i)=x_i.$

Definition 1. Any system of vectors that satisfies (1) is called an orthonormal system. An orthonormal system is called complete if for any $latex x\in R^n$ we have the decomposition (3). Exercise 1 shows that our system of unit vectors is complete orthonormal. A complete orthonormal system is called an orthonormal basis.

Analyzing a linear mapping

Exercise 2. Let $latex A$ be a matrix of size $latex n\times k.$ Suppose you don't know the elements of $latex A$ but you know the products $latex (Ax)\cdot y$ for all $latex x,y.$ How can you reveal the elements of $latex A$ from $latex (Ax)\cdot y?$ How do you express $latex Ax$ using the elements you define?

Solution. Let us partition $latex A$ into rows and suppose the elements $latex a_{ij}$ are known. Let us try unit vectors as $latex x,y:$

(4) $latex (Ae_j)\cdot e_i=\left(\begin{array}{c}A_1e_j \\... \\A_ne_j\end{array}\right)\cdot e_i=A_ie_j=a_{ij}.$

Using (2) and (4) one can check that $latex (Ax)\cdot e_i=A_ix=\sum_ja_{ij}x_j.$ Hence, from (3) we have the answer to the second question:

(5) $latex Ax=\sum_i[\left(Ax\right)\cdot e_i]e_i=\sum_i\left(\sum_ja_{ij}x_j\right)e_i=\sum_{i,j}x_ja_{ij}e_i.$

The above calculation means that when $latex a_{ij}$ are unknown, we can define them by $latex a_{ij}=(Ae_j)\cdot e_i$ and then the action of $latex A$ on $latex x$ will be described by the last expression in (5).

We know that a mapping generated by a matrix is linear. The converse is also true: a linear mapping is given by a matrix:

Exercise 3. Suppose a mapping $latex f:R^k\rightarrow R^n$ is linear: $latex f(ax+by)=af(x)+bf(y)$ for any numbers $latex a,b$ and vectors $latex x,y.$ Then there exists a matrix $latex A$ of size $latex n\times k$ such that $latex f(x)=Ax$ for all $latex x.$

Proof. Based on (4) in our case put $latex a_{ij}=f(e_j)\cdot e_i$. Applying (3) to $latex f(x)$ we get

(6) $latex f(x)=\sum_i[f(x)\cdot e_i]e_i$

(the summation index $latex j$ is replaced by $latex i$ on purpose). Plugging (2) in (6)

$latex f(x)=\sum_i\left[f\left(\sum_jx_je_j\right)\cdot e_i\right]e_i=$ (both $latex f$ and scalar product are linear)

$latex =\sum_i\left[\left(\sum_jx_jf\left(e_j\right)\right)\cdot e_i\right]e_i=\sum_{i,j}x_j(f\left(e_j\right)\cdot e_i)e_i$$latex =\sum_{i,j}x_ja_{ij}e_i=Ax.$

The last equation is the definition of $latex A$.

Exercise 4. An inverse of a linear mapping is linear (and given by a matrix by Exercise 3).

Proof. Let $latex f(x)=Ax$ be a linear mapping and suppose its inverse $latex f^{-1}$ in the general sense exists. Then $latex f^{-1}(Ax)=x$ for all $latex x.$ Let us put $latex x=ay+bz$ for arbitrary numbers $latex a,b$ and vectors $latex y,z.$ Then we have $latex f^{-1}(A(ay+bz))=ay+bz$ or, using linearity of $latex A,$

$latex f^{-1}(aAy+bAz)=ay+bz.$

Putting $latex Ay=u,$ $latex Az=v$ we get

$latex f^{-1}(au+bv)=af^{-1}(u)+bf^{-1}(v).$

Thus, $latex f^{-1}$ is linear.

Remark. In all of the above it is important that $latex e_j$ are unit vectors. For a different basis, the results drastically change.

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