26
Jul 18

Is the inverse of a linear mapping linear?

Is the inverse of a linear mapping linear?

Orthonormal basis

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

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

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

(2) x=\sum_jx_je_j.

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

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

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

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

Solution. Let us try unit vectors as x,y:

(4) (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}.

One can check that ( Ax)\cdot e_i=A_ix=\sum_ja_{ij}x_j. Hence, from (3) we have the answer to the second question:

(5) 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.

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

Proof. Based on (4) in our case put a_{ij}=f(e_j)\cdot e_i.

Applying (3) to f(x) we get

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

(the summation index j is replaced by i on purpose). Plugging (1) in (6)

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

=\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= (applying (5))

=\sum_{i,j}x_ja_{ij}e_i=Ax.

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

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

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

Putting Ay=u, Az=v we get

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

Thus, f^{-1} is linear.

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