Math 217: §2.4 Invertible linear maps and matrices Professor Karen Smith

φ Definitions: Let X and Y be any sets, and X −→ Y any mapping. φ The mapping X −→ Y is injective if for all y ∈ Y , there is at most one x ∈ X such that φ(x) = y. Put differently, no two x in the source map to the same y in the target under φ. φ The mapping X −→ Y is surjective (or onto) if for all y ∈ Y , there is some x ∈ X such that φ(x) = y. φ The mapping X −→ Y is invertible (or bijective) if for each y ∈ Y , there is a unique x ∈ X such that φ(x) = y. ψ When φ is invertible, we can define the inverse mapping Y −→ X to be the map sending each y to that unique x with φ(x) = y.

A. For each linear mapping below, consider whether it is injective, surjective, and/or invertible. If it is invertible, give the inverse map.

3 3 1. The linear mapping R → R which scales every vector by 2.

1 Solution note: This is surjective, injective, and invertble. The inverse scales by 2 .  1  2 0 0 1 The matrix of the inverse is 0 2 0. 1 0 0 2

3 3 2. The linear mapping R → R which rotates every vector by θ around the x-axis. Solution note: Invertible (hence surjective and injective). The inverse rotates by −θ.

2 2 3. The mapping R → R defined by projection onto a line L. Solution note: Not surjective, since the image is the line L. Not injective, since all points on a given line perpendicular to L have the same image. Not invertible.

2 2 4. The mapping R → R defined by reflection over the line L. Solution note: This is invertible (so injective and surjective). It is its own inverse!

1 5 5. The shear 2 → 2 defined by multiplication by the matrix . What is the matrix of R R 0 1 the inverse map?

1 −5 Solution note: Invertible. The inverse is . 0 1

n m B. The point is: For a linear transformation φ : R → R we can say • φ is surjective if and only if for all ~y in the target, the equation φ(~x) = ~y has at least one solution, and

• φ is injective if and only if for all ~y in the target, the equation φ(~x) = ~y has at most one solution, and • φ is bijective if and only if for all ~y in the target, the equation φ(~x) = ~y has exactly one solution.

1. Rephrase the three bullet points above in terms of solving a system of linear equations in- volving the matrix of φ. Solution note: Let A be the matrix of T . Then T is surjective if and only if for n all ~y ∈ R , the system of linear equations A~x = ~y has at least one solution (ie. is n consistent). Then T is injective if and only if for all ~y ∈ R , the system of linear equations A~x = ~y has at most one solution. Then T is invertle if and only if for all n ~y ∈ R , the system of linear equations A~x = ~y has exactly one solution.

1 4 1 3 3 2. Let T : R → R be the linear transformation given by left multiplication by 0 1 1 . Use 0 1 1 0 row-reduction to determine whether or not there is an vector ~x such that T (~x) = 2 . 1   x1 Solution note: We want to know whether or not there is an ~x = x2 such that x3     x1 0 T (x2) = 2 . That is, we want to know if the system x3 1     x1 0 A x2 = 2 x3 1

1 4 1 0 has a solution or not. The augmented matrix is 0 1 1 2 , which can be row- 0 1 1 1 1 4 1 0  reduced to 0 1 1 2  by replacing row 3 by ”row 3 minus row 2”. This is not yet 0 0 0 −1 in row reduce echelon form, but already we see that we have an inconsistent system, since the last row stands for the equation 0x1 + 0x2 + 0x3 = −1, which obviously has no solutions. So there is no solution to this system, and this means there is no ~x such 0 0 that T (~x) = 2 . This means that 2 is not in the image of T and that T is not 1 1 surjective.

  x1 + x2 + x3 3 3 C. Consider T : R → R given by T (~x) =  x2 . x1 + x3 1. Is T linear? If so, what is its matrix A? Why do you think some people call the matrix the “coefficient matrix” of T ? 1 1 1 Solution note: Yes, T is linear. Its matrix is A = 0 1 0 , which is the matrix of 1 0 1 coefficients of linear forms defining the map.

2. What are the source and target of T ?

3 Solution note: The source and target space are all of R .

3 3. Explain why ~b ∈ R is in the image of T if and only if the system of linear equations A~x = ~b is consistent. Solution note: This is really just repeating what we did in problem B. Again: Suppose 3 3 that ~b ∈ R is in the image of T . This means that there is some ~x in the source R     x1 + x2 + x3 b1 ~ ~ such that T (~x) = b. For that ~x, we have  x2  = b2 = b showing that x1 + x3 b3 ~x is a solution to A~x = ~b. Conversely, if A~x = ~b is consistent, then there is vector ~x     x1 + x2 + x3 b1 ~ ~ which satisfies  x2  = b2 showing that A~x = T (~x) = b, so b is in the x1 + x3 b3 image of T .

3 4. The image of T is a plane of R . Find the equation of this plane. Also, write the image in set builder notation. Hint: Use what you know about solving linear equations and row reduction.

3 Solution note: The image is any vector T (~x) in R ; this will be any vector of the  T form ~y = x1 + x2 + x3 x2 x1 + x3 . Thus the image will be any ~y for which we can find an x1, x2 and x3 solving this equality. Row reducing the augmented matrix, we have     1 1 1 y1 1 0 1 y1 − y2 0 1 0 y2 → 0 1 0 y2  , 1 0 1 y3 0 0 0 −y1 + y2 + y3 3 so the image is {~y ∈ R y3 = y1 − y2}. Geometrically, this is a plane through the 3 tr origin in R with normal vector [1 − 1 − 1] .

5. Is T is invertible? Is its matrix invertible? Explain. Also, without computing, why is the rank less than 3? Solution note: No! We just verified that the image of the map T is a plane, which is 3 NOT the whole target R . Any ~y in the target which does not satisfy y3 6= y1 − y2 will not be hit by T , that is, we have no solution to T (~x) = ~y. So T is not surjective, hence it can’t be invertible (since invertible means SURJECTIVE AND INJECTIVE. 0 1 Alternatively, we can also say that T is not injective: note that T (0) = T (0) = 0 1 0 0 . So there are two different vectors taken to the same vector by T , and it is not 0 injective. Hence T is not invertible. So also the matrix of T is not invertible. Its rank can not be three, because if the rank of a 3 × 3 matrix is 3, then it is invertible. 

D. 1. True or False: A linear transformation is invertible if and only its matrix is invertible. Explain. 2. Explain how to find the inverse of A using row reduction (or to tell that no such inverse exists). 1 2 4 3. Demonstrate your technique by finding the inverse of 0 1 2 . 0 0 π

n n Solution note: 1. TRUE. Suppose φ : R → R is invertible, with matrix A. Let n n φ : R → R be the inverse, with matrix B. We know that the composition φ ◦ ψ and ψ ◦ φ are both the identity maps. On the other hand, the compositions have matrices AB and BA respectively, so A is invertible, with inverse B. Conversely, if A is the matrix of φ and we know that A is invertible. Then the map ψ given by left multiplication by the inverse matrix B is the inverse of φ. Indeed, the compositions will have matrix AB and BA which are both the identity matrix, hence define the identity map. 2). To find the inverse of an n × n matrix A we form the n × 2n matrix [AIn] and row reduce until we have either a row of zeros in the first n × n spot (in which case A is not invertible) OR the identity matrix, in which case the second n × n matrix will be A−1. See the book: Theorem 2.4.5. 3) Row reduce: 1 2 4 1 0 0 0 1 2 0 1 0 . 0 0 π 0 0 1 We get 1 0 0 1 −2 0  0 1 0 0 1 −2/π . 0 0 1 0 0 1/π 1 −2 0  So the inverse is 0 1 −2/π . 0 0 1/π