Math 308 Discussion Problems #4 Chapter 4 (after 4.3)

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1 Math 38 Discussion Problems #4 Chapter 4 (after 4.3) () (after 4.) Let S be a plane in R 3 passing through the origin, so that S is a two-dimensional subspace of R 3. Say that a linear transformation T : R 3 R 3 is a reflection about S if T (v) = v for any vector v in S and T (n) = n whenever n is perpendicular to S. Shown: Solid arrows v S, w a generic vector, n perpendicular to S, and the reflections of each vector as dashed arrows. (Note that T ( v) = v and T ( n) = n.) Let T be the linear transformation given by T (x) = Ax, where A is the matrix This linear transformation is the reflection about a plane S. Find a basis for S. Solution. From the setup, S is the set of vectors v satisfying the equation A v = v. Using Method 2 from Worksheet 3, Problem #6, we can write this as A v = I v (A I) v = where I is the 3 3 identity matrix. This shows that for every vector v in S, (A I) v =, i.e. S is the null-space of the matrix A I = R 3+R 2 R 3 R 2R 2 R From this reduced echelon form, we see that S is the set of solutions of 2x + x 2 x 3 =. To obtain a basis we solve the equation, for example setting x 2 = s, x 3 = t to we find that v = s + t 2 2 s = s t t.

2 Thus, a basis for S = null(a I) is the pair of vectors 2 2,. (2) (after 4.2) (Geometry Question) In this problem we continue the Geometry Problem from Chapter 2, but now we work in R 3. Consider the infinite linear system given by the equations ax + by = where you should think of these having a z variable with zero coefficient. (a) Describe the solution space of the above system. Solution. Each linear equation represents a plane in R 3. One can see directly that each plane contains the z-axes, i.e. the line x = y =, since for every z the point is a solution of every equation ax + by =. Moreover, for any point not z in the z-axis, you can find a value of a and b such that the point does not satisfy the equation ax + by =. This shows that the solution space of the above system consists exactly of the z-axis. (b) How many linearly independent solutions are there in this solution space? (i.e., what is the dimension of this solution space?) Solution. All the vectors in the solution space are linearly dependent since they all live in the z-axis. This implies that there is at most linearly independent solution. (c) Write down a basis of the solution space. Solution. We already know that the solution space is the z-axis therefore a basis will be given by e 3. (d) Express this solution space as the kernel of a finite matrix. What is the smallest size matrix that will do the job? Solution. We observe that given any two equations ax + by = and cx + dy = that represent different planes, i.e. such that (c, d) is not a multiple of (a, b), the set of common solutions is precisely the z-axis; you can for example reduce to echelon form the corresponding augmented matrix and see that the system is equivalent to x = and y =. This shows that the solution space can be written as the kernel of any 2 3 matrix whose third column contains only zeros. Two examples are ( ) ( 2 2 This is the smallest size for a matrix to have the desired kernel. If you consider a matrix with only one row, its kernel will be a subspace of R 3 of dimension at least 2 (since the rank can be at most ). (e) If we keep on doing this example in higher and higher dimensional space, what happens to the dimension of the solution space? Solution. The equation ax + bx 2 = in R n will have n 2 free variables and it will be always consistent since the zero vector is always a solution. Moreover each single equation has a solution space that contains S, the span of {e 3,..., e n } since every vector with x = x 2 = will always be a solution. This shows that the solution set of the infinite system will coincide precisely with S. In particular it will )

3 have dimension n 2. This agrees with the three dimensional case in which the solution space has dimension 3 = 2, and with the two dimensional case in which the solution space consists of a point, and therefore has dimension 2 2 =. (3) (after 4.2) Expand the set w + x + y + z =., to be a basis for the subspace Solution. The subspace S is a 3-dimensional plane in R 4 (the system of equations defining S is automatically in echelon form since it only has equation; we see right away that it has 3 free variables). So, a basis for S must consist of three vectors. We already have two vectors which are clearly linearly independent. Therefore we just need one more vector in S that isn t in the span of the first two. We can find vectors in S by plugging in simple values (like zero) for some of our free variables: w x y = z,,... The second one is actually one of the given basis vectors. So, let s try one of the others. We just need to check that our overall set of vectors is linearly independent:. Since every column of the echelon form has a leading term, the three original vectors are linearly independent. Therefore (since we know S has dimension 3) they are a basis for S. (4) (after 4.3) Find an invertible n n matrix A and an n n matrix B such that rank(ab) rank(ba), or explain why such matrices cannot exist. Solution. To solve this problem we will use the correspondence between matrices and linear transformation. Recall that to any n n matrix M we can associate a linear transformation T M : R n R n defined by T M ( x) = M x. Then we can relate rank and nullity of the matrix M to kernel and image of the linear transformation. In particular the nullity of M corresponds to the dimension of the kernel of T M and the column space of M corresponds to the image of T M. With this equivalence we have that the rank of M corresponds to the dimension of the image of T M. Observe that in general AB BA so we might expect that the rank of AB and the rank of BA will be different without any assumptions on A and B Now consider T A, T B : R n R n the linear transformations associated to A and B respectively, i.e. T A ( x) = A x and T B ( x) = B x. Then the linear transformation associated to AB is T AB = T A T B, i.e. the linear transformation defined by T AB ( x) = T A (T B ( x)). Similarly the linear transformation associated to BA is T BA = T B T A. We will show first that the linear transformation T AB has the same kernel as T B. For this notice that since A is invertible, it has nullity and therefore T A is injective, i.e. T A ( x) = only if x =. Therefore T AB ( x) = T A (T B ( x)) = if and

4 only if T B ( x) =, i.e. x is in the kernel of T B (i.e. the null space of B). Since rank +nullity = n we see that rank(ab) = rank(b). Now consider the image of T BA. Since A is invertible we have that T A is invertible, in particular it is surjective. This implies that the image of T A (i.e. the span of the columns of A) is the whole R n. Now what is the image of T BA? This is equivalent to ask for which vectors w R n there exists x R n such that T BA ( x) = w. If w is in the image of T B, by definition of the image, there exists a vector v R n such that T B ( v) = w. On the other hand the fact that T A is surjective means that for every vector v R n there exists a vector x R n such that T A ( x) = v. If we combine these two things we see that for every vector w in the image of T B there exists a vector x R n such that T BA ( x) = w. This shows that the dimension of the image of T BA is equal to the dimension of the image of T B, so in particular rank(ba) = rank(b). This shows that rank(ab) = rank(b) = rank(ba) so for any choice of A and B where A is invertible, the two ranks coincide. In particular there cannot exist a n n invertible matrix A and a n n matrix B such that rank(ab) rank(ab). CAUTION: this is not true if we don t assume A is invertible. For example for this choice of A and B ( ) ( ) A = B = then AB is the zero matrix while BA is not zero, so their ranks are different. However, in this case neither A or B are invertible. (5) (after 4.3) Find a 3 4 matrix A with nullity 2 and with col(a) = span, 4 3, , or explain why such a matrix can t exist. Solution. By the rank-nullity theorem, A will have to have rank 2. Since there are three vectors given, this seems unlikely to be possible, but it will be possible if the three given vectors are linearly dependent. Let s check: By the columns method from class, this means that the first two vectors are linearly independent, and the third one is a linear combination of the first two. So, to find A, we should pick four columns in the subspace we ve described (so that the resulting columns still span this subspace and nothing more). We should make sure to include two linearly independent columns, so that col(a) is not smaller than desired. One easy option is A = An even simpler one is A =

5 A more complicated solution would be to take the fourth column to be another linear combination of the two basis vectors we found. For instance, their sum: A = (6) (after 4.3) Find a 3 3 matrix A and a 3 3 matrix B, each with nullity, such that AB is the matrix, or explain why such matrices cannot exist. Solution. We will follow the following steps: (a) Let A, B be matrices (not necessarily 3 3). Then AB = if and only if col(b) null(a). To show an if and only if statement, we show both implications: ( ): Suppose AB =. We have to show that, if v is any vector from col(b), then v null(a). If the columns of B are b,..., b m, a vector in col(b) is always of the form v = x b + + x m bm = B x. To show that v null(a), we have to show that A v =. Since v = B x, we get: A v = AB x = (matrix) x =. Since A v =, we see that v null(a). ( ): Suppose col(b) null(a). Then we know A b i = for every column b i of B. To show that AB =, we use the definition of matrix multiplication: AB = A b A b 2... A b m = =. Now suppose A, B are size 3 3. (b) Construct an example where A, B both have rank, and AB =. There are many examples. To make A have rank, we can make all its rows (or columns) be proportional. For instance, A = 3 3 has rank. We should repeat this for B, but when picking the columns, we should make sure that col(b) null(a). For instance, we can see directly that null(a), so we could just set B =.