2/21/2002, IMAGE AND KERNEL Math 21b, O. KnillHOMEWORK: 3.1 10,22,34,38*,44,48*IMAGE. If T : Rn→ Rmis a linear transformation, then {T (~x) | ~x ∈ Rn} is called the image of T . IfT (~x) = A~x, then the image of T is also called the image of A. We write im(A) or im(T ).EXAMPLES.1) If T (x, y, z) = (x, y, 0), then T (~x) = Axyz=1 0 00 1 00 0 0xyz. The image of T is the x − y plane.2) If T (x, y)(cos(φ)x − sin(φ)y, sin(φ)x + cos(φ)y) is a rotation in the plane, then the image of T is the wholeplane.3) If T (x, y, z) = x + y + z, then the image of T is R.SPAN. The span of ~v1, . . . , ~vkare vectors in Rnis the set of all combinations c1~v1+ . . . ck~vk.PROPERTIES.The image of a linear transformation ~x 7→ A~x is the span of the column vectors of A.The image of a linear transformation contains 0 and is closed under addition and scalar multiplication.KERNEL. If T : Rn→ Rmis a linear transformation, then the set {x | T (x) = 0 } is called the kernel of T .If T (~x) = A~x, then the kernel of T is also called the kernel of A. We write im(A) or im(T ).EXAMPLES. (The same examples as above)1) The kernel is the z-axes. Every vector (0, 0, z) is mapped to 0.2) The kernel consists only of the point (0, 0, 0).3) The kernel consists of all vector (x, y, z) for which x + y + z = 0. The kernel is a plane.PROPERTIES.The kernel of a linear transformation contains 0 and is closed under addition and scalar multiplication.IMAGE AND KERNEL OF INVERTIBLE MAPS. A linear map ~x 7→ A~x, Rn7→ Rnis invertible if and onlyif ker(A) = {~0} if and only if im(A) = Rn.HOW DO WE COMPUTE THE IMAGE? The rank of rref(A) is the dimension of the image. The columnvectors of A span the image.EXAMPLES. (The same examples as above)1)100and010span the image.2)cos(φ)− sin(φ)andsin(φ)cos(φ)span the image.3) The 1D vector1 spans theimage.HOW DO WE COMPUTE THE KERNEL? Just solve A~x =~0. Form rref(A). For every column withoutleading 1 we can introduce a free variable si. If ~x is the solution to A~xi= 0, where all sjare zero except si= 1,then ~x =Pjsj~xjis a general vector in the kernel.EXAMPLE. Find the kernel of the linear map R3→ R4, ~x 7→ A~x with A =1 3 02 6 53 9 1−2 −6 0. Gauss-Jordanelimination gives: B = rref(A) =1 3 00 0 10 0 00 0 0. We see one column without leading 1 (the second one). Theequation B~x = 0 is equivalent to the system x + 3y = 0, z = 0. After fixing z = 0, can chose y = t freely andobtain from the first equation x = −3t. Therefore, the kernel consists of vectors t−310. In the book, youhave a detailed calculation, in a case, where the kernel is 2 dimensional.domaincodomainkernelimageWHY DO WE LOOK AT THE KERNEL?• Useful to understand linear maps. To which de-gree are they non-invertible.• Helpful to understand quantitatively to whichextent linear equations can not be solved. IfAx = b and y is in the kernel, then also A(x +y) = b, so that x + y solves the system also.WHY DO WE LOOK AT THE IMAGE?• A solution Ax = b can be solved if and only if bis in the image of A.• Knowing about the kernel and the image is use-ful in the similar way that it is useful to knowabout the domain and range of a general mapand to understand the graph of the map.In general, the abstraction helps to understand topics like error correcing codes (Problem 53/54 in the book),where two matrices H, M with the property that ker(H) = im(M) appear. The encoding x 7→ M x is robustin the sense that adding an error e to the result M x 7→ Mx + e can be corrected: H(M x + e) = He allows tofind e and so M x. This allows to recover x = P M x with a projection P .PROBLEM. Find ker(A) and im(A) for the 4 × 3 matrix A = [5, 1, 4].ANSWER. A · ~x = A~x = 5x + y + 4z = 0 shows that the kernel is a plane with normal vector [5, 1, 4] throughthe origin. The image is the codomain, which is R.PROBLEM. Find ker(A) and im(A) of the linear map x 7→ v × x, (the cross product with v.ANSWER. The kernel consists of the line spanned by v, the image is the plane orthogonal to v.PROBLEM. Fix a vector w in space. Find ker(A) and image im(A) of the linear map from R6to R3given byx, y 7→ [x, v, y] = (x × y) · w.ANSWER. The kernel consist of all (x, y) such that their cross product orthogonal to w. This means that theplane spanned by x, y contains w.PROBLEM Find ker(T ) and im(T ) if T is a composition of a rotation R by 90 degrees around the z-axes withwith a projection onto the x-z plane.ANSWER. The kernel of the projection is the y axes. The x axes is rotated into the y axes and therefore thekernel of T . The image is the x-z plane.PROBLEM. Can the kernel of a square matrix A be trivial if A2= 0, where 0 is the matrix containing only 0.ANSWER. No: if the kernel were trivial, then A were invertible and A2were invertible and different from 0.PROBLEM. Is it possible that a 3 × 3 matrix A satisfies ker(A) = R3without A = 0?ANSWER. No, if A 6= 0, then A contains a nonzero entry and therefore, a column vector which is nonzero.PROBLEM. What is the kernel and image of a projection onto plane Σ : x − y + 2z = 0?ANSWER. The kernel consists of all vectors orthogonal to Σ, the image is the plane Σ.PROBLEM. Given two square matrices A, B and assume AB = BA. You know ker(A) and ker(B). What canyou say about ker(AB)?ANSWER. ker(A) is contained in ker(BA). Similar ker(B) is contained in ker(AB). Because AB = BA, thekernel of AB contains both ker(A) and ker(B). (It can be bigger: A = B =0 10 0.)PROBLEM. What is the kernel of the partitioned matrixA 00 Bif ker(A) and ker(B) are known.ANSWER. The kernel consists of all vectors (~x, ~y), where ~x in ker(A) and ~y ∈
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