Chapter SSM: Linear Algebra. 5. Find all x such that A x = , so that x 1 = x 2 = 0.

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1 Chapter Find all x such that A x : Chapter, so that x x ker(a) { } Find all x such that A x ; note that all x in R satisfy the equation, so that ker(a) R span( e, e ) 5 Find all x such that A x 5 ; x x ; x x x x x ker(a) span t t 7 Find all x such that A x Since rref(a) I we have ker(a) { } 9 Find all x such that A x Solving this system yields ker(a) { } Solving the system A x we find that ker(a) span Solving the system A x we find that ker(a) span,, 5 y Fact, the image of A is the span of the columns of A: ( ) im(a) span,,, t 6

2 Section Since any two of these vectors span all of R already, we can write ( ) im(a) span, ( ) 7 y Fact, im(a) span, R (the whole plane) 9 Since the four column vectors of A are parallel, we have im(a) span R y Fact, im(a) span, 7 9, 5 6 8, a line in It is hard to tell by inspection whether this span is a plane in R or all of R ; we need to find out whether the third column vector is a linear combination of the first two This shows that the third column vector is not contained in the span of the first two, so that im(a) R im(t ) R and ker(t ) { }, since T is invertible (see Summary 8) 5 im(t ) R and ker(t ) { }, since T is invertible (see Summary 8) 7 Let f(x) x x x(x ) x(x )(x + ) Then im(f) R, since lim f(x) and lim x f(x) x but the function fails to be invertible since the equation f(x) has three solutions, x,, and 9 Use spherical coordinates (see any good text on multivariable calculus): f 65 sin(φ) cos(θ) φ sin(φ) sin(θ) θ cos(φ)

3 Chapter The plane x + y + z is spanned by the two vectors and, for example Therefore, A does the job There are many other correct answers The plane is the kernel of the linear transformation T 5 kernel(t ) { x : T ( x) v x } the plane with normal vector v x y z x + y + z from R to R im(t ) R, since for every real number k there is a vector x such that T ( x) k, for example, x k v v v 7 A, A, A, so that ker(a) span( e ), ker(a ) span( e, e ), ker(a ) R, and im(a) span( e, e ), im(a ) span( e ), im(a ) { } 9 a If a vector x is in ker(), that is, x, then x is also in ker(a), since A( x) A( x) A : ker() ker(a) Exercise 7 (with A ) illustrates that these kernels need not be equal b If a vector y is in im(a), that is, y A x for some x, then y is also in im(a), since we can write y A( x): im(a) im(a) Exercise 7 (with A ) illustrates that these images need not be equal, so that ker(a) span a rref(a) im(a) span 6 8 span 66

4 Section Note that im(a) and ker(a) are perpendicular lines b A A If v is in im(a), with v A x, then A v A x A x v Figure : for Problem c c Any vector v in R can be written uniquely as v v + v, where v is in im(a) and v is in ker(a) (See Figure ) Then A v A v + A v v (A v v by part b, A v since v is in ker(a)), so that A represents the orthogonal projection onto im(a) span Using our work in Exercise as a guide, we come up with the following procedure to express the image of an n m matrix A as the kernel of a matrix : If rank(a) n, let be the n n zero matrix If r rank(a) < n, let be the (n r) n matrix obtained by omitting the first r rows and the first m columns of rrefa I n 5 As we solve the system A x, we obtain r leading variables and m r free variables The general vector in ker(a) can be written as a linear combination of m r vectors, with the free variables as coefficients (See Example, where m r 5 ) 7 im(t ) L and ker(t ) L 9 If v and w are in ker(t ), then T ( v + w) T ( v) + T ( w) +, so that v + w is in ker(t ) as well 67

5 Chapter If v is in ker(t ) and k is an arbitrary scalar, then T (k v) kt ( v) k, so that k v is in ker(t ) as well 5 We need to find all x such that A x If A x, then x is in ker(a) { }, so that x Since ker() { }, we can conclude that x It follows that ker(a) { } 5 a Using the equation + (or ), we can write the general vector x in ker(h) as x p + r + s x p + q + s x p + q + r x x p p + q + r + s x 5 q x 6 r x 7 s v v b ker(h) span( v, v, v, v ) by part (a), and im(m) span( v, v, v, v ) by Fact, so that im(m) ker(h) M x is in im(m) ker(h), so that H(M x) v v Not a subspace, since W does not contain the zero vector W im 5 6 is a subspace of R, by Fact We have subspaces { }, R, and all lines and planes (through the origin) To prove this, mimic the reasoning in Example 7 Yes; we need to show that W contains the zero vector We are told that W is nonempty, so that it contains some vector v Since W is closed under scalar multiplication, it will contain the vector v, as claimed 9 These vectors are linearly dependent, since v m v + v + + v m 68

6 Section Linearly independent, since the two vectors are not parallel, and therefore not redundant ( ) Linearly dependent, since the second vector is redundant 5 Linearly dependent y Fact 8, since we have three vectors in R, at least one must be redundant We can perform a straightforward computation to reveal that v v + v 7 Linearly independent The first two vectors are clearly not redundant, and since rref 6 I, the last vector is also not redundant Thus, the three vectors turn out to be linearly independent 9 Linearly dependent First we see that is not redundant, because it is first, and non-zero However,, so it is redundant and are clearly not redundant, but + + 5, so it is 5 redundant Certainly, since the second vector equals the first, the second is redundant So v v, v v, revealing that is in ker(a) The first column is, so it is redundant v, so is in ker(a) 5 The third column equals the first, so it is redundant and v v, or v + v v Thus, is in ker(a) 7 A basis of im(a) is,, by Fact 69

7 Chapter 9 The three column vectors of A span all of R, so that im(a) R We can choose any two of the columns of A to form a basis of im(a); another sensible choice is e, e The two column vectors of the given matrix A are linearly independent (they are not parallel), so that they form a basis of im(a) im(a) span( e, e, e ), so that e, e, e is a basis of im(a) 5 If v i is a linear combination of the other vectors in the list, v i c v + + c i v i + c i+ v i+ + + c n v n, then we can subtract v i from both sides to generate a nontrivial relation (the coefficient of v i will be -) Conversely, if there is a nontrivial relation c v + + c i v i + + c n v n, with c i, then we can solve for vector v i and thus express v i as a linear combination of the other vectors in the list 7 No; as a counterexample, consider the extreme case when T is the zero transformation, that is, T ( x) for all x Then the vectors T ( v ),, T ( v m ) will all be zero, so that they are linearly dependent 9 Yes; the vectors are linearly independent The vectors in the list v,, v m are linearly independent (and therefore non-redundant), and v is non-redundant since it fails to be in the span of v,, v m To show that the columns of are linearly independent, we show that ker() { } Indeed, if x, then A x A, so that x (since A I m ) y Fact 8, rank() # columns m, so that m n and in fact m < n (we are told that m n) This implies that the rank of the m n matrix A is less than n, so that the columns of A are linearly dependent (by Fact 8) Consider a linear relation c v + c ( v + v ) + c ( v + v + v ), or, (c + c + c ) v + (c + c ) v + c v Since there is only the trivial relation among the vectors v, v, v, we must have c + c + c c + c c, so that c and then c and then c, as claimed 5 Yes; if A is invertible, then ker(a) { }, so that the columns of A are linearly independent, by Fact 8 7 y Fact 8, the rank of A is Thus, rref(a) 9 L im 7

8 Section To write L as a kernel, think of L as the intersection of the planes x y and y z, that is, as the solution set of the system x y y z Therefore, L ker There are other solutions 5 a Consider a relation c v + + c p v p + d w + + d q w q Then the vector c v + + c p v p d w d q w q is both in V and in W, so that this vector is : c v + + c p v p and d w + + d q w q Now the c i are all zero (since the v i are linearly independent) and the d j are zero (since the w j are linearly independent) Since there is only the trivial relation among the vectors v,, v p, w,, w q, they are linearly independent b In Exercise 5 we show that V + W span( v,, v p, w,, w q ), and in part (a) we show that these vectors are linearly independent 5 The zero vector is in V, since v for all v in V If w and w are both in V, then ( w + w ) v w v + w v + for all v in V, so that w + w is in V as well If w is in V and k is an arbitrary constant, then (k w) v k( w v) k for all v in V, so that k w is in V as well x x 55 We need to find all vectors x in R 5 such that x x +x +x +x +5x 5 x x 5 5 These vectors are of the form x a b c 5d 5 x a x b a + b + c + d x c x 5 d 7

9 Chapter The four vectors to the right form a basis of L ; they span L, by construction, and they are linearly independent, by Fact 5 57 We will begin to go through the possibilties for j until we see a pattern: j : Yes, because is in ker(a) (the first column is ) j : No, this would just be a multiple of the second column, and only if the jth component is zero j : Yes, since is in ker(a) At this point, we realize that we are choosing the redundant columns Thus, j can also be 6 and 7, because, and are in ker(a) 5 Clearly the second column is just three time the first, and thus is redundant Applying the notion of Kyle Numbers, we see:, so is in the ker(a) No other vectors belong in our list, so a basis of 6 the kernel is ( ), and a basis of the image is ( ) 7

10 Section The two columns here are independent, so there are no redundant vectors ( Thus, ) is a basis of the kernel, and the two columns form a basis of the image:, 5 The first two vectors are non-redundant, but the third is a multiple of the first We see:, so a basis of the kernel is, and a basis of the image consists 6 ( ) of the non-redundant columns, or, 7 We immediately see fitting Kyle numbers for one relation: Now, since the second column is redundant, we remove it from further inspection and keep a zero above it: However, in this case, there are no more redundant vectors Thus, a basis of the kernel is (, and a basis of the image is, ) 9 The second column is redundant, and we can choose Kyle numbers as follows:, but the third column is non-redundant Thus, a basis of the kernel is, while a basis of the image is, Here it is clear that only the third column is redundant, since it is equal to the first Thus, a basis of the kernel is, and, is a basis of the image Here we first see, then, so both the second and third columns are redundant, and a basis of the kernel is, This leaves ( ) to be a basis of the image 7

11 Chapter 5 We quickly find that the third column is redundant, with the Kyle numbers, then see that the fourth column is also redundant, Thus, a basis of our kernel is,, while, is a basis of our image 7 For this problem, we again successively use Kyle Numbers to find our kernel, investigating the columns from left to right We initially see that the first column is redunant:, then the third column:, followed by the fifth column: Thus,,, is a basis of the kernel, and (, ) is a basis of the image 9 We see that the third column is redundant, and choose Kyle numbers as follows: 5 5, then we see that the fourth column is also redundant, 5 7

12 Section Thus, 5, the image rref is a basis of the kernel, and,, is a basis of, which we can use to spot a vector in the kernel: Since the third column is the only redundant one, this forms a basis of the kernel, and implies that the third column of A is also redundant Thus, a basis of im(a) is, rref 6 8 Using the method of Exercises 7 and 9, we find the kernel:, then So a basis of ker(a) is, The non-redundant column vectors of A form a basis of im(a):, rref We will emulate Exercise to find the 9 kernel: 75

13 Chapter 5 5 5, then 5 So a basis of ker(a) is, 9 and a basis of im(a) is,, 9 7 Form a matrix A with the given vectors as its columns We find that rref(a) I, so that the vectors do indeed form a basis of R, by Fact 9 9 x x x ; let x s and x t Then the solutions are of the form x x s t s s + t x t Multiplying the two vectors by to simplify, we obtain the basis Proceeding as in Exercise 9, we can find the following basis of V :,,, Now let A be the matrix with these three vectors as its columns Then im(a) V by Fact, and ker(a) { } by Fact 8, so that A does the job A We can write V ker(a), where A is the n matrix A c c c n Since at least one of the c i is nonzero, the rank of A is, so that dim(v ) dim(ker(a)) n rank(a) n, by Fact 7 A hyperplane in R is a line, and a hyperplane in R is just a plane 76

14 Section 5 We need to find all vectors x in R n such that v x, or v x + v x + + v n x n, where the v i are the components of the vector v These vectors form a hyperplane in R n (see Exercise ), so that the dimension of the space is n 7 Since dim(ker(a)) 5 rank(a), any 5 matrix with rank will do; for example, A 9 Note that ker(c) { }, by Fact 7a, and ker(c) ker(a) Therefore, ker(a) { }, so that A is not invertible We can choose a basis v,, v p of V, where p dim(v ) dim(w ) Then v,, v p is a basis of W as well, by Fact c, so that V W span( v,, v p ), as claimed dim(v + W ) dim(v ) + dim(w ), by Exercise 5b 5 Note that im(a) span( v,, v p, w,, w q ) V, since the w j alone span V To find a basis of V im(a), we omit the redundant vectors from the list v,, v p, w, w q, by Fact Since the vectors v,, v p are linearly independent, none of them are redundant, so that our basis of V contains all vectors v,, v p and some of the vectors from the list w,, w q 7 Using the terminology suggested in the hint, we need to show that u,, u m, v,, v p, w,, w q is a basis of V +W Then dim(v +W )+dim(v W ) (m+p+q)+m (m+p)+(m+q) dim(v ) + dim(w ), as claimed Any vector x in V + W can be written as x v + w, where v is in V and w is in W Since v is a linear combination of the u i and the v j, and w is a linear combination of the u i and w j, x will be a linear combination of the u i, v j, and w k ; this shows that the vectors u,, u m, v,, v p, w,, w q span V + W To show linear independence, consider the relation a u + + a m u m + b v + + b p v p + c w + + c q w q Then the vector a u + + a m u m + b v + + b p v p c w c q w q is in V W, so that it can be expressed uniquely as a linear combination of u,, u m alone; this implies that the b i are all zero Now our relation simplifies to a u + + a m u m + c w + + c q w q, which implies that the a i and the c j are zero as well (since the vectors u,, u m, w,, w q are linearly independent) 9 The nonzero rows of E span the row space, and they are linearly independent (consider the leading ones), so that they form a basis of the row space:,, 5 a All elementary row operations leave the row space unchanged, so that A and rref(a) have the same row space 77

15 Chapter b y part (a) and Exercise 5, dim(row space of A) dim(row space of rref(a)) rank(rref(a)) rank(a) 5 Using the terminology suggested in the hint, we observe that the vectors v, A v,, A n v are linearly dependent (by Fact 8), so that there is a nontrivial relation c v + c A v + + c n A n v We can rewrite this relation in the form (c I n + c A + + c n A n ) v The nonzero vector v is in the kernel of the matrix c I n + c A + + c n A n, so that this matrix fails to be invertible 55 If rank(a) n, then the n non-redundant columns of A form a basis of im(a) R n, so that the matrix formed by the non-redundant columns is invertible (by Fact 9) Conversely, if A has an invertible n n submatrix, then the columns of form a basis of R n (again by Fact 9), so that im(a) R n and therefore rank(a) dim(im(a)) n 57 As in Exercise 56, let m be the smallest positive integer such that A m In Exercise 56 we construct m linearly independent vectors v, A v,, A m v in R n ; now m n by Fact 8 Therefore A n A m A n m A n m, as claimed 59 Prove Fact d: If m vectors v,, v m span an m-dimensional space V, then they form a basis of V We need to show that the vectors v i are linearly independent We will argue indirectly, assuming that the vectors are linearly dependent; this means that at least one of the vectors v i is redundant, say v p ut then V span( v,, v p,, v m ) span( v,, v p, v p+,, v m ), contradicting Fact b 6 a Note that rank(), so that dim(ker()) 5 rank() and dim(ker(a)) since ker() ker(a) Since ker(a) is an subspace of R 5, dim(ker(a)) could be,, or 5 It is easy to give an example for each case; for example, if A and, then A and dim(ker(a)) 78

16 Section b Since dim(im(a)) 5 dim(ker(a)), the possible values of dim(im(a)) are,, and, by part a 6 a y Exercise 9b, im(a) im(a), and therefore rank(a) rank(a) b Write v v m and A A v A v m If r rank(), then the r nonredundant columns of will span im(), and the corresponding r columns of A will span im(a), by linearity of A y Fact b, rank(a) dim(im(a)) r rank() Summary: rank(a) rank(a), and rank(a) rank() 65 Let v,, v 6 be the columns of matrix A Following the hint, we observe that v 5 v + 5 v + 6 v, which gives the relation v + 5 v + 6 v v 5 Thus the vector 5 x 6 is in the kernel of matrix A Since x fails to be in the kernel of matrix, the two kernels are different, as claimed 67 Exercise 66 shows that if two matrices A and of the same size are both in rref and have the same kernel, then A Apply this fact to A and rref(m) , so x 7 5, so x, so x This may not be as obvious as Exercises and, but we can find our coefficients simply 5 by solving the matrix

17 Chapter 7 We need to find the scalars c and c such that c + c Solving c a linear system gives c, c Thus x c 9 We can solve this by inspection: Note that our first coefficient must be because of the first terms of the vectors Also, the second coefficient must be due to the last terms However, v + v + of v and v We can also see this by attempting to solve, which turns out to be inconsis- tent Thus, x is not in V Proceeding as in Example, we find x Thus, we reason that x is not in the span Here, we quickly see that since x c +c +c, c must equal We find c similarly, since x () + c + c Finally, now that x () + ( ) + c, c must be zero So +, and x 5 This may be a bit too difficult to do by inspection Instead we reduce to 8 8, 5 revealing that x 8 v v + 5 v, and x y inspection, we see that in order for x to be in V, x v + v v (by paying attention to the first, second, and fourth terms) Now we need to verify that the third terms work out So, () + () () 5 x 8

18 Section Thus x is in V, and x 9 a S, and we find the inverse S to be equal to Then S AS b Our commutative diagram: x c + c T x c c a b c So, c d c c c c T ( v ) T ( v ) a S T ( x) A x c A c + c c T T ( x) c + c A c, and we quickly find c, and we find the inverse S to be equal to 7 Then S AS 7 7 b Our commutative diagram:

19 Chapter x c + c T T ( x) A x c A + c A 7 c + c 7c c 7c x T T ( x) c a b c 7c 7 So,, and we quickly find c d c c T ( v ) T ( v ) a S, and we find the inverse S to be equal to Then S AS 5 6 b Our commutative diagram: x c + c T T ( x) A x c A + c A c + c c c c x T T ( x) c c a b c c So,, and we quickly find c d c c 5 5 c T ( v ) T ( v ) 6 6 c 8

20 Section 5 We will use the commutative diagram method here (though any method suffices) x c c + c x c c c c c c + 6c T T ( x) A x c + c 5 c + c 7 ) ( c ( + + c + 6 ( c c ) + (c + 6c ) c c T T ( x), so 7 We use a commutative diagram: x c + c + c x c c c 6 T c c 9c, so 9 c 9 Let s build column-by-column : T ( v ) T ( v ) T ( v ) c + 6c T ( x) A x ) c A + c A + c A c c 8 T T ( x) 8 9c

21 Chapter We can use a commutative diagram to see how this works: x c v + c v + c v x c c c c c c, so c c Here we will build column-by-column: T ( v ) T ( v ) T ( v ) T T ( x) v x c ( v v ) + c ( v v ) + c ( v v ) c ( v ) + c ( ) + c ( v ) c v c v T T ( x) c c ( v v ) v ( v v ) v ( v v ) v v, since all three are perpendicular unit vectors So, 5 Using another commutative diagram: x c v + c v + c v x c c c T T ( x) c T ( v ) + c T ( v ) + c T ( v ) c ( v ( v v ) v ) + c ( v ( v v ) v )+ c ( v ( v v ) v ) c ( v v ) + c ( v ) + c ( v ) c v + ( c + c ) v + c v T T ( x) 8 c c + c c

22 Section So This is a shear along the second term 7 We want a basis ( v, v ) such that T ( v ) a v and T ( v ) b v for some scalars a a and b Then the -matrix of T will be T ( v ) T ( v ), which b is a diagonal matrix as required Note that T ( v) v v for vectors parallel to the line L onto which we project, and T ( w) w for vectors perpendicular to L Thus, ( we can pick) a basis where v is parallel to L and v is perpendicular, for example,, 9 Using the same approach as in Exercise 7, we want a basis, v, v, v such that T ( v ) a v, T ( v ) b v and T ( v ) c v First we see that if v, then T ( v ) v Next we notice that if v and v are perpendicular to v, then T ( v ) v and T ( v ) v So we can pick v and v, for example We will use the same approach as in Exercises 7 and 9 Any basis with vectors in the plane and one perpendicular to it will work nicely here! So, let v, v be in the plane v can be, and v (note that these must be independent) Then v should be perpendicular to the plane We will use v perpendicular to the plane because all vectors perpendicular to So, our basis is:,, y definition of coordinates (Definition ), x the coefficient vector This is + ( ) lie in the plane 5 If v, v is a basis with the desired property, then x v + v, or v x v Thus we can make v any vector in the plane that is not parallel to x, and then let v x v 85

23 Chapter For example, if we choose v 7 y Fact, we have A SS, then v a b c d 9 From the picture, u + v w, so that w u v, ie, w d c b a 5 Let ( v, v,, v m ) Then, let x a v + a v + + a m v m and y b v + b v + + b m v m Then x + y a v + a v + + a m v m + b v + b v + + b m v m (a + b ) v + (a + b ) v + + (a m + b m ) v m a + b a + b a m + b m a a a m + b b b m x + y 5 y Definition, we have x S x 7 x and x 55 y Definition, we have x x R and x R x, ie, P }{{} P 58 x R, so that x 57 If we can find a basis ( v, v, v ) such that the -matrix of A is, then A must be similar to ecause of the entries in the matrix, it is required that A v v, A v v and A v v So, all we need for our basis is to pick independent v, v in the plane, and v perpendicular to the plane x y x y x y 59 First we find the matrices S such that, or z t z t z t x y x x + y y y The solutions are of the form S, where y and t z t z z + t t are arbitrary constants Since there are invertible solutions S (for example, let y t ), the matrices and are indeed similar 86

24 Section x y x y 6 We seek a basis v, v z such that the matrix S v t v satisfies z t 5 9 x y x y the equation Solving the ensuing linear system 7 z t z t gives S z z z t t We need to choose z and t so that S will be invertible For example, if we let z 6 and 9 9 t, then S, so that v 6, v 6 x y p q x y x y p q 6 First we find the matrices S such that, z t q p z t z t q p px qz py qt px qy qx + py or, If q, then the solutions are of the qx + pz qy + pt pz qt qz + pt t z form S, where z and t are arbitrary constants Since there are invertible z t p q p q solutions S (for example, let z t ), the matrices and are indeed q p q p similar (If q, then the two matrices are equal) 65 a If S I n, then S AS A b If S AS, then SS A If we let R S, then R R A, showing that is similar to A a a a 67 The matrix we seek is T T + bc bc ad c c ac + cd a + d 69 The matrix of the transformation T ( x) A x with respect to the basis, is D Thus S 6 AS D for S 7 a Note that AS S If x is in ker(), then A(S x) S x S, so that S x is in ker(a), as claimed b We use the hint and observe that nullity () dim(ker ) p dim(ker A) nullity(a), since S v,, S v p are p linearly independent vectors in ker(a) Reversing the roles of A and shows that, conversely, nullity(a) nullity(), so that the equation nullity(a) nullity() holds, as claimed 87

25 Chapter 7 a y inspection, we can find an orthonormal basis v v, v, v of R : v v 6 8, v, v 8 6 Figure : for Problem 7a b Now T ( v ) v, T ( v ) v and T ( v ) v (see Figure ), so that the matrix of T with respect to the basis v, v, v is Then A SS S AS, where S A Thus A 77 Let S be the n n matrix whose columns are e n, e n,, e Note that S has all s on the other diagonal and s elsewhere: { if i + j n + s ij otherwise Also, S S Now, S AS SAS; the entries of are b ij s i,n+ i a n+ i,n+ j s n+ j,j a n+ i,n+ j Answer: b ij a n+ i,n+ j is obtained from A by reversing the order of the rows and of the columns True or False 88

26 True or False F; It s a subspace of R T, by Summary 9 5 T, by Summary 9 7 T; We have the nontrivial relation u + v + w 9 T, by Fact T, by Fact 8 F; The number n may exceed 5 T, by Fact 6, parts b and c 7 T, by Definition 9 T; Check that F; We are unable to find an invertible matrix S as required in the definition of similarity T, by Definition (all vectors in R are linear combinations of e, e, e ) 5 T, since A v A 7 F; Suppose v v Then T ( v ) T ( v ) e cannot be e 9 T, since A (A)A A T; Consider any basis v, v, v of V Then k v, v, v is a basis as well, for any nonzero scalar k F; Consider, but 5 F; Let V span( e ) and W span( e ) in R, for example 7 T; Consider the linear transformation with matrix w w n v v n 9 F; Note that R isn t even a subset of R A vector in R, with two components, does not belong to R T; Let A, for example, with ker(a) im(a) span( e ) T; Matrix S AS is invertible, being the product of invertible matrices 89

27 Chapter 5 T; Pick three vectors v, v, v that span V Then V im v v v 7 T; Pick a vector v that is neither on the line nor perpendicular to it Then the matrix of the linear transformation T ( x) R x with respect to the basis v, R v is, since R(R v) v 9 T; Note that A( C), so that all the columns of matrix C are in the kernel of A Thus C and C, as claimed 5 F; Suppose such a matrix A exists Then there is a vector v in R such that A v but A v As in Exercise 58a we can show that vectors v, A v, A v are linearly independent, a contradiction (we are looking at three vectors in R ) 5 F; Think about this problem in terms of building such an invertible matrix column by column If we wish the matrix to be invertible, then the first column can be any column other than (7 choices) Then the second column can be any column other than or the first column (6 choices) For the third column, we have at most 5 choices (not or the first or second columns, as well as possibly some other columns) For some choices of the first two columns there will be other columns we have to exclude (the sum or difference of the first two), but not for others Thus, in total, fewer than matrices are invertible, out of a total 9 5 matrices Thus, most are not invertible 9

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