Department of Aerospace Engineering AE602 Mathematics for Aerospace Engineers Assignment No. 4

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1 Department of Aerospace Engineering AE6 Mathematics for Aerospace Engineers Assignment No.. Decide whether or not the following vectors are linearly independent, by solving c v + c v + c 3 v 3 + c v : v, v, v 3, v. Decide also if they span R, by trying to solve c v + + c v,,,. To check for dependency or independency, we use c v + c v + c 3 v 3 + c v...() If c c c 3 c is the only way to solve the above equation, then the vectors are linearly independent. If any of the c i, then the given vectors are linearly dependent. Solving the equation () c + c + c 3 + c we get four equations with four unknowns as follows c + c c + c c + c 3 c 3 + c From the above set of equations we get c c c 3 c Hence with c ; c ; c 3 ; c, the equation () goes to zero. That is c v + c v + c 3 v 3 + c v Note tat c, c, c 3, c can take any other values also. Therefore the given vectors v, v, v 3, v are linearly independent.

2 Now trying to solve c v + + c v,,, We have c + c...() c + c...(3) c + c 3...() c 3 + c...(5) From () and (3), we get c c and c c 3 and ence c c 3 Substituting c c 3 in equation (3), we get c 3 + c...(6) Equations (5) and (6) contradict each other and hence the vectors do not span R.. Decide the dependence or independence of (a),,,,,, 3,, ; Equation to be solved is, c v + c v + c 3 v 3 c + c + c 3 3 This is like solving Ac Solving Ac is equivalent to solving Uc where "U" is the upper triangular matrix form of "A" Let A 3 R R - R ; R3 R3 - *R

3 3 5 R3 R3 + R 3 7 U Now solving Uc c + c + c c 3 c 3 c c 3 c c + c + 3c 3 c c c c 3 and hence the given vectors are independent. (b) v v, v v 3, v 3 v, v v for any vectors v, v, v 3, v ; Solving the equation, c v v + c v v 3 ) + c 3 (v 3 v + c ( v v )...() If c c c 3 c, then equation () gets satisfied That is v v + v v 3 + v 3 v + v v Hence the vectors v v, v v 3, v 3 v, v v are linearly dependent for any vectors v, v, v 3, v 3

4 (c),,,,,,,,, x, y, z, for any numbers x, y, z. Solution: To find dependency or independency we have to solve c + c + c 3 + c x y z c + c + c x...() c + c 3 + c y...() c 3 + c z...(3) Case : When xyz. Then from equations (), () and (3), we have c c c 3 but c can assume any value and hence the vectors are not linearly independent. Case : When either x or y or z not equal to zero (or) x, y & z all are not equal to zero From equations (), () and (3), z c 3 c...() y (c +c 3 ) c...(5) x (c +c ) c...(6) If either x or y or z is not equal to zero, then for x or y or z to have a finite value (which is not equal to zero), c must not be equal to zero. If it is zero then the value tends to infinity. Hence for the case also the given vectors are not linearly dependent. Hence the given set of vectors are always linearly dependent for any existing non-infinite x, y, z.

5 .3 Prove that if any diagonal element of is zero, the rows are linearly dependent. T a b c d e f The equation to be solved for determining dependency or independency is, c a b c + c d e + c 3 f c a...() c b + c d...() c c + c e + c 3 f...(3) CASE : a, d, f From equation (), If a, c need not be zero (it can assume any value). Hence the rows are linearly dependent. CASE : a, d, f From (), we have c From (), we have c d Now since d, c need not be zero which proves that the rows are again linearly dependent. CASE 3: a, d, f From (), we have c From (), since d, we have c From (3), since f, c 3 can assume any value. Hence for the present case the rows are again linearly dependent. 5

6 . Is it true that if v, v, v 3 are linearly dependent, then also the vectors w v + v, w v + v 3, w 3 v + v 3, are linearly independent? (Hint: Assume some combination c w + c w + c 3 w 3, and find which c i are possible.) If v, v, v 3 are linearly dependent, then in the equation c v + c v + c 3 v 3 there is a c i which is not equal to zero...() To check the dependency of vectors w v + v, w v + v 3, w 3 v + v 3, we consider the equation c w + c w + c 3 w 3 c (v + v ) + c (v + v 3 ) + c 3 (v + v 3 ) (c + c )v + (c + c 3 )v + (c + c 3 )v 3 C v + C v + C 3 v 3 From (), if there is a non-zero c i, then there should definitely be a non-zero C i since C i is composed of c i. If there is such a non-zero C i, then vectors w, w, w 3 are also linearly dependent. Hence it is not true that if v, v, v 3 are linearly dependent, then the vectors w v + v, w v + v 3, w 3 v + v 3, are linearly independent..5 Describe geometrically the subspace of R 3 spanned by (a),,,,,,,, ; Here the x and z components are zero and only the y component is present. Hence the given vectors span a line (y-axis) in R 3. (b),,,,,,,, ; Here x component is zero but y and z components are present. Hence these vectors span the yz -plane in R 3. 6

7 (c) all six of these vectors. Which two form a basis? In all the six vectors, the x-component is zero. Only the y and z components are present. Hence these six vectors span the yz -plane in R 3. The vectors (,,) and (,,) form the basis because the other vectors in the column space can be expressed as a linear combination of these two vectors. For example + (d) all vectors with positive components. The vectors with positive components can be multiplied by any scalar (positive or negative) so that they span the whole of R 3..6 To decide whether b is in the subspace spanned by w,, w l, let the vectors w be the columns of A and try to solve Ax b. What is it the result for (a) w,,, w,,, w 3,,, b 3,, 5 ; For the above system Ax b to have a solution, the vector "b" must lie in the column space of "A". Otherwise the system will not have a solution. "A" matrix is formed by the vectors "w" as follows A Hence Ax b is x x 3 5 Applying Gauss elimination to the "A" matrix R R-R 7

8 Interchanging rows and 3 U Here the third pivot is equal to zero and hence the system does not have a solution. As already stated if there is no solution then it means that the vector "b" does not lie in the column space of "A". Hence b is not in the subspace spanned by w, w, w 3. (b) w,,, w, 5,, w 3,,, w,,, and any b? Ax 5 x x b b b 3 Applying Gauss elimination R R-*R x x b b b 3 b There are pivots in column, and 3. Hence x, x and are basic variables whereas is a free variable. b 3 b 3 / x b b x + x b x 5b b 8

9 x x 5b b b b b 3 / Whatever be the values of b, b, b 3, there is definitely going to be a solution available as we can clearly see from the above equation. Hence definitely vector "b" is in the subspace spanned by vectors w, w, w 3, w (which is the column space in R 3 )..7 By locating the pivots, find a basis for the column space of U 3 Express each column that is not the basis as a combination of the basic columns. Find also a matrix A with this echelon form U, but a different column space. Given U 3 which is in the echelon form. Here the columns and 3 contain the pivots ( and ) respectively. Hence columns and 3 are the basis for the column space of "A". Basis for column space are and...() Now columns that are not the basis can be expressed as a linear combination of the basis columns as follows: v v 3 v That is 3 Also v v 3 v By looking into "U" we can see that A for which when we apply the Gaussian 9

10 elimination, we get back the given "U" matrix. Again here columns and 3 are the basis for the column space as the remaining two columns ( and ) can be expressed as a linear combination of the basis columns. Now the basis for column space are and 8 which are different from the basis obtained in (). Hence the above matrix A is a matrix with this echelon form U, but a different column space..8 Find the dimension and construct a basis for the four subspaces associated with each of the matrices A and U 8. SOLUTION : First let us consider the matrix A 8 Here m; n; r (where r is the rank of the matrix A) Column space R(A): Dimension of R(A) is "r" which is equal to. Basis for R(A) is column contains the pivot ( is the pivot). because in the echelon form of "A", only the second Row space R(A T ): Dimension of R(A T ) is "r" which is equal to. Basis for R(A T ) is because it is the only non-zero row in the echelon form of "A". Null space N(A): Dimension of N(A) is "n-r" which is equal to 3. Basis for null space are found by solving Ax where the solution vector "x" forms the null space of "A". Solving Ax is the same as solving Ux x x

11 x + x x x x where x,, are the free variables When x,, When x,, When x,, x x x x x x Basis of null space of "A" are, and Left Null space N(A T ): Dimension of N(A T ) is "m-r" which is equal to For finding the left null space of "A", we solve A T y 8 y y y y

12 Here y is the free variable because in the echelon form of "A T ", only the first column contains a pivot. U is the echelon form When y, y Therefore basis for left null space is Now considering the matrix U Column space: Dimension is. Basis is the column containing the pivots which is Row space: Dimension is. Basis is the non-zero row which is Null space: Dimension is 3. Basis are, and Left Null space: Dimension is. Basis is obtained by solving U T y y y Here again y is the free variable because in the echelon form of "U T ", only the first column contains a pivot.

13 When y, basis of left null space is.9 Find the dimension and a basis for the four fundamental subspaces for both A and U. First considering the matrix A Here m3; n; r (since in the echelon form, the last row becomes zero and there are two non-zero rows) Column space: Dimension is. "U" matrix. Basis are and since these are the columns which contains pivots in Row space: Dimension is. Basis are and Null space: dimension is. Basis are obtained by solving Ax or Ux x x Here x & x are basic variables and & are the free variables x and x 3

14 x x + When, When, x x x x Therefore the basis are and Left null space: dimension is. Solving A T y y y y 3 Here y 3 is the free variable and y y 3 & y When y 3, the basis becomes Now considering the matrix U Column space: Dimension is. Basis are and

15 Row space: Dimension is. Basis are and Null space: Dimension is. Basis are and Left null space: Dimension is. Solving U T y y y y 3 Here y 3 is the free variable ; y, y, Hence assigning y 3, we get te basis as. Describe the four subspaces in 3-dimensional space associated with A. Column space: Here m3; n3; r Dimension is. Basis are and The column space spans the x-y plane in the space R 3. 5

16 Row space: Dimension is. Basis are and The row space spans the y-z plane in the vector space R 3 Null space: Dimension is. Solving Ax x x Basis for null space is The null space spans the line x-axis in the vector space R 3. Left null space: Dimension is. Solving A T y y y y 3 y, y (y 3 is te free variable) Therefore Basis is The left null space spans the line (z-axis) in the vector space R 3. Find the rank of A and write the matrix as A uv T : 3 A and A. 6 6

17 a) A 3 6 R3 R3-*R 3 which is the echelon form form A uv T Rank of "A" is. All matrices with rank can be expressed in the Therefore A 3 b) A R R-R which is the echelon form. form A uv T Hence rank is. All matrices with rank can be expressed in the A (). Find a left-inverse and/or a right inverse (when they exist) for A and M and T a b a. a) A Here r; m; n3 rm implies right inverse exists for "A" AC I where C is the right inverse and C A T (AA T ) 7

18 AA T (AA T ) C A T (AA T ) 3 3 is the right inverse of "A" 3 3 b) M R R-R R3 R3-R which is the echelon form Here r; m3; n rn implies there exists a left inverse for "M" BM I where B is the left inverse and is equal to B (M T M) M T M T M (M T M) inverse of "M" B (M T M) M T is the left 8

19 c) T a b a and they both are the same. Here rmn. Hence both left and right inverse exist det(t) a T a b a a T a b a a is the left and right inverse of "T" 9