Questions and Answers #3 Matrix Transformations

Questions and Answers Sheet 3 Matrix Transformations Question #1 For the matrix A below, find a nonzero vector in Nul A and a nonzero vector in Col A. 5 −9 2 3  A =  −8 −9 −11 21  4 −3 −17 27  Find a nonzero vector in Nul A.  −3  2 A=  0   1 Answer: Consider the following matrix, 5 −9 2 3  A =  −8 −9 −11 21  4 −3 −17 27  Find a non-zero vector in Nul A. Use row reduced method.  1 3/ 2 A =  −8 −9  4 −3 1 1 R1 → R1  0 2 0 5 / 2 −9 / 2  −11 21  −17 27  3 / 2 5 / 2 −9 / 2  3 9 −15  −9 −27 45  1 3 / 2 5 / 2 −9 / 2  R2 → R2 + 8 R1 , R3 → R3 − 4 R1  0 3 9 −15  0 0 0 0  1 0 −2 3  R3 → R3 + 3R2  0 1 3 −5 0 0 0 0  1 R1 → R1 − R2 2 The matrix has 2 pivot columns and 2 free columns. Then, AX=0  x1  1 0 −2 3    0  0 1 3 −5  x2  = 0   x    0 0 0 0   3  0  x4  𝑥1 − 2𝑥3 + 3𝑥4 = 0 ⇒ 𝑥1 = 2𝑥3 − 3𝑥4 𝑥2 + 3𝑥3 − 5𝑥4 = 0 ⇒ 𝑥2 = −3𝑥3 + 5𝑥4 So, the general solution is,  x1 = 2 x3 − 3x4   x2 = −3x3 + 5 x4  x and x are free variables 4  3 Then, a vector parametric solution is,  x1   2 x3 − 3x4   2 x3   −3x4   2   −3  x   −3x + 5 x   −3x   5 x   −3 5 3 4 3  2 =  = +  4  =   x3 +   x4  x3     x3   0   1  0 x3             x4 1  x4     x4   x4   0  Therefore, the required non-zero vectors in Nul A are, (2,-3,1,0),(-3,5,0,1) Find a non zero vector in Col A observe the equation (1), the first two columns of the matrix 1 0 −2 3  0 1 3 −5 are pivot columns   0 0 0 0  these columns in the original matrix define the Column Space of the matrix. 2 3     Column Space = p −8 + q −9 where p and q are any real numbers.      4   −3 Therefore, the required nonzero vectors in Col A are, (2,-8,4),(3,-9,3) Question #2 With T defined by T(x)=Ax, find a vector x whose image under T is b, and determine whether x is unique.  1 −5 −7  −2 A , b =  −2 −3 7 5    Answer: Solve T(x)=b for x, that is solve Ax=b to find x. The augmented matrix is  1 −5 −7 | −2 A A b =    −3 7 5 | −2 Reduce matrix to reduced row echelon form R right  R2  −3 7 5  1 −5 −7 | −2  1   −3 7 5 | −2  R + 1 R → R ~  0 − 8 − 16   2 1 2 3 3 3  5 −2  3  −3 7   − 8 R2 → R2 8 16 8 0 − − −  3 3 3  R1 − 7 R2 → R1  −2  8 −  3  −3 0 −9 −9  ~  0 1 2 1  −3 0 −9 −9  1  1 0 3 3  0 1 2 1  − 3 R1 → R1 ~ 0 1 2 1     Therefore, we get 𝑥1 + 3𝑥3 = 3 ⇒ 𝑥1 = 3 − 3𝑥3 𝑥2 + 2𝑥3 = 1 ⇒ 𝑥2 = 1 − 2𝑥3 That is  x1   3 − x3   3  −3       x =  x2  = 1 − 2 x3  = 1  + x3  −2  x3   x3  0  1  Result: x is not unique. Question #3 Determine if the columns of the matrix form a linearly independent set. Justify each answer.  1 4 −3 0   −2 −7 5 1     −4 −5 7 5 Answer: The given matrix  1 4 −3 0   −2 −7 5 1     −4 −5 7 5 The columns of the matrix form a linearly independent set if equation Ax=0 has only trivial solution Reduce matrix in reduced echelon form 4 1  −2 −7   −4 −5   −4 −5   0 −9  2  0 0    −4 −5   0 −9  2  0 0   −4 −5   0 −9 2  0 0   −4 −5 0 1   0 0    −4 −5 7 5  −3 0   R1right  R3  9 3 3  5 1 0 − − R2 − 1/ 2 R1 → R2  2 2 2 7 5   1 1  0  0 − 3 3       −4 −5 7 7 5  5     3 3  R3 + 1/ 4 R1 → R3  9 3 3 − 0 − − 2 2  R3 + 11/18R2 → R3  2 2 2   1 1 1 1   0  − 0 − 3 3  3 3    7 5   −4 −5 7 5   −3R → R   3 3 3 3  0 −9 0 0  −  2 2  R2 − 3 / 2 R3 → R2  2  0  1 1  0 1 − 1    − 3 3  7 5  R − 7 R → R  −4 −5 0 12  3 1 0 0 1 ~  0 1 0 0   −2 / 9 R2 → R2  0 0 1 −1 1 −1 0 12  1 0 0 −3 R1 + 5R2 → R1   0 0 ~ 0 1 0 0  1/ 4 R1 → R1 1 −1 0 0 1 −1 Thus. we get equation Ax=0 has not only trivial solution, the columns of the matrix form a linearly dependent set. Hence, the columns of the matrix form a linearly dependent set Result: The columns of the matrix form a linearly dependet set Question #4 Use the definition of Ax to write the matrix equation as a vector equation, or vice versa. 4  −5 7 6  −1 3  −8  −8 x1   + x2   + x3   =   7  −5 0 0          −4  1  2   −7  Answer: From the Theorem 3 the matrix equation is Ax=b where A in 𝑚 × 𝑛 matrix 𝑏 ∈ 𝑅𝑚 . We have 4  −5 7 6  −1 3  −8  −8 x1   + x2   + x3   =   7  −5 0 0          −4  1  2   −7  then the matrix equation of the given vector equation is  x1  4   −5 x2   7 x3   6   4 −5 7   − x   3  x   −8 x   −8   −1 3 −8 3 2  1 +  + =    x1  7   −5 x2   0   0   7 −5 0            − x1   x2   2  x3   −7   −4 1 2  6  x1     x  =  −8  2  0   x3    7 Hence, the matrix equation is  4 −5 7  6 x   1  −1 3 −8      x2  =  −8  7 −5 0     0     x3    − 4 1 2   7  4 −5 7  6  −1 3 −8  x1   −8  x  =   Result:   7 −5 0   2   0     x3     −4 1 2  7 Question #5 Find an orthogonal basis for the column space of each matrix.  3 −5 1  1 1 1    −1 5 −2     3 −7 −8  Answer: Use Gram-Schmidt process to orthogonalize column vectors of A 3 1 v1 = x1 =    −1   3 x v v2 = x2 − 2 1 v1  v1  −5   3   −5   6   1  1         −40  1   1   2   3    v1 = − = + =  5  20  −1  5   −2  3             −7   3   −7   6   −1 1 3 1 1 1   x v x v 30 −10  3  v3 = x3 − 3 1 v1 − 3 2 v2 =   −   − =  −2  20  −1 20  3  v1  v1 v2  v2       8 3  −1  1   9 / 2   1/ 2   −3  1   3/ 2   3/ 2   1  + =  =  −  −2   −3 / 2   3 / 2   1           8   9 / 2   −1/ 2   3  Orthogonal basis for column space of A:  3   1   −3 1 3 1 {  ,   ,  }  −1  3   1         3   −1  3  Question #6 Find an explicit description of Nul A by listing vectors that span the null space. 1 5 −4 −3 1  A = 0 1 −2 1 0  0 0 0 0 0  Answer: First find the general solution of Ax=0 in terms of the free variables. Since 1 5 −4 [ A0] = 0 1 −2 0 0 0 1 R1 → R1 − 5R2 0 0 −3 1 0  1 0 0  0 0 0  0 6 −8 1 0  1 −2 1 0 0  0 0 0 0 0  the general solution is 𝑥1 = −6𝑥3 + 8𝑥4 − 𝑥5 𝑥2 = 2𝑥3 − 𝑥4 with 𝑥3 , 𝑥4 and 𝑥5 free. So  x1   −6  8  −1 x  2  −1 0  2       x =  x3  = x3  1  + x4  0  + x5  0           x4  0 1 0  x5   0   0   1  and a spanning set for Nul A is  −6   8   −1  2   −1  0        { 1  ,  0  ,  0 }       0 1 0  0   0   1  Result:  −6   8   −1  2   −1  0        { 1  ,  0  ,  0 }       0 1 0  0   0   1  Question #7 Assume that A is row equivalent to B. Find bases for Nul A and Col A. 1 2 A= 1  3 1 0 B= 0  0 2 −5 11 −3 4 −5 15 2  2 0 4 5  6 −5 19 −2  2 0 4 5 0 5 −7 8  0 0 0 −9   0 0 0 0 Answer: Since B is a row echelon form of A, we see that the first, third and fifth columns of A are its pivot columns. Thus a basis fo Col A is 1   −5  −3  2   −5  2  {  ,   ,  } 1   0   5         3   −5  −2  To find a basis for Nul A, we find the general solution of Ax=0 in terms of the free variables. Since it is row equivalent to B we can simply get reduced row echelon form of B: 1 R → R2 0 4 5 2 5  5 −7 8  1 R3 → R3 −9 0 0 −9   0 0 0 1 0  0  0 2 0 0 0 1  0  0  0 2 0 0 5 0 0 0 0 4 −7 5 0 0 5 1  R → R − 5R  1 3  0 1 0  8  R → R2 − R3  1 2 5 0  0  0 2 0 0 1 0 0 0 0 4 −7 5 0 0 5  0  1  0  7 to get: 𝑥1 = −2𝑥2 − 4𝑥4 , 𝑥3 = 𝑥4 , 𝑥5 = 0, with 𝑥2 and 𝑥4 free. So 5  x1   −2   −4  x  1  0   2     x =  x3  = x2  0  + x4 7 / 5        x4  0  1   x5   0   0  And a basis for Nul A is  −2   −4  1  0      { 0  , 7 / 5}     0  1   0   0  Result: Basis for Col A is: 1   −5  −3  2   −5  2  {  ,   ,  } 1   0   5         3   −5  −2  Basis for Nul A is  −2   −4  1  0      { 0  , 7 / 5}     0  1   0   0  Question #8 Let W be the subspace spanned by the given vectors. Find a basis for 1 0  −1 1   w1 = ,w =    3  2  −2       −2  1 Answer: Let 1 0  −1 1   A = [ w1 , w2 ] =   3 −2     −2 1  We know that 𝑊 = (𝑐𝑜𝑙(𝐴)) = 𝑛𝑢𝑙(𝐴𝑇 ) so let's we find null space of 𝐴𝑇 We see that  y1  y  y =  2  W  y3     y4  if 𝑦1 + 𝑦3 − 𝑦4 = 0 and 𝑦2 − 2𝑦3 + 𝑦4 = 0 . Then  − y3 + y4   −1 1  −1  1  2y − y  2  −1  2   −1 3 4      W ={ } = { y3 +y } = span(   ,   )  y3  1 40 1 0           0 1 0 1  y4  Result:  − y3 + y4   −1 1  −1  1  2y − y  2  −1  2   −1 3 4      W ={ } = { y3 +y } = span(   ,   )  y3  1 40 1 0           0 1 0 1  y4  Question #9 If we have a matrices a b  1   A= and e12 ( ) =    c d  0 1  then by doing product a a + b  a + c b + d   we can interpret that right multiplication by Ae12 ( ) =  and e (  ) A = 12   c d   c c + d   𝑒12 to A gives a column-operation: add 𝜆 -times first column to the second column. In similar way, left multiplication by 𝑒12 (𝜆) to A gives row-operation on A. Is there any conceptual (not computational, if any) way to see that elementary row and column operations on a matrix can be expressed as multiplication by elementary matrices on left or right, accordingly? Answer: Here conceptual and computational ideas go hand in hand. We can see this by looking at the multiplication of A with 𝑒12 (𝜆) in some detail. We have 1    1 0   0   0 1 0 1  = 0 1  + 0 0  = I +  0 0         It is the position 12 of the blue marked 1 which determines selected row resp. column of A. We obtain 0 Ae12 ( ) = A( I +   0 0 [e12 ( ) A = ( I +   0 1 0 a  ) = A+   ] 0 0 c  1 c d  )A = A +     0 0 0  Question #10 Assume that T is a linear transformation. Find the standard matrix of T. 𝑇: 𝑅𝟚 ⇒ 𝑅𝟚 rotates points (about the origin) through - 𝜋/4 radians (clockwiese). [Hint: 𝑇(𝑒1 ) = (1/√2, −1/√2).] Answer: Given: points are rotated about the origin through − 𝜋 4 radians. General rotation matrix for rotation about the origin through 𝜃 rafians: cos T =  sin −sin  𝜋 . Replace 𝜃 by − and evaluate:  4 cos    cos = − 4 T =  sin −   4  1   4 = 2    1 cos −   − 4   2 − sin − The finally answer is  1  2   1  − 2 1  2  1  2   1  2  1  2 