1.  (4+3+3 points) Let

A = [ ⃗a1

⃗a2 ⃗a3  ] =  l 0(5)

3

−8

−2(9)

(a) Explain if {⃗a1 ,⃗a2 ,⃗a3 } is linearly independent.

(b) Explain if {⃗a1 ,⃗a2 } is linearly independent.

(c) Explain if A is invertible.

Solutions:

(a) Reduce A to row echelon form,

A ∼ l−0(5) 「 0

3

− 2

− 9(1)

This show that there are 3 pivots so the rank of A is T = 3. which is the number n of columns of A . Hence the set of columns {⃗a1 ,⃗a2 ,⃗a3 } is linearly independent.

(b) The set {⃗a1 ,⃗a2 } is a part of the set {⃗a1 ,⃗a2 ,⃗a3 } that is linearly independent by (a), it is also linearly independent.

(c) A is invertible since it has full rank 3 (T = n = 3) by the Invertible Matrix Theorem.

2.  (10 points) (a) Find A − 1  for

A =「 0

−1(a)

0

− b

(b) Find X in AX =  l 2(1) −2(0) 」

Solutions: (a) Apply elementary row operations

「 0     0     1   0   0   1 ~「 0     0   1   0   0   1 ~「 0   0

A − 1  =  l 」

「 0   0     1

(b) Let B =  「(l) . Then X = A − 1 B =  「(l) 1 + 2 − .

3.  (8+2 points) Given

2

A = 「 −6(1)

−5(4)

− 2

−4(4)

(a) Find the LU factorization of A .

(b) Explain how to use (a) to quickly find |A| with almost no more computation. Solutions: (a)

A =  l   21 − 」 l 0(2) 」 l 0(2) −7(4) − 5(2) 」

−6 − 2      4              0 − 14    10              0      0      0

To find L , perform the following operations on the first matrix below, divide the first column by the circled number (pivot) 2,

divide the second column by circled number (pivot) 7,

「(l) 6(1) − 1(7)4   1 ❀ 「(l) − − 2(1)   1 = L

Therefore,

A = LU =  l 1(1)         1(0)   0(0) 」 l 0(2)   −7(4) − 5(2) 」

(b) |A| = |LU| = |L||U| = 0.

4.  (4+3+1+1 points) Let T : R4 −→ R3  be defined by

T  \图图( l''「 」地地l)图图) =  「(l) l(」)

(a) Find the matrix A for T under the standard bases and find its reduced row echelon form.

(b) Find a basis for each of the column space C(A), row space Row (A), and null space N(A).

(c) Is the transformation T one-to-one? Justify your answer.

(d) Is the transformation T onto? Justify your answer.

(b) Using the RREF, a basis for C(A) is {⃗a1 ,⃗a2 ,⃗a3 } ,

a basis for Row (A) is {(1, 0, 0, − 1), (0, 1, 0, − 1), (0, 0, 1, − 1)} ,

and a basis null space N(A) is {(1, 1, 1, 1)} ,

where all vectors are column vectors.

(c) First explanation: the null space N(A) is non-trivial, so T is not one-to-one.

Second explanation: Matrix A is of size m × n = 3 × 4. From the RREF of A, the rank r = 3 < n, therefore T is not one-to-one.

(d) From the RREF of A, the rank r = 3 = m, therefore T is onto.

5.  (4+3+3 points) (a) Compute the determinant |A|, where

l 1 − 2     5   2 」

A =

「 2     0     3   5

(b) Use part (a) to justify that A − 1  exists.

(c) Find the following values

|AT A − 1 | =

|A− 1 AT A| =

| − A| =

Solutions: (a) Expand along second row of A, then along the last row of the resulting matrix,

' 1 − 2   2 '

(b) From (a), since |A| 0, A − 1  exists by the Invertible Matrix Theorem.

(c) |AT A − 1 | = |AT ||A− 1 | = |A||A− 1 | = 1

|A− 1 AT A| = |A− 1 ||AT ||A| = |A| = 12

| − A| = ( − 1)4 |A| = 12

6.  (2+6+2 points) Let

A = [ ⃗a1

⃗a2 ⃗a3  ] =  l1(1)   1(0)

(a) Explain that {⃗a1 ,  ⃗a2 ,  ⃗a3 } is a basis for the column space C(A).

(b) Construct an orthogonal basis {⃗v1 ,  ⃗v2 ,  ⃗v3 } using {⃗a1 ,  ⃗a2 ,  ⃗a3 } .

(c) Find an QR Factorization for A .

Solutions: (a)  |A|   =  1 0,  by invertible matrix theorem,  the set of columns {⃗a1 ,  ⃗a2 ,  ⃗a3 } is linearly independent, and is therefore a basis for C(A) by definition (of a basis).

(b) We apply GSP (Gram-Schmidt Process using what we called the Beautiful Formula for the projections in class) to the basis in (a), we obtain (omitting the algebra) an orthonormal basis {⃗u1 ,⃗u2 ,⃗u3 } with

⃗u1  = ( , , ),⃗u2  = ( , , ),⃗u3  = (0, , ),

each being a column vector.

(c) Since Q = [ ⃗u1 ⃗u2 ⃗u3  ],

7.  (10 points) Let

A =  l 0(4)

「 1

2(0) 」 =  l   0(2) 」

1 , 「 11

(1) Find a least square solution of A⃗x = .

(2) Find the distance from to the column space C(A).

Solutions:

(1) We solve the system of normal equations AT Ax∗ = , where AT A = ] , =  「(l) 1

There is only one least square solution for this problem (though not true in general):

x∗ = [2(1) ]

(2) The distance from to the column space C(A) is the same as the least square error,

i.e.,

∥ − Ax∗ ∥ = 「(l) 1 − 「(l)

[2(1) ] = ^84

8.  (10 points) For the quadratic form

Q(⃗x) = 2x1(2) − 6x2(2) + 6x1 x2 ,

find an orthogonal change of variables to transform Q(⃗x) into a new form without cross terms, and write down the new form explicitly.

Solutions: Note that Q(⃗x) = ⃗xT A⃗x with (symmetric) matrix for Q(⃗x) given by

A = [3(2) −6(3) ] ,

which has eigen-values λ 1  = − 7 and λ2  = 3.

A basis for the eigen-space N(A− λ 1 I) is [ −]3(1) , which is normalized as ⃗u1  = [ 0(0)   ]. A basis for the eigen-space N(A − λ2 I) is [1(3) ], which is normalized as ⃗u2  = [ 0(0)   ].

(Note:  for symmetric matrix A  with larger size n, one may need to apply Gram-

Schmidt Process to obtain orthonormal bases for some eigen-spaces N(A − λI).)

Hence the orthogonal matrix P  =[ ⃗u1 ⃗u2  ] =  [ 0(0)

change of variables is ⃗x = P⃗y, which gives

Q(⃗x) = Q(P⃗y) = ⃗yT PT AP⃗y = ⃗yT  [ −0(7)   3(0) ] ⃗y = − 7y1(2) + 3y2(2) .

9.  (13 points) Find a SVD (Singular Value Decomposition) of

A =  l    0(1)   1(1) 」

「−1   1

Solutions: AT A = [0(2)   3(0) ]. Characteristic equation |AT A − λI| = (λ − 3)(λ − 2).    Eigen value λ1  = 3, with normalized eigen-vector ⃗v1  = [1(0) ]. Eigen value λ2  = 2, with normalized eigen-vector ⃗v2  = [0(1) ]. V = [ ⃗v1 ⃗v2  ]= [1(0)   0(1) ].

Singular values σ1  =^λ1  =^3, σ2  =^λ2  =^2. Σ =  l 」

「 0      0 .

Next ⃗u1  = =  「(l) , ⃗u2  = =  「(l) − , and ⃗u3  can be found by solving ⃗y

from system of equations ⃗u1(T)⃗y = 0, ⃗u2(T)⃗y = 0, or equivalently AT ⃗y = 0, i.e.,

A basis for the solutions space is ⃗y =

l     1 」

「 − 1(2) .