HELP with details! Consider the linear function f : Find the matrix A such that

Question

HELP with details!

Consider the linear function f : Find the matrix A such that f(x) = Ax, for any vector x R3. Compute the rank, a basis for the range, the nullity, and a basis for null space for the following matrices: Consider the matrices Answer the following questions. Compute, if possible, AB and BA. Compute AT and BT. Compute BTAT, and verify that BTAT = (AB)T. Consider a real matrix A MmXn with m>n. The matrix A is of full rank, i.e. rank(A) = n. Find the null space of A. Show that, for any vector x G Mn, xTATAx = |Ax| Show that the matrix ATA is invertible. (Hint: show the null space n(ATA) = {0}.) Find the rank of matrices A, ATA and AAT, for Consider the following Mat lab commands A = [10, 2, 3; 4, 10, 6; 7, 8, 10]; B = A - diag([9,5,l]); Write the matrix B. What is the rank of B? Give a basis for TZ(B). What is the nullity of B? Give a basis for A/"(B). True or False, no explanation is required If m = n and the columns of matrix A G MnXn are linearly independent, then the rows of the matrix are also linearly independent. The dimension of the null space, or nullity, for matrix A is equal to the nullity of AT. The linear system Ax = b can have exactly two solutions. In this question, we will study the linear system Ax = b, where A G RnXn, x G Rn,and b G Mn. If A is invertible, which of the following statements are true. The number of solutions to the linear system Ax = b depends on the particular choice of right-hand side vector b. The range of matrix A, R(A) = Mn. The null space of matrix A, N(A) = {0}

Explanation / Answer

4) The column of A is the image of (1 0 0)^T , (0 1 0)^T and (0 0 1)^T respectively, so :
A=
2 1 1
-2 1 2
-1 1 3

5)

(a) Let's transform into row echelon form
A=
1 3 0 2
1 3 1 6
0 0 1 4

First : R2 <- R2-R1
1 3 0 2
0 0 1 4
0 0 1 4

Second: R3 <- R3-R2
R=
1 3 0 2
0 0 1 4
0 0 0 0

So rank(A)=2 (because one line is zero)

We have to find two independent column for the basis of the range.
The first and 3rd column are clearly independent since a(1,1,0)+b(0,1,1)=(a,a+b,b)=0 => a=b=0
So the basis for the range is (1,1,0)^T and (0,1,1)^T

The nullity is 4-2=2 (rank theorem)
Solving RX=0 (from row echelon matrix) for some X=(a b c d)^T gives :
a+3b+2d=0 => d=-1/2(a+3b)
c+4d=0    => c = -4d = 2(a+3b)

So the vectors looks like (a,b,2(a+3b),-(a+3b)/2) = 1/2a(2,0,4,-1)+1/2b(0,2,12,-3)
So a basis for the null space is (2,0,4,-1)^T and (0,2,12,-3)^T

(b)
A=
2 1 0
-2 1 2
-1 1 3

Since it's a 3x3 matrix, let's compute the determinant, if it is not zero, then rank(A)=3

det(A)=-2*(2*1-(-1)*1)+3*(2*1-(-2)*1)= 6, so rank(A)=3
A basis for the range is (2,-2,-1)^T,(1,1,1)^T,(0,2,3)^T (or the natural basis of R^3, since it's invertible)
The nullity is 3-3=0, so null space is {(0,0,0)^T}

(c)
A=
1 2 3
4 5 6
7 8 9

det(A)=1*(5*9-6*8)-4*(2*9-8*3)+7*(2*6-5*3)=0, so A is not invertible and rank(A)<=2
So rank(A)=2 since the first two column aren't colinear (so linearly independant)
The basis for the range is (1,4,7)^T and (2,5,8)^T
The nullity is 3-2=1

AX=0 =>
a+2b+3c=0 (1)
4a+5b+6c=0 (2)
7a+8b+9c=0 (3)

(2)-(1) : 3(a+b+c)=0 => c=-a-b
(2) => 4a+5b+6(-a-b)=0 => -2a=b , so c=-a+2a=a

So a+b+c=0 and the kernel has vectors like (a,-2a,a) = a(1,-2,1)
The basis for the null space is (1,-2,1)

(6)
(a)
A is 2x3 and B is 3x4 so AB is possible (and will be 2x4), but BA is not possible

AB =
-3 -5 -9 -11
22 34 42 54

(b)
A^T =
0 4
3 -6
1 2

B^T=
4 -1 0
5 -2 1
6 -3 0
7 -4 1

(c)
B^TA^T =
-3 22
-5 34
-9 42
-11 54

Whis is indeed (AB)^T

7)
(a) rank(A)+nullity(A)=n, so null space is trivial (0)
(b)
||Ax||^2 = <Ax,Ax> by definition
But <Ax,Ax>=(Ax)^T Ax = x^TA^TAx (since (Ax)^T=x^TA^T and <x,y>=x^Ty by definition)
So x^TA^Tx=||Ax||^2
(c) Suppose A^TAX = 0 (X is an element of nullspace of A^TA)
Then x^TA^TAx = 0
But we just proved that ||Ax||^2 = x^TA^TAx
So ||Ax||^2 = 0
Since ||.||is a norm, then Ax=0
Since A is a full rank matrix, then x=0
So the null space of A^TA is trivial
So A^A is invertible


8)
Row rank is the same as column rank, so rank <=2, we see that first two rows are obviously independent (not colinear), so rank(A)=2

A^TA=
2 -1
-1 9
rank(A)=2 since det(A^TA)=2*9-1*1=17 (!=0)

AA^T=
2 2 1
2 4 4
1 4 5

det(AA^T)=2*(4*5-4*4)-2*(2*5-4*1)+1*(2*4-4*1)=0 , so rank(A)<=2
The first two column aren't colinear, so linearly independent, hence rank(A)=2


9)
(a)
A=
10 2 3
4 10 6
7 8 10

diag[9,5,1]=
9 0 0
0 5 0
0 0 1

B =
1 2 3
4 5 6
7 8 9
This is the matrix of exercise 5c)
(b)Same as exercice 5)c)
(c)Same as exercice 5)c)

10)
(a) True (row rank = column rank, so if row/col is linearly independent, both are)
(b) False AX=0 doesn't imply A^TX=0
(c) False (either unique, none or infinite)
(d)
i)   False, the solution has unique solution
ii) True since A is invertible, image of A R(A)=R^n
iii) True since A is invertible, null space is trivial

(This was worth more than 1500pts ....)

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