Let A AT and of 5 Times 5 Show that det A 0 Let A be a 4 T

Let A = -A^T and of 5 Times 5. Show that det (A) = 0. Let A be a 4 Times 6 matrix with dim null (A) = 2. Find dim col (A) and dim row (A). Is the set of matrices H = [v_1, v_2, v_3] with v_1 = (a, 0, a)^T, v_2 = (0, a, 0)^T and v_3 = (a, 0, a)^T, with a any real number, a subspace of the

Solution

4.

a)

We know that determinant is invariant to taking transpose. ie

det(A)=det(A^T)

Also we know for two square matrices, A,B

det(AB)=det(A)=det(B)

det(AA^T)=det(A*(-A))=-det(A)^2

det(AA^T)=det(A)det(A^T)=det(A)det(A)=det(A)^2

So, det(A)^2=-det(A)^2

So, det(A)=0

b)

For a matrix of size:mxn by rank nullity theoremL

dim null(A)+ rank (A)=n

rank(A)=dim row(A)

n=6, dim null(A)=2

Hence, dim row(A)=4

dim col(A)=dim row(A)=4

5.

We denote the transpose by: \' ie x^T=x\'

In H a is the parameter

1.

0 belongs to H. This can be seen by setting:a=0

2.

Let, A and B be two matrices in H with parameters,a and b

A+B has columns

(a,0,a)\'+(b,0,b)\'=(a+b,0,a+b)\'

(0,a,0)\'+(0,b,0)\'=(0,a+b,0)\'

(a,0,a)\'+(b,0,b)\'=(a+b,0,a+b)\'

Hence, A+B is also in H with parameter, a+b

3.

Let c be a real number and A be in H with parameter, a

Sol columns of cA are:

c(a,0,a)\'=(ca,0,ca)\'

c(0,a,0)\'=(0,ca,0)\'

c(a,0,a)\'=(ca,0,ca)\'

So, cA is also in H with parameter, ca

Hence, H is a subspace.