I need process of answer Let L be the linear transformation from R2 to R2 given

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I need process of answer

Let L be the linear transformation from R2 to R2 given by the following equation: L(x, y)T = (x + 2y, x + 3y)T. Let M denote the matrix of L with respect to the standard bases for R2 and R2. Determine the matrix M. Let epsilon = {u1, u2} = {(2, 1)T (5, 3)T} be a new basis for R2. Find the matrix A = [L] for L with respect to this new basis. Check that M and A have the same trace and the same determinant. Let A be an n x n matrix. Define the trace of A by the formula tr(A)= That is, the trace of a matrix is the sum of the of the diagonal entries of the matrix. Recall that for n x n matrices A and B, tr(AB) = tr(BA). Prove that for two n x n matrices A and B, tr(A+B) = tr(A)+tr(B). Prove that for any n x n matrix A, tr(A) = tr(AT). Show that there do not exist n x n(n 0)matrices A and B such that AB - BA = In where In is the n x n identity matrix. (Hint, take the trace of the left-hand side and take the trace of the right- that they cannot be equal.) Let G be the undirected graph with adjacency matrix Draw a diagram of the graph G. Compute the characteristic polynomial of A. State a specific formula for A3 that is implied by the Cayley-Hamilton Theorem. Check by direct calculation that this formula is true. Use the formula to determine A10, and explain the consequences for walks of length 10 on the graph G.

Explanation / Answer

3)

a) Just by reading the coeffcicients in the defininition of L we can say that :

M =
1 2
1 3

b)
E=
2 5
1 3

det(E)=6-5=1

So the inverse of the matrix
E^(-1) =
3 -5
-1 2

T(2,1) = (4,5) and E^(-1)(4,5)=(-13 6)^T
T(5,3) = (0,1) and E^(-1)(11,14)=(-37,17)

So the matrix is :
A=
-13 -37
6   17

trace(A)=17-13=4 and det(A) = -17*13+37*6=1
trace(M)=3+1 = 4 and det(M)=3-2=1

Both have same trace and determinant.

4)

(a) the diagonal elements at position ii of the matrix A+B is aii+bii
So tr(A+B)=sum(i=1..n)(aii+bii)=sum(i=1..n)aii+ sum(i=1..n)bii = tr(A)+tr(B)

(b) the diagonal elements of A and A^T are the same, so the sum is unchanged and : tr(A^T) = tr(A)

(c) tr(AB-BA)=tr(AB)-tr(BA)=tr(AB)-tr(AB)=0 , but tr(In) = 1+1..+1 = n
Hence if n!=0, you cannot find matrices A,B such that AB-BA = In

5)
I let you draw G.
By expanding by the first row the determinant we have :

det(A-xI) = (1-x)( -x*(1-x)-1 ) - (1-x) = -x^3+2x^2+x-2

According to cayley-hamilton theorem this polynom verifies P(A)=0 => A^3 = 2A^2+A-2I

Let's check :

A^2 =
2 1 1
1 2 1
1 1 2

A^3 =
3 3 2
3 2 3
2 3 3

2A^2+A =
5 3 2
3 4 3
2 3 5

2A^2+A-2I = A^3 => this is true
3 3 2
3 2 3
2 3 3


A^3 = 2A^2+A-2I
A^4 = A*(2A^2+A-2I) = 2A^3+A^2-2A = 2(2A^2+A-2I)+A^2-2A=5A^2-4I
A^6 = (2A^2+A-2I)^2 = 4A^4+A^2+4I+4A^3-8A^2-4A = 4(5A^2-4I)+A^2+4I+4(2A^2+A-2I)-4A^2-4A
A^6 = (20+1+8-8)A^2 + (-8+4-16)I = 21A^2-20I
A^10 = A^6*A^4 = (5A^2-4I)(21A^2-20I) = 105A^4-84A^2-100A^2+80I = 105(5A^2-4I)-184A^2+80I
A^10 = (105*5-184)A^2 +(80-420)I = 341A^2-340I

Hence :
A^10 =
342 341 341
341 342 341
341 341 342

The ij-th entry of the 10-th power of A tells you the number of walks of length 10 from vertex i to vertex j
So they are 341 number of walks of length 10 from vertex i to vertex j (i!=j) and 342 from vertex i to i (same)

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