Bases and Coordinates: B1 and B2 for [R][/3] - Homework Statement

[Eleni]

Homework Statement

Let B1={([u][/1]),([u][/2]),([u][/3])}={(1,1,1),(0,2,-1),(1,0,2)} and
B2={([v][/1]),([v][/2]),([v][/3])}={(1,0,1),(1,-1,2),(0,2,1)}
a) Show that B1 is a basis for [R][/3]
b) Find the coordinates of w=(2,3,1) relative to B1
c)Given that B2 is a basis for [R[/3], find the transition matrix [P[/B1→B2]
d) Use the transition matrix [P[/B1→B2] to find the coordinates of w relative to B2
e) What is the relationship between [P[/B1→B2] and [P[/B2→B1]? Find [P[/B2→B1]
f) Suppose [[x]][/B2]= (0, 3, -1). Use the appropriate transition matrix to find [[x]][/B1]

The Attempt at a Solution

a) if B1 is a basis for [R][/3] then there should be a unique solution for all the coefficients where;
[x][/1] [u][/1] + [x][/2] [u][/2] + [x][/3][u][/3]=(0,0,0)

the system of equation gives the matrix of the form;

1 0 1 1 0 0
1 2 0 0 1 0
1 -1 2 →RREF = 0 0 1
solving for Ax=0 gives [x][/1] [u][/1] + [x][/2] [u][/2] + [x][/3][u][/3]=(0,0,0)
Thus a unique solution exists for all R(X)∈[R][/3] and B1 is a basis for [R][/3]

b) To find w relative to B1 we use the equation from part a) and sub in w values.

[[w]][/B1] = [x][/1] (1,1,1) + [x][/2] (0,2,-1) + [x][/3](1,0,2)=(2,3,1)

This is matrix form yeilds;

1 0 1 2 1 0 0 3
1 2 0 3 0 1 0 0
1 -1 2 1→RREF = 0 0 1 -1
∴ [x][/1] = 3
[x][/2] = 0
[x][/3] = -1

So the coordinate vector [[w]][/B1] = (3, 0, -1)

c) The transition matrix [P[/B1→B2] =

1 1 0 1 0 1 1 0 0 2/3 2/3 1/6
1 -1 2 1 2 0 0 1 0 1/3 -2/3 5/6
0 2 1 1 -1 2 →RREF = 0 0 1 1/3 1/3 1/3

∴ The transition matrix [P[/B1→B2] = 2/3 2/3 1/6
1/3 -2/3 5/6
1/3 1/3 1/3
d) To find [[w]][/B2] we will callculate [[P[/B1→B2]][w]

2/3 2/3 1/6 2 1 0 0 5
1/3 -2/3 5/6 3 0 1 0 -2
1/3 1/3 1/3 1 →RREF = 0 0 1 0

∴ The coordinate vector of [[w]][/B2] = (5,-2,0)

e)What is the relationship between [P[/B1→B2] and [P[/B2→B1]? Find [P[/B2→B1]

I am unsure about this one. I am sure there must be a theorem or rule regarding this but I can't find it in my textbook.

f) I am guessing that once I have found [P[/B2→B1] then I can find [[x][/B1] the same as I found [[w]][/B2] but with the transition matrix from [P[/B2→B1] in part e)I am not 100% confident in my answers part parts a) through to d) so if you can see errors please correct me. However I am predominantly concerned with parts e) and f)

I thank you in advance for any advice and help.

 

[Ray Vickson]

Homework Statement

Let B1={([u][/1]),([u][/2]),([u][/3])}={(1,1,1),(0,2,-1),(1,0,2)} and
B2={([v][/1]),([v][/2]),([v][/3])}={(1,0,1),(1,-1,2),(0,2,1)}
a) Show that B1 is a basis for [R][/3]
b) Find the coordinates of w=(2,3,1) relative to B1
c)Given that B2 is a basis for [R[/3], find the transition matrix [P[/B1→B2]
d) Use the transition matrix [P[/B1→B2] to find the coordinates of w relative to B2
e) What is the relationship between [P[/B1→B2] and [P[/B2→B1]? Find [P[/B2→B1]
f) Suppose [[x]][/B2]= (0, 3, -1). Use the appropriate transition matrix to find [[x]][/B1]

The Attempt at a Solution

a) if B1 is a basis for [R][/3] then there should be a unique solution for all the coefficients where;
[x][/1] [u][/1] + [x][/2] [u][/2] + [x][/3][u][/3]=(0,0,0)

the system of equation gives the matrix of the form;

1 0 1 1 0 0
1 2 0 0 1 0
1 -1 2 →RREF = 0 0 1
solving for Ax=0 gives [x][/1] [u][/1] + [x][/2] [u][/2] + [x][/3][u][/3]=(0,0,0)
Thus a unique solution exists for all R(X)∈[R][/3] and B1 is a basis for [R][/3]

b) To find w relative to B1 we use the equation from part a) and sub in w values.

[[w]][/B1] = [x][/1] (1,1,1) + [x][/2] (0,2,-1) + [x][/3](1,0,2)=(2,3,1)

This is matrix form yeilds;

1 0 1 2 1 0 0 3
1 2 0 3 0 1 0 0
1 -1 2 1→RREF = 0 0 1 -1
∴ [x][/1] = 3
[x][/2] = 0
[x][/3] = -1

So the coordinate vector [[w]][/B1] = (3, 0, -1)

c) The transition matrix [P[/B1→B2] =

1 1 0 1 0 1 1 0 0 2/3 2/3 1/6
1 -1 2 1 2 0 0 1 0 1/3 -2/3 5/6
0 2 1 1 -1 2 →RREF = 0 0 1 1/3 1/3 1/3

∴ The transition matrix [P[/B1→B2] = 2/3 2/3 1/6
1/3 -2/3 5/6
1/3 1/3 1/3
d) To find [[w]][/B2] we will callculate [[P[/B1→B2]][w]

2/3 2/3 1/6 2 1 0 0 5
1/3 -2/3 5/6 3 0 1 0 -2
1/3 1/3 1/3 1 →RREF = 0 0 1 0

∴ The coordinate vector of [[w]][/B2] = (5,-2,0)

e)What is the relationship between [P[/B1→B2] and [P[/B2→B1]? Find [P[/B2→B1]

I am unsure about this one. I am sure there must be a theorem or rule regarding this but I can't find it in my textbook.

f) I am guessing that once I have found [P[/B2→B1] then I can find [[x][/B1] the same as I found [[w]][/B2] but with the transition matrix from [P[/B2→B1] in part e)I am not 100% confident in my answers part parts a) through to d) so if you can see errors please correct me. However I am predominantly concerned with parts e) and f)

I thank you in advance for any advice and help.

Please, please, please try to avoid the horrible notation [u][/1], etc. If you mean u₁ just use plain ASCII---like u_1-- or else use the "x₂" button on the input panel at the top of the input page, which is what I used to get u₁. Better still, use LaTeX, and write $u_1$ or ${\bf u}_1$, etc.

I won't even try to read your material---it hurts my eyes to do so.

However, for the last question (about the relationship between the matrices $R = P[B_1 \to B_2]$ and $S = P[B_2 \to B_1]$, just think about what these matrices accomplish. What must happen to a vector ${\bf x} = (a,b,c)$ in basis $B_1$ when you apply $R$ to it, then apply $S$ to the first result? In other words, look at $B_1 \to B_2 \to B_1$ and ask yourself what you must get.

BTW: if you want to see what was the LaTeX code I wrote, just right-click on a displayed expression and choose the menu item "Show math as → Tex commands". You will see it is an order of magnitude easier to write than what you were typing.

 

[Eleni]

Please, please, please try to avoid the horrible notation [u][/1], etc. If you mean u₁ just use plain ASCII---like u_1-- or else use the "x₂" button on the input panel at the top of the input page, which is what I used to get u₁. Better still, use LaTeX, and write $u_1$ or ${\bf u}_1$, etc.

I won't even try to read your material---it hurts my eyes to do so.

However, for the last question (about the relationship between the matrices $R = P[B_1 \to B_2]$ and $S = P[B_2 \to B_1]$, just think about what these matrices accomplish. What must happen to a vector ${\bf x} = (a,b,c)$ in basis $B_1$ when you apply $R$ to it, then apply $S$ to the first result? In other words, look at $B_1 \to B_2 \to B_1$ and ask yourself what you must get.

BTW: if you want to see what was the LaTeX code I wrote, just right-click on a displayed expression and choose the menu item "Show math as → Tex commands". You will see it is an order of magnitude easier to write than what you were typing.

My apologies for the absolutely horrible notation. I used the "x₂" button but deleted Sub and replaced it with my values (as I have never used this type of button before.) I was very disappointed to have typed out the entire question, post it and then find that it did not make sense as it had when I typed it in the draft box. That will teach me to preview my work before hand. Apologies again and thank you for pointing me to latex I will try and make sure my posts make more sense in the future

 

[Ray Vickson]

Please, please, please try to avoid the horrible notation [u][/1], etc. If you mean u₁ just use plain ASCII---like u_1-- or else use the "x₂" button on the input panel at the top of the input page, which is what I used to get u₁. Better still, use LaTeX, and write $u_1$ or ${\bf u}_1$, etc.

I won't even try to read your material---it hurts my eyes to do so.

However, for the last question (about the relationship between the matrices $R = P[B_1 \to B_2]$ and $S = P[B_2 \to B_1]$, just think about what these matrices accomplish. What must happen to a vector ${\bf x} = (a,b,c)$ in basis $B_1$ when you apply $R$ to it, then apply $S$ to the first result? In other words, look at $B_1 \to B_2 \to B_1$ and ask yourself what you must get.

BTW: if you want to see what was the LaTeX code I wrote, just right-click on a displayed expression and choose the menu item "Show math as → Tex commands". You will see it is an order of magnitude easier to write than what you were typing.

When you write u, followed by a click on the x₂ button you will see on your screen u[S UB][/S UB] (remove the spaces!), and with your cursor situated between the "]" and the "[". Just type whatever you want in that space and then use your mouse (or arrow keys) to go past the last "]".