What is the approach for showing linear independence in a basis?

[shaon0]

1. Homework Statement
See Attachment. Help with b) c) will be appreciated.

3. The Attempt at a Solution

For the third question, my approach is sub in values for a and b which correspond to the co-effs. of the given basis. Then assemble a matrix from them. Eg: f[(5,0);(1,3)]=[(1,1);(2;-1)] which gives me [(-17,4);(15,-3)] for c).

Any help is appreciated.

 

[vela]

1. Homework Statement
See Attachment. Help with b) c) will be appreciated.

3. The Attempt at a Solution

For the third question, my approach is sub in values for a and b which correspond to the co-effs. of the given basis. Then assemble a matrix from them. Eg: f[(5,0);(1,3)]=[(1,1);(2;-1)] which gives me [(-17,4);(15,-3)] for c).

Any help is appreciated.

I'm not sure how you got your final matrix.

What you want to do is apply the function to the basis vectors. If you do this with the first basis vector, you get
$$f(1+x) = \begin{pmatrix}5 \\ 0\end{pmatrix}$$This may be, in part, what you intended to say, but what you wrote above is backward. Now you want to express the resulting vector in terms of the given basis of R². That is, you want to find the constants a₁₁ and a₂₁ such that
$$\begin{pmatrix}5 \\ 0\end{pmatrix} = a_{11} \begin{pmatrix}2 \\ 3\end{pmatrix} + a_{21} \begin{pmatrix}1 \\ 2\end{pmatrix}$$If you work this out, you'll find a₁₁=10 and a₂₁=-15. These two constants will be the first column of the matrix
$$A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = \begin{pmatrix} 10 & a_{12} \\ -15 & a_{22} \end{pmatrix}$$which represents the function f relative to the two bases. Do the same thing with the second basis vector to find the second column of A.

 

[shaon0]

I'm not sure how you got your final matrix.

What you want to do is apply the function to the basis vectors. If you do this with the first basis vector, you get
$$f(1+x) = \begin{pmatrix}5 \\ 0\end{pmatrix}$$This may be, in part, what you intended to say, but what you wrote above is backward. Now you want to express the resulting vector in terms of the given basis of R². That is, you want to find the constants a₁₁ and a₂₁ such that
$$\begin{pmatrix}5 \\ 0\end{pmatrix} = a_{11} \begin{pmatrix}2 \\ 3\end{pmatrix} + a_{21} \begin{pmatrix}1 \\ 2\end{pmatrix}$$If you work this out, you'll find a₁₁=10 and a₂₁=-15. These two constants will be the first column of the matrix
$$A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = \begin{pmatrix} 10 & a_{12} \\ -15 & a_{22} \end{pmatrix}$$which represents the function f relative to the two bases. Do the same thing with the second basis vector to find the second column of A.

Wow! Thanks a lot, it's so much clearer now :) Would you happen to know what to do for 6b). I've tried the same approach you suggested for the other question but didn't get the right answer.

 

[vela]

You do the same thing. Show us your work so we can see where you're the problem lies.

 

[shaon0]

You do the same thing. Show us your work so we can see where you're the problem lies.

-2[1,5]+5[1,6]=[3,-4] and 1[1,5]+2[1,6]=[3,13] Basis: {[3,-4];[3,13]}

 

[vela]

I'm not sure what that has to do with the matrix for the transformation.

 

[shaon0]

I'm not sure what that has to do with the matrix for the transformation.

T[(1,5);(1,6)]=[(-2,5);(1,2)].
a1+5a2=-2;a1+6a2=1;a3+5a4=5 and a3+6a4=2 where T[(a1,a3);(a2;a4)]
[(1,5);(1,6)|(-2,1)] and [(1,5);(1,6)|(5,2)] which gives me, T=[(-13,28);(11,-23)] which is not correct.

 

[vela]

Please explain your steps and use standard notation. I still can't figure out what you're doing. For example, what does "T[(1,5);(1,6)]=[(-2,5);(1,2)]" mean?

 

[shaon0]

Please explain your steps and use standard notation. I still can't figure out what you're doing. For example, what does "T[(1,5);(1,6)]=[(-2,5);(1,2)]" mean?

Sorry vela, I've attached my working on this msg.

 

[vela]

Suppose you have two vectors $\vec{x}, \vec{y} \in \mathbb{R}^2$ such that $\vec{y} = T(\vec{x})$. Then you can write
\begin{align*}
\vec{x} = x_1 \hat{e}_1 + x_2\hat{e}_2 \\
\vec{y} = y_1 \hat{e}_1 + y_2\hat{e}_2
\end{align*}where $\hat{e}_1=\begin{pmatrix}1 \\ 0 \end{pmatrix}$ and $\hat{e}_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ are the standard basis vectors for $\mathbb{R}^2$. When you say T has the matrix $\begin{pmatrix} -2 & 1 \\ 5 & 2 \end{pmatrix}$ with respect to this basis, it means that
$$\begin{pmatrix} y_1 \\ y_2 \end{pmatrix} = \begin{pmatrix}-2 & 1 \\ 5 & 2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$$
In this problem, you're given a second basis, $\hat{b}_1 = \begin{pmatrix} 1 \\ 5\end{pmatrix}$ and $\hat{b}_2 = \begin{pmatrix} 1 \\ 6 \end{pmatrix}$. As before, you can express $\vec{x}$ and $\vec{y}$ as a linear combination of them:
\begin{align*}
\vec{x} = c_1 \hat{b}_1 + c_2 \hat{b}_2 \\
\vec{y} = d_1 \hat{b}_1 + d_2 \hat{b}_2
\end{align*}Obviously c₁ and c₂ aren't going to be the same numbers as x₁ and x₂, and d₁ and d₂ won't be equal to y₁ and y₂. However, because $\vec{y} = T(\vec{x})$, c₁, c₂, d₁, and d₂ will be related by a matrix equation
$$\begin{pmatrix} d_1 \\ d_2 \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}$$It is this matrix, $\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$, that you're looking for, the matrix of T relative to the basis B.

Now consider what happens when you choose c₁=1 and c₂=0. If you think about it a bit, you should be able to solve for the first column of the matrix. Conceptually, you're doing exactly the same thing you did in the previous problem.

 

[shaon0]

Suppose you have two vectors $\vec{x}, \vec{y} \in \mathbb{R}^2$ such that $\vec{y} = T(\vec{x})$. Then you can write
\begin{align*}
\vec{x} = x_1 \hat{e}_1 + x_2\hat{e}_2 \\
\vec{y} = y_1 \hat{e}_1 + y_2\hat{e}_2
\end{align*}where $\hat{e}_1=\begin{pmatrix}1 \\ 0 \end{pmatrix}$ and $\hat{e}_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ are the standard basis vectors for $\mathbb{R}^2$. When you say T has the matrix $\begin{pmatrix} -2 & 1 \\ 5 & 2 \end{pmatrix}$ with respect to this basis, it means that
$$\begin{pmatrix} y_1 \\ y_2 \end{pmatrix} = \begin{pmatrix}-2 & 1 \\ 5 & 2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$$
In this problem, you're given a second basis, $\hat{b}_1 = \begin{pmatrix} 1 \\ 5\end{pmatrix}$ and $\hat{b}_2 = \begin{pmatrix} 1 \\ 6 \end{pmatrix}$. As before, you can express $\vec{x}$ and $\vec{y}$ as a linear combination of them:
\begin{align*}
\vec{x} = c_1 \hat{b}_1 + c_2 \hat{b}_2 \\
\vec{y} = d_1 \hat{b}_1 + d_2 \hat{b}_2
\end{align*}Obviously c₁ and c₂ aren't going to be the same numbers as x₁ and x₂, and d₁ and d₂ won't be equal to y₁ and y₂. However, because $\vec{y} = T(\vec{x})$, c₁, c₂, d₁, and d₂ will be related by a matrix equation
$$\begin{pmatrix} d_1 \\ d_2 \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}$$It is this matrix, $\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$, that you're looking for, the matrix of T relative to the basis B.

Now consider what happens when you choose c₁=1 and c₂=0. If you think about it a bit, you should be able to solve for the first column of the matrix. Conceptually, you're doing exactly the same thing you did in the previous problem.

Ok, thanks Vela.

 

[2slowtogofast]

for 6A it says show the vectors are a basis of R^2. Since the vectors are 2-d is it enough just to show they are linearly ind then that shows they are a basis. I thought there was somthing that said

a set of n-dimensional vectors is a basis of R^n if the set is linearly IND.

I am not posting this as a solution I am asking as a question

 

[vela]

Yes, that's right, though what you said wasn't quite right. There's a theorem that says: if you have a set of n linearly independent vectors (not a set of n-dimensional vectors) from an n-dimensional space, that set of n vectors is a basis. So in this problem, it's enough to show that the vectors are linearly independent to prove they are a basis for R².