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I am working on a question and was looking for some help. The question is

Suppose that $\mathbf{v}_1 = (1,2)$, $\mathbf{v}_2 = (2,-1)$ and that the basis $\beta$ is $\beta = \left \langle \mathbf{v}_1 , \mathbf{v}_2 \right \rangle$. Let $T$ be the linear transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$ given by $T(\mathbf{v}_1)=\mathbf{v}_1$ and $T(\mathbf{v}_2)= \mathbf{0}$.

(a) Write down the matrix for $T$ in the new basis $\beta$. (You should be able to do this directly from the definition of $T$.)

(b) Use this to write down the matrix for $T$ in the standard basis.

So, the matrix to convert a vector from $\beta$ to the standard basis is

$$ \begin{bmatrix} 1 & 2\\ 2 & -1 \end{bmatrix} .$$

As such, the matrix to convert a vector from the standard basis to $\beta$ is

$$ \begin{bmatrix} 1 & 2\\ 2 & -1 \end{bmatrix}^{-1} = \begin{bmatrix} 1/5 & 2/5 \\ 2/5 & -1/5 \end{bmatrix} $$

Converting $\mathbf{v}_1$ and $\mathbf{v}_2$ into $\beta$ is done in the following manner,

$$ \begin{bmatrix} 1/5 & 2/5 \\ 2/5 & -1/5 \end{bmatrix} \begin{bmatrix} 1\\ 2 \end{bmatrix} = \begin{bmatrix} 1\\ 0 \end{bmatrix} $$

$$ \begin{bmatrix} 1/5 & 2/5 \\ 2/5 & -1/5 \end{bmatrix} \begin{bmatrix} 2\\ -1 \end{bmatrix} = \begin{bmatrix} 0\\ 1 \end{bmatrix}. $$

Which, in the transformation, gives $T_{\beta} (1,0) = (1,0)$ and $T_{\beta}(0,1)=(0,0)$. So the matrix for $T$ in the basis $\beta$ is

$$ A= \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}. $$

Now, I am not sure if what I have done is correct. In any case, how do I proceed with (b)? Thanks for any help. I think I am a bit mixed up.

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2 Answers 2

up vote 1 down vote accepted

What you have right now is this:

$$(\text{$A$ in $\beta$ basis})(\text{input vector in $\beta$ basis}) = \text{output vector in $\beta$ basis}$$

Let me just write this as

$$A v= v'$$

where $v, v', A$ are all understood to be in the $\beta$ basis.

You have the matrix $B$ that converts from $\beta$ basis to standard basis. Multplying by that on the left will convert the output vector to standard basis. But you also want to input a vector in standard basis. You can accomplish this by inserting a $BB^{-1}$ between $A$ and $v$ like so:

$$B(AB^{-1}[Bv]) = Bv'$$

$Bv$ is an input vector in the standard basis. $Bv'$ is an output vector in the standard basis. Therefore, $BAB^{-1}$ converts $A$ from $\beta$ basis to standard basis.

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The answer you've gotten for (a) is correct, but you did a lot of unnecessary work. If $T(v_1) = v_1$ then you know the first column of $A$ contains the coefficients in the basis expansion of $v_1$. Since $v_1$ is in your basis this expansion is just $v_1$ and so the first column is $\begin{bmatrix} 1 \\ 0\end{bmatrix}$. Similarly the second column must be zero.

Now once you've gotten $A$ you change basis, so just compute $BAB^{-1}$ where $B$ is the change of basis matrix you've already found.

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