Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.