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Suppose $U$ has a basis $e_1$, $e_2$, $e_3$ and $V$ has a basis $f_1$, $f_2$ and that $T:U \rightarrow V$ is a linear transformation whose matrix $A$ with respect to the given bases is $A=\begin{pmatrix} 0 & 1 & 2\\ 3 & 4 & 5 \end{pmatrix}$.

The first part of the questions asks to show that $e_1$, $e_1$ + $e_2$, $e_1$ + $e_2$ + $e_3$ forms a basis of $U$ and that $f_1$, $f_1$ + $f_2$ forms a basis of $V$. I've shown that. Now there are two things I need help with.

  1. Write a matrix $A'$ of $T$ with respect to $e_1$, $e_1$ + $e_2$, $e_1$ + $e_2$ + $e_3$ and $f_1$, $f_1$ + $f_2$.

  2. Write a change of basis matrix $P$ from $e_1$, $e_2$, $e_3$ to $e_1$, $e_1$ + $e_2$, $e_1$ + $e_2$ + $e_3$.

$$P=\begin{pmatrix} 1 & -1 & 0\\ 0 & 1 & -1\\ 0 & 0 & 1 \end{pmatrix}$$

The answer is given, but I do not know how to work it out. Could you help me? Also, are $e_1$, $e_2$, $e_3$ and $f_1$, $f_2$ supposed to be standard bases? Thank you for your time.

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  • $\begingroup$ vectors that form a basis are independent with respect to each other. So with linear combinations of basis vectors, you create a new basis. So if e1, e2, e3 are a basis, so are e1, e1+e2, e1+e3 I think this answers part of your question $\endgroup$ Commented May 30, 2013 at 20:52

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I'm not sure it's necessary to say the bases are standard at all.

One of the big problems I perceive with writing out matrices as arrays is that people lose the ability to think about them abstractly--they no longer know what the matrices mean. I think this is part of the way matrices are taught, so it's not your fault.

Let's look at the matrix $A$. What does it mean? Well, any matrix can be thought of as a linear function. You can equivalently describe $A$ as so:

$$A(e_1) = 3 f_2, \quad A(e_2) = f_1 + 4 f_2, \quad A(e_3) = 2 f_1 + 5 f_2$$

You'll notice each of these is a column, but with the bases written out.

How does this approach help? It makes problem (1) really trivial. Let $f_2' = f_1 + f_2$. Then $f_2 = f_2' - f_1$. Similarly, let $e_2' = e_1 + e_2$ and $e_3' = e_1 + e_2 + e_3$. Now, you should be able to evaluate the components. Consider $A(e_2')$:

$$\begin{align*}A(e_2') &= A(e_1) + A(e_2) \\ &= 3f_2 + f_1 + 4 f_2 \\ &= f_1 + 7 f_2 \\ &= f_1 + 7 (f_2'- f_1) \\&= -6 f_1 + 7 f_2'\end{align*}$$

All it takes to evaluate that is simple substitution.

Now, for (2), what is it we want to do when we use the change of basis matrix? We want to feed in a vector in the new (primed) basis, convert it back to the old basis, use the map $A$, and the convert back to the primed basis. The change of basis map should then take a basis vector in the primed basis and convert it to a basis vector in the original, unprimed basis. So like this:

$$P(e_1') = e_1, \quad P(e_2') = e_2, \quad P(e_3') = e_3$$

I'll handle only one of these cases, the last one. You should observe that $e_3' = e_2' + e_3$, or $e_3 = -e_2' + e_3'$. That basically does it--this is exactly the third column of $P$ that you're given. All you have to do for this part is express $e_1, e_2, e_3$ as linear combinations of $e_1', e_2', e_3'$. Since $e_1 = e_1'$, you only really have to do 2 equations, at that.

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  • $\begingroup$ In the first part for $A(e_2')$ why does it need to be expressed as $−6f_1+7f'_2$ instead of $f_1 + 7 f_2$? $\endgroup$ Commented May 30, 2013 at 21:13
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    $\begingroup$ The problems says "express in terms of $f_1$ and $f_1 + f_2$." I defined $f_2' = f_1 + f_2$ for convenience. $\endgroup$ Commented May 30, 2013 at 21:17
  • $\begingroup$ Thank you for your answer. $\endgroup$ Commented May 30, 2013 at 21:20

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