Thread: Change of coordinate matrix, new basis

1. Change of coordinate matrix, new basis

I've been working on the following problem:

"Let $V$ be a finite-dimensional vector space over a field $F$, and let $\beta=\{x_1, ..., x_n\}$ be an ordered basis for $V$. Let $Q$ be an $n\times n$ invertible matrix with entries from $F$. Define $x_j'=\displaystyle\sum_{i=1}^n Q_{ij}x_i$ for $1 \le j \le n$, and set $\beta'=\{x_1', ..., x_n'\}$. Prove that $\beta'$ is a basis for $V$ and hence that $Q$ is the change of coordinate matrix changing $\beta'$-coordinates into $\beta$-coordinates."

Can someone point me in the right direction with this? Obviously I need to show that $\beta'$ is linearly independent and spans $V$. I know there are n vectors in $\beta'$ so it only remains to show that its elements are linearly independent, correct? I can't seem to figure it out though.

2. Re: Change of coordinate matrix, new basis

Originally Posted by AlexP
I've been working on the following problem:

"Let $V$ be a finite-dimensional vector space over a field $F$, and let $\beta=\{x_1, ..., x_n\}$ be an ordered basis for $V$. Let $Q$ be an $n\times n$ invertible matrix with entries from $F$. Define $x_j'=\displaystyle\sum_{i=1}^n Q_{ij}x_i$ for $1 \le j \le n$, and set $\beta'=\{x_1', ..., x_n'\}$. Prove that $\beta'$ is a basis for $V$ and hence that $Q$ is the change of coordinate matrix changing $\beta'$-coordinates into $\beta$-coordinates."

Can someone point me in the right direction with this? Obviously I need to show that $\beta'$ is linearly independent and spans $V$. I know there are n vectors in $\beta'$ so it only remains to show that its elements are linearly independent, correct? I can't seem to figure it out though.
Well, you know that $Q$ is invertible and thus $Q(\beta)$ is a basis, isn't $Q(\beta)=\beta'$?

3. Re: Change of coordinate matrix, new basis

How are you defining $Q(\beta)$?

4. Re: Change of coordinate matrix, new basis

Originally Posted by AlexP
How are you defining $Q(\beta)$?
The image of the basis, i.e. if we take the matrix of the transformation $Q$ with respect to $Q$ then isn't $x'_j=Q(x_j)$?

5. Re: Change of coordinate matrix, new basis

Originally Posted by AlexP
How are you defining $Q(\beta)$?
in general, if A is a set, and T is a linear transformation (which every matrix is), then T(A) is the set of all T(a) where a is an element of A.

so in this case, $Q(\beta) = \{Q(x_1),Q(x_2),\dots,Q(x_n)\}$. the trick is to show all of these elements are distinct (which you

should be able to do just from the fact that Q is invertible, and hence 1-1). do these n vectors span V? well, they have to, since Q

is invertible, and thus onto (so we can write any vector w in V as Q(v) for some other vector v in V, and we can write v as a linear

combination of the xj, and thus write w = Q(v) as the same linear combination of the Q(xj), since Q is linear).

but, if Q(β) spans V, and has the same cardinality as a basis (this is where the distinctness comes in), it has to be a basis, too,

because it is a minimal spanning set.

**********************

you can also show the linear independence directly:

suppose $c_1x'_1 +\dots +c_nx'_n = 0$. then

$c_1x'_1 +\dots +c_nx'_n = c_1Q(x_1)+\dots +c_nQ(x_n)=$

$Q(c_1x_1)+\dots +Q(c_nx_n) = Q(c_1x_1+\dots +c_nx_n) = 0$.

since Q is 1-1, $c_1x_1+\dots +c_nx_n = 0$, which means by the

linear independence of the xj that $c_1 = \dots = c_n = 0$.

this is a recurring theme in linear algebra:

injective/1-1 ~~ linear independence
surjective/onto ~~ spanning

6. Re: Change of coordinate matrix, new basis

I completely understand how the invertibility of $Q$ results in a new basis by taking $Q(\beta)$, but I didn't think that $Q$ was acting as left multiplication on $x_i$. I feel this is somewhat dumb, but am I really interpreting $x'_j=\displaystyle\sum_{i=1}^nQ_{ij}x_i$ for $1 \le j \le n$ incorrectly? I don't see how it amounts to left multiplication by $Q$ on an element of $\beta$. This must be where my real confusion lies.