Change of coordinate matrix, new basis

**AlexP** · July 11th 2011, 07:23 PM

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.

**Drexel28** · July 11th 2011, 07:41 PM

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 $Q(\beta)=\beta'$ ?

**AlexP** · July 11th 2011, 08:10 PM

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

**Drexel28** · July 11th 2011, 08:12 PM

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 $x'_j=Q(x_j)$ ?

**Deveno** · July 13th 2011, 12:10 PM

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

**AlexP** · July 13th 2011, 08:39 PM

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.

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