Change of coordinate proof

**Jameson** · November 4th 2008, 06:51 PM

Let V be a finite-dimensional vector space over F, and let $B=(x_1 , ...x_{n})$ be an ordered basis for V. Let Q be an n by n invertible matrix with entries from F.

Define $x'_{j}= \sum_{i=1}^{n} Q_{ij}x_{i}$ , where $1 \le j \le n$

and set $B'=(x'_{1}, ...x'_{n})$ .

Prove that B' is a basis for V and that Q in the change of coordinate matrix from B to B'.

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I've been thinking about this for a while. For B' to be a basis for V, it must be linearly independent and span(V). I just can't show it's linearly independent for the life of me. I've tried thinking of several ways to represent the relation of B' to QB' but nothing is working well. Should I make matrix representations of B and B'?

I need a little push.

**ThePerfectHacker** · November 4th 2008, 07:53 PM

Originally Posted by Jameson

Let V be a finite-dimensional vector space over F, and let $B=(x_1 , ...x_{n})$ be an ordered basis for V. Let Q be an n by n invertible matrix with entries from F.

Define $x'_{j}= \sum_{i=1}^{n} Q_{ij}x_{i}$ , where $1 \le j \le n$

Say that $\sum_{j} a_j x_j' = 0 \implies \sum_j a_j \sum_i Q_{ij}x_i = 0 \implies \sum_i \sum_j a_j Q_{ij} x_i = 0$
Therefore, $\sum_j a_j Q_{ij} = 0$ for each $i$ by linear independence on $(x_1,...,x_n)$ .... [1]
Let $\bold{c}_1,...,\bold{c}_n$ be the coloumns of matrix $Q$ in order.

Then, $\sum_j a_j\bold{c}_j = \bold{0}$ where $\bold{0}$ is a coloumn of zeros by [1].
But this means that $a_j = 0$ since $Q$ is invertible (and so its coloumns are linearly independent).

Since $(x'_1,...,x_n')$ is a linearly independent set of $n$ elements it means it must also be a basis.

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