Given a finite dimensional vector space V of dimension n with ordered bases $B$ and $B^{'}$ how can I go about proving that the transition matrix from $B$ and $B^{'}$ is invertible?
I tried expressing the determinant of this matrix but do not have concrete numbers to show it is non-zero.
Is there an easy way of proving this?
Best Answer
You don't need the numbers at all. Let $B=\{v_{1},\cdots,v_{n}\}$ and $B'=\{w_{1},\cdots,w_{n}\}$ be two basis of $V$. For every $v_{i}\in B$, we can write $v_{i}=\sum_{j=1}^{n}\alpha_{ij}w_{j}$. Thus we can build a matrix $M=(\alpha_{ij})_{ij}$ (it is the transition matrix from $B$ to $B'$). Note that because $B'$ is basis so it is linearly independent. From here, if we reduce matrix $M$ using row elementary operation, we can get the diagonal matrix which the entries in diagonal are all not zeros, so $\mathrm{det}(M)\ne 0$.