# Finding matrix representation of linear transformation with respect to permuted basis

I'm struggling with understanding how I can figure out exactly what the following matrix representation looks like:

Let $$V$$ be an $$n$$-dimensional vector space. Let $$\mathcal{B} = (b_1,\ldots,b_n)$$ be an ordered basis for $$V$$. Now fix a permutation $$\sigma$$ of the set $$\{1,\ldots, n\}$$. Let $$\mathcal{C} = (b_{\sigma(1)},b_{\sigma(2)},\ldots,b_{\sigma(n)})$$, and let $$T : V \to V$$ be such that for each $$i$$, $$T(b_i) = b_{\sigma(i)}$$. Give $$[T]_{\mathcal{C}}^{\mathcal{B}}$$.

I think my main struggle is not quite understanding where I'm stuck. How do I know exactly where the ones and zeroes will be in $$[T]_{\mathcal{C}}^{\mathcal{B}}$$? Any hints or advice is appreciated. Thanks in advance!

• Recall that the columns of a transformation matrix are the images of the basis vectors. – amd Sep 11 at 9:20

$$T : V \to V$$ is the linear transformation that permutes an input vector, in this case, basis vectors from $$B$$.
Let $$\sigma$$ be a permutation of degree $$n$$, then $$T$$ is the $$n\times n$$ permutation matrix associated to $$\sigma$$, call it $$T_{\sigma}$$ and is constructed under the rule $$T_{(i,j)}=1 \iff \sigma(i)=j$$ and $$T_{(i,j)}=0 \iff \sigma(i)\neq j$$. Thus $$T\cdot b_i = b_{\sigma(i)}$$