# Matrix is computed are permuted

what happens to the matrix of a linear transformation on a finite dimensional vector space when the elements of the basis with respect to which the matrix is computed are permuted among themselves?justify your answer with an example.

• Is the question complete? – sai-kartik May 26 at 15:26
• what happens to the matrix of a linear transformation on a finite dimensional vector space when the elements of the basis with respect to which the matrix is computed are permuted among themselves? justify your answer with example – Gunjan ojha May 26 at 15:51
• Didn’t you ask this same question just the other day? – amd May 27 at 0:55

Let our ordered basis for $$\mathcal{V}$$ be $$B_1 = \{\beta_1, \beta_2, \cdots, \beta_n\}$$. Then a matrix which represents a linear transformation $$T$$ from $$\mathcal{V}$$ to some other vector space $$\mathcal{W}$$ is $$M = \begin{bmatrix} T(\beta_1) & T(\beta_2) & \cdots & T(\beta_n)\end{bmatrix}$$ In other words, the columns of the matrix $$M$$ representing $$T$$ are the outputs of $$T$$ on the basis elements, in order. Hence, changing the order of the basis will change the order of the matrix's columns in the same way. Everything else remains unchanged. So for example if I reverse the order of the basis (i.e. the new basis is now $$B_2 = \{\beta_n, \beta_{n - 1}, \cdots, \beta_1\}$$, then the new matrix $$M'$$ representing $$T$$ is $$M' = \begin{bmatrix} T(\beta_n) & T(\beta_{n-1}) & \cdots & T(\beta_1)\end{bmatrix}$$ Note that this matrix's output is still in the original basis of $$\mathcal{W}$$. So for example, if we have a linear transformation $$T: \mathcal{V} = \mathbb{R}^2 \to \mathcal{W} = \mathbb{R}^3$$ represented by $$M = \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6\end{bmatrix}$$ with respect to the standard basis of $$\mathcal{V}$$, then if we switch the order of the basis elements for $$\mathcal{V}$$ to $$\{\hat{e}_2, \hat{e}_1\}$$, then the new matrix $$M'$$ representing $$T$$ with respect to this new basis has the two columns switched in the same fashion, i.e. $$M' = \begin{bmatrix} 2 & 1 \\ 4 & 3 \\ 6 & 5\end{bmatrix}$$