# The matrix P is the transition matrix from what basis B to the basis B'

The Matrix

$$P = \begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 3 \\ 3 & 0 & 1 \end{bmatrix}$$ is the transition matrix from what basis B to the basis B' = {(1,0,0),(1,1,0),(1,1,1)} for R3? I'm having a hard time with this one, any help is appreciated, thanks :)

• Is that the matrix you want? – user99914 Apr 6 '15 at 22:38

let the basis $B = \{u, v, w\}.$ then $$Pu = 1(1,0,0)^T+3(1,1,1)^T \to u = P^{-1}(4,3,3)^T = \pmatrix{0.1&-0.1&0.3\\0.9&0.1&-0.3\\-0.3&0.3&0.1}\pmatrix{4\\3\\3}=\pmatrix{1\\3\\0.}$$

can you find $v, w?$

$\mathbf{Hint}:$Recall that the transition matrix has the coordinates of the basis you are looking for as its columns, that is suppose your basis is B=${[(x_1,y_1,z_1),(x_2,y_2,z_2),(x_3,y_3,z_3)]}={(b_1,b_2,b_3)}$

$$1(1,0,0)+0(1,1,0)+3(1,1,1)=(4,3,3)$$ $$1(1,0,0)+1(1,1,0)+0(1,1,1)=(2,1,0)$$

what next?

• Hey. If you don't mind me asking. Is (4,3,3) , (2,1,0) and ( 4,4,1) the basis B? I arrived at the answer – Zhi J Teoh Nov 11 '15 at 4:01

Possibly, the best approach to the matter is in this way.
Given a matrix $\bf P$, you can see it as converting the standard base ($\bf I$) into the vertical vectors of $\bf P$. $$\begin{array}{l} {\bf P} = {\bf P}\;{\bf I}\quad \Rightarrow \quad \left( {{\bf p}_1 |{\bf p}_2 |{\bf p}_3 } \right) = {\bf P}\left( {{\bf u}_1 |{\bf u}_2 |{\bf u}_3 } \right)\quad \Rightarrow \quad {\bf p}_k = {\bf P}\;{\bf u}_k \\ \quad \Downarrow \;{\rm if}\,{\bf P}\,{\rm invertible} \\ {\bf I} = {\bf P}^{\, - \,{\bf 1}} \;{\bf P} = {\bf P}\;{\bf P}^{\, - \,{\bf 1}} \\ \end{array}$$

If the vectors of $\bf P$ also constitute a basis, that implies that the matrix is invertible and that you can reverse the process, getting
${\bf P}^{\, - \,{\bf 1}}$ converts ${\bf P}$ to ${\bf I}$, and
${\bf P}$ converts ${\bf P}^{\, - \,{\bf 1}}$ to ${\bf I}$.

So, for any matrix $\bf X$ and any invertible $\bf P$ we can write $${\bf X} = {\bf I}\;{\bf X} = {\bf P}\;{\bf P}^{\, - \,{\bf 1}} \;{\bf X} = {\bf P}\;\left( {{\bf P}^{\, - \,{\bf 1}} \;{\bf X}} \right)$$

Coming to your case, the above simply translates into $$\bbox[lightyellow] { {\bf X} = \left( {\begin{array}{*{20}c} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array}} \right) = {\bf P}\;\left( {{\bf P}^{\, - \,{\bf 1}} \;{\bf X}} \right) = \left( {\begin{array}{*{20}c} 1 & 1 & 0 \\ 0 & 1 & 3 \\ 3 & 0 & 1 \\ \end{array}} \right)\left( {\frac{1}{{10}}\left( {\begin{array}{*{20}c} 1 & 0 & 3 \\ 9 & {10} & 7 \\ { - 3} & 0 & 1 \\ \end{array}} \right)} \right) }$$