Finding Invertible Matrices for Canonical Matrix Question

[franky2727]

given A={(1,2,1),(2,4,2),(3,6,3)} findr and invertable matrices Q and P such that Q^-1AP={(I_r,0),(0,0)} where each zero denotes a matrix of zeros not necessarily the same size

paying special attension to the order of the vectors write down the bases of R³ with respect to which Q^-1AP represents the mapping x->Ax

i think i can do the first part getting row opps of r3-3r1 and r2-2r1 and then column opps of c2-2c1 and c3-c1 giving me {(1,0,0),(0,0,0),(0,0,0) and therefore r=1

then i do I₃ with the same row opps giving Q^-1={(1,0,0),(-2,1,0),(-3,0,1)) giving Q=(Q^-1)^-1 = {(1,0,0),(1/2,1,0),(1/3,0,1)

same with the column opps on P gives me {1.-2.-1),(0,1,0),(0,0,1)}=P

i believe this is right but have not done it in a while and may be messing up the method so a check wouldn't go a miss, also i don't know how to do the second part of the question, it looks slightly familia with the getting vectors in the right order but i can't remember where to start so help here would be aprichiated thanks. on a side note this is revision not homework so feal free to splurt it all out :P

 

[HallsofIvy]

It's not difficult to check it yourself is it?

If Q^-1 and P are as you say then you want to calculate
Q^{-1}AP= \begin{bmatrix}1 & 0 & 0 \\ -2 & 1 & 0 \\ -3 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 2 & 1 \\ 2 & 4 & 2\\ 3 & 6 & 3\end{bmatrix}\begin{bmatrix}1 & -2 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}

Is that equal to
\begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}?

 

[franky2727]

ah ye, silly me, what about the second part? no ideas where to start there

 

[franky2727]

paying special attension to the order of the vectors write down the bases of R³ with respect to which Q^-1AP represents the mapping x->Ax

how is this done or even started?

 

[franky2727]

what does this " with respect to which Q^-1AP represents the mapping x-> Ax mean?