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Ok, so I have a question which is:

$$ A = \left ( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array} \right ) $$

Find non-singular matrices P,Q such that $PAQ$ is a diagonal matrix of 1's and 0's, with the 1's appearing before the 0's, via a change of basis.

Now I can do this simply with row/column operations to get that: $$P=I$$ and $$Q= \left ( \begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{array} \right ) $$

But I am unsure how to do this via a change of basis?

I can see that $null(A)=span(\{(1,1,-1)\})$ and $range(A)=span(\{(1,0,0),(0,1,0)\})$ and that a change of basis matrix to this basis is given by:

$$ \left ( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & -1 \end{array} \right ) $$

and I can see that this does the job for Q, but I'm not really sure what I have done here (I've just blindly changed basis and this matrix has popped out!)

any help as to how I should be going about this and what it is I am actually doing would be great.

Thanks very much for any help

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Do you know about eigenvectors and eigenvalues? –  TMM Apr 7 '12 at 13:10
    
@TMM Yeah, so if I have a matrix with dim(V) distinct eigenvalues then I would have a basis such that the matrix is diagonal (which is a basis of eigenvectors), but here as we are in $\mathbb{R}$ we need not have 3 and in fact only have 2 distinct eigenvalues, i think I must be missing something? –  hmmmm Apr 7 '12 at 13:23
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1 Answer 1

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The procedure in general is as follows:

  • Compute a basis $\mathcal B_1$ of $\textrm{null}(A)$.
  • Extend this basis to a basis of all the space via $\mathcal B_1 \cup \mathcal B_2$.
  • Now $\mathcal B_3=\{Av: v\in \mathcal{B}_2\}$ is a basis of $\textrm{range}(A)$.
  • Extend $\mathcal B_3$ to a basis of all the space via $\mathcal B_3 \cup \mathcal B_4$.

The two bases you should consider are $\mathcal B_1 \cup \mathcal B_2$ and $\mathcal B_3 \cup \mathcal B_4$ (try to understand why).

In your case $\mathcal B_1=\{(1,1,-1)\}$, you extended this randomly to a basis adding $\{(1,0,0),(0,1,0)\}$. The magic is that $A(1,0,0)=(1,0,0)$ and $A(0,1,0)=(0,1,0)$, so that $\mathcal B_3=\mathcal B_2$. The last step is to complete $\mathcal B_3$ to a basis, which can be done adding $(0,0,1)$.

The final result is: the first basis is the one you wrote down and the second basis is the canonical basis (hence $P=I$).

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