# find new base vectors based on old ones and the given matrices in both bases

I've got the following matrix $A$ for an endorphism within a base $v_1, v_2, v_3$

$$A = \left( \begin{array}{ccc} 0 & 0 & -1 \\ 1 & 0 & -3 \\ 0 & 1 & -3 \\ \end{array} \right)$$

I need to find the base vectors $v^'_1, v^'_2, v^'_3$ for which the matrix of this endorphism looks like follows (this is the Jordan matrix of this endorphism as I worked out)

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

So I know there must exist a matrix $T$ so that $AT = TA^'$. But how does this help to get the base vectors $v^'_1, v^'_2, v^'_3$ based on $v_1, v_2, v_3$?

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So far I found only solutions on the internet which use generalized eigenvectors and nilpotent etc. which are all terms we haven't learnt yet.. So I somehow need another way to find the base vectors.. So just if I knew matrix $A$ has another looking $B$ in another base independent of Jordan form etc. –  Johann Jokum May 1 '12 at 19:11

The columns of the matrix $T$ express the new basis (not "base") vectors $v_i'$ in terms of the old basis vectors $v_j$ via $v_i'=\sum_jT_{ji}v_j$. It's not quite correct to say that you're looking for "the" basis vectors $v_i'$ for which the matrix of the endomorphism is $A'$, since there are infinitely many such bases; in particular, if you have one such basis, you can get another one by scaling all basis vectors by the same factor, which just introduces a constant factor on either side of the condition $AT=TA'$. Here's how you can find one such basis.

From the first column of $A'$ you can see that $a(v_1')=-v_1'$ (where $a$ is the endomorphism). This yields an equation for the first column $c_1$ of $T$, which contains the coefficients of $v_1'$ expressed in terms of the $v_j$:

$$Ac_1=-c_1\;.$$

The solutions of this equation are the multiples of $\pmatrix{1\\2\\1}$; we can use any of them, e.g. $\pmatrix{1\\2\\1}$ itself.

The second column of $A'$ tells you that $a(v_2')=v_1'-v_2'$, and thus $Ac_2=c_1-c_2$. Again we can make an arbitrary choice since to any solution $c_2$ we can add any multiple of $c_1$ to get another one. Let's use $c_2=\pmatrix{1\\1\\0}$. The third column of $A'$ yields $a(v_3')=v_2'-v_3'$, and thus $Ac_3=c_2-c_3$. One solution is $c_3=\pmatrix{1\\0\\0}$ (again up to an arbitrary multiple of $c_1$), so in total we have

\begin{align} v_1'&=v_1+2v_2+v_3\;,\\ v_2'&=v_1+v_2\;,\\ v_3'&=v_1\; \end{align}

and

$$T=\pmatrix{1&1&1\\2&1&0\\1&0&0}\;.$$

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