Suppose that

1) we do not know the theory of lambda-matrix;

2) we can not see the transition matrix directly (use elementary tranformations).

How can we deduce that the following two matrices are similar? (What we only know is that they have the same eigenvalues, eigenvectors, rank, determinant)

$$ \begin{pmatrix} 1&1&0\\ 0&0&1\\ 0&0&0 \end{pmatrix},\quad \begin{pmatrix} 1&1&1\\ 0&0&1\\ 0&0&0 \end{pmatrix} $$

  • $\begingroup$ It is not clear what you mean by the theory of lambda-matrix (such a theory does not exist to my knowledge). You might be talking about Jordan normal forms (just a wild guess, no real connection with anything called lambda other than eigenvalues, which apparently you do know about). $\endgroup$ – Marc van Leeuwen Dec 7 '13 at 11:11

Both $A$ and $B$ have characteristic polynomial $\lambda^2 (\lambda - 1)$, and thus eigenvalues $0$ (with algebraic multiplicity $2$) and $1$ (with algebraic multiplicity $1$). Since $A (A-I)$ and $B(B-I)$ are both nonzero, they both have a nontrivial Jordan block (which must be of size $2$) for eigenvalue $0$. So both have Jordan canonical form $$ \pmatrix{0 & 1 & 0\cr 0 & 0 & 0\cr 0 & 0 & 1\cr}$$ and thus they are indeed similar.

  • $\begingroup$ When OP said we don't know the theory of lambda-matrix, I think they were saying not to use the characteristic polynomial. $\endgroup$ – Travis Bemrose Nov 5 '13 at 23:11

It is not clear what you are asking. If you want to know this for these concrete matrices, then you are probably going to see the transition matrix in the end, maybe violating point 2. Or do you want to show by pure reasoning that any two $2\times2$ matrices with the same two eigenvalues, eigenvectors, rank and determinant as these matrices are similar?

Let me first list what apparently you know but did not say explicitly: eigenvalues are $0,1$, each with a $1$-dimensional eigenspace spanned respectively by $(1,-1,0)$ and by $(1,0,0)$, rank$~2$ and (therefore) determinant$~0$. From these informations alone one cannot tell whether the characteristic polynomial is $x(x-1)^2$ or $x^2(x-1)$ (there are easy examples of both cases), and a fortiori not whether the matrices are similar, so the project of my second question above is not realisable.

So assuming you just want to find a concrete similarity, but without any guesswork, you could proceed as follows. Since in fact it is the eigenvalue $0$ that is a double root of the characteristic polynomial for both matrices, the kernel of $E_0(M)=(M-0I)^2$ is a characteristic subspace of dimensions$~2$ in both cases, containing the eigenspace for$~0$, and it should be made to match between the two. For the first matrix$~A$ this subspace $E_0(A)$ is generated by the mentioned eigenvector $(1,-1,0)$ and for instance $(0,1,-1)$ (which is mapped by$~A$ to that eigenvector); for the second matrix$~B$, the subspace $E_0(B)$ is generated by the same eigenvector $(1,-1,0)$ and for instance $(0,2,-1)$ (which is mapped by$~B$ to that eigenvector). We see that our last gnerator behaves the same in both cases, so transformed to the basis $(1,-1,0),(0,1,-1),(1,0,0)$ the first matrix takes the same form as the second matrix transformed to the basis $(1,-1,0),(0,2,-1),(1,0,0)$, namely $$ \pmatrix{0&1&0\\0&0&0\\0&0&1}, $$ and in particular the matrices are similar. To check you can find an invertible $P$ such that $PAP^{-1}=B$ by composing the change of basis matrices appropriately: $$ P=\pmatrix{1&0&1\\-1&2&0\\0&-1&0}\pmatrix{1&0&1\\-1&1&0\\0&-1&0}^{-1} =\pmatrix{1&0&0\\0&1&-1\\0&0&1}. $$ Indeed you can check that conjugation by $P$ transforms $A$ into $B$; it is so simple (in fact $PA=A$) that you might even have seen it directly (in spite of you point$~2$).


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