Any matrix has eigen value either 0 or 1?

Question

Any matrix is similar to a block identity matrix,(i.e.a matrix which has a block of identity matrix in the upper left part and all other entries are zero) ,but we know that similar matrices have same eigen values,,does not that imply that any matrix can have eigen value only 0 or 1? But obviously a matrix can have eigen values other than 0 or 1. Surely,I am trapped in a paradox. Can anyone help me out? Thanks in advance.

Answer 1

The statement seems to confuse two different concepts of matrix equivalence. These two notions arise naturally from the representation of two different abstract linear-algebraic objects as square matrices, namely, (1) linear transformations and (2) symmetric, bilinear forms.

Suppose we have a linear transformation from $\Bbb C^n$ to itself with matrix representation $A$ with respect to some basis. Changing the basis, say, with a change-of-basis matrix $P$, gives a new matrix representation, $P^{-1} A P$. In particular, two $n \times n$ matrices $A, B$ can both encode a given linear transformation iff there is some $P$ such that $$B = P^{-1} A P ,$$ in which case we say that $A$ is similar to $B$. The Jordan normal form gives an essentially unique representative of each similarity class of matrices. If two matrices are similar, their eigenvalues, as well as the algebraic and geometric multiplicities thereof.

On the other hand, suppose we have a bilinear symmetric form on $\Bbb C^n$, again with a matrix representation $A$ (since the form is symmetric, so is $A$). Changing the basis, again by $P$, gives a new matrix representation for the bilinear form, namely $P^\top A P$, and we declare $n \times n$ matrices $A, B$ to be congruent if $$B = P^\top A P .$$ One can show by explicit an construction change of basis that for any bilinear symmetric form on $\Bbb C^n$ there is a basis with respect to which the form has the matrix representation $$\pmatrix{I_m&\\&0_{n - m}} .$$ So, the nonnegative integer $m$ is the only invariant of a congruence class of symmetric complex matrices. In particular, eigenvalues are not preserved under congruence---only the number of nonzero eigenvalues, counting multiplicity, is. (Over other fields the story can be more complicated: For example, for real matrices we must replace this statement with Sylvester's Law of Intertia, which gives rise the notion of signature of a real symmetric bilinear form.)

Answer 2

If I understand your assertion correctly, it’s equivalent to saying that for a linear transformation $T:V\to W$, it’s always possible to choose bases for $V$ and $W$ such that the matrix of $T$ with respect to these bases has the form $$\left[\begin{array}{c|c}I_r & 0 \\ \hline 0&0\end{array}\right],$$ where $r$ is equal to the rank of $T$. This of course also holds for the special case of an automorphism $T:V\to V$, but the catch is that the “input” and “output” bases might not be the same, which is what’s required for similarity. In other words, for any square matrix $A$ you can find invertible square matrices $S$ and $T$ such that $S^{-1}AT$ has the above form, but in general $S\ne T$, so this will not be a similarity.