# Any matrix has eigen value either 0 or 1?

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.

• Please try to see what happens for diagonal matrices, and real triangular matrices with diagonal entries bigger than 1. – P Vanchinathan Mar 1 at 7:06
• "Any matrix is similar to a block identity matrix"..... no, this is not true at all. – Morgan Rodgers Mar 1 at 7:07
• We know that any square matrix of order n represents a linear map from a n dimensional vector space to itself,,now if we choose the basis of the kernel and extend it to a basis of the given vector space,then with respect to the new basis the matrix of the same linear map gives me a upper left identity block.I know I am trapped in a paradox,,please help me out. – Soumyadip Sarkar Mar 1 at 7:13

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.)

• Actually,I am a beginner to linear algebra,so i don't understand Jordan normal form or symmetric Bilinear form.I am very sorry for that,but thanks for your time,,can you explain in a bit layman language why the paradox is arising? – Soumyadip Sarkar Mar 1 at 7:35
• There's no paradox: It's just not true that any (say, $n \times n$ complex) matrix is similar to a matrix of the form $\pmatrix{I&\\&0}$. It is true that any such symmetric matrix is congruent to one of that form, but similarity and congruence are two different notions of equivalence that apply to two different classes of matrices. For symmetric matrices, similarity is a much stronger condition than congruence. – Travis Mar 1 at 7:39
• To see this concretely, fix $\lambda \in \Bbb C$ and notice that for any invertible matrix $P$, conjugating $\lambda I$ by $P$ gives $P (\lambda I) P^{-1} = \lambda P P^{-1} = \lambda I$, so the only matrix similar to $\lambda I$ is $\lambda I$ itself. On the other hand, the only eigenvalue of $\lambda I$ is $\lambda$. Conversely, if $\lambda \neq 0$ and we choose a square root $\mu$ of $\lambda$ and take $Q = \mu^{-1} I$, then $Q^\top (\lambda I) Q = (\mu^{-1} I) (\lambda I) (\mu^{-1}) = I$, so all of the matrices $\lambda I$, $\lambda \neq 0$ are congruent. – Travis Mar 1 at 7:44
• But,say matrix A represents a linear map from a n dimensional vector space to itself,,say basis of the kernel is {u1,u2,......,uk} and we extend this to a basis of the given vector space say {w1,w2,........wm;u1,u2,.......,uk},,now if T(wi)=wi for all i=1,2,....,m,,then with respect to that basis it will have block identity form,,but that is not the case always,,,I got it,,,thanks. – Soumyadip Sarkar Mar 1 at 7:51
• You're welcome, I hope you found the explanation useful. – Travis Mar 1 at 7:54

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.

• You are absolutely correct,,that's where I got it wrong,,anyway I learnt something properly as I made mistake,,thanks for your time,sir. – Soumyadip Sarkar Mar 4 at 14:40