3
$\begingroup$

I have these general wondering about matrices but I don't know to proceed with a proof or a counter example. Suppose that $A$ (dimension $n\times n$) is a real symmetric matrix.

  1. If $A$ has $n$ eigenvalues that are all $1$'s, does $A$ equal the identity matrix?
  2. If $A$ has $n$ eigenvalues that are all $0$'s, does $A$ equal the zero matrix?

Can someone elucidate things for me please?

Edit: I learned/can look up diagonalization theorems for real matrices.

$\endgroup$
  • $\begingroup$ Compute $Av$ for a given $v$ and express $v$ in the eigenbasis. $\endgroup$ – djechlin Dec 3 '14 at 1:55
  • 3
    $\begingroup$ Do you know a theorem about "diagonalizing" a real symmetric matrix? $\endgroup$ – GEdgar Dec 3 '14 at 1:56
  • $\begingroup$ Let $N=\begin{pmatrix}0&1\\0&0\end{pmatrix}$. Take $1+N$ for the first, and $N$ for the second. $\endgroup$ – Pedro Tamaroff Dec 3 '14 at 2:05
4
$\begingroup$

The answer to both is yes.

Hint: All symmetric matrices are diagonalizable. That is, $A$ is similar to a diagonal matrix with the eigenvalues of $A$ on the diagonal.

$\endgroup$
1
$\begingroup$

A real symmetric matrix is orthogonally diagonalisable.So there exists an orthogonal matrix $P$ such that $P^{-1}AP=D$ where $D$ is a diagonal matrix.Then since $1$ is the only eigen value of $A$ , $0$ is the only eigen value of $A-I$ so $det(A-I)=0$. Again $A=PDP^{-1}$

$\implies det((PDP^{-1})-I)=0$

$\implies det(P(D-I)P^{-1})=0$

$\implies det(D-I)=0$

Since $D$ is a diagonal matrix $D=I$. Consequently $A=PIP^{-1}=I$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.