Matrix with only eigenvalues = 1

Question

I have a matrix with the following properties:

1. The matrix is real, symmetric, positive semidefinite.
2. All the eigenvalues of the matrix are $1$.

Is the only matrix that satisfies these properties the identity matrix?

Answer 1

If $A$ is real symmetric, it is diagonalizable by the spectral theorem. As all eigenvalues are equal to $1$, $A$ is equivalent to the identity matrix . But the only matrix equivalent to the identity is the identity itself.

Answer 2

$A$ has real eigenvalues, $1$ is the only one, so $B = A-I$ is symmetric and nilpotent. If $B^2x =0$, then $\|Bx\|^2 = x^TB^2x =0$, so $Bx=0$. This implies $B=0$.

Answer 3

Yes, the matrix is $I$.

For if $A$ is real symmetric matrix, there exists a real orthogonal matrix $O$,

$OO^T = O^TO = I, \tag 1$

such that

$O^TAO = \text{diag}(\mu_1, \mu_2, \ldots, \mu_n), \tag 2$

where the $\mu_i$ are the eigenvalues of $A$; here $\text{diag}(\mu_1, \mu_2, \ldots, \mu_n)$ is the matrix with the $\mu_i$ along the main diagonal and zeroes elsewhere. Since we are given that every

$\mu_i = 1, \tag 3$

we have

$\text{diag}(\mu_1, \mu_2, \ldots, \mu_n) = I; \tag 4$

then (2) becomes

$O^TAO = I, \tag 5$

from which

$A = IAI = (OO^T)A(OO^T) = O(O^TAO)O^T$ $= OIO^T = OO^T = I. \tag 6$