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If a have a matrix say $A$ that is orthogonally diagonalizable (i.e. it can be written as $\lambda_1u_1u_1^T+ \lambda_2 u_2u_2^T+\dotsc \lambda_nu_nu_n^T$ , where the $u_i$ are the eigenvectors of the matrix $A$ and $\lambda_i$ are the eigenvalues).

I am just wondering is $A$ a projection matrix?

By definition, a matrix say $B$ is a projection matrix iff $B^2=B$

But it seems $A^2$ not necessarily equal $A$.

Is it correct that $A$ which is orthogonally diagonalizable not necessarily a projection matrix itself?

john_w
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    The scalar matrices $\lambda I$ are all orthogonally diagonalizable, but $(\lambda I)^2 = \lambda I$ only for $\lambda = 0, 1$. – Travis Willse Jul 08 '16 at 08:32
  • Because when I write A as A = UDU^T and then multiply A by itself i.e. AA = UDU^T UDU^T, it seems that it becomes UD^2 U^T . Where U^T is U transpose. – john_w Jul 08 '16 at 08:34
  • My question is: Is it a projection matrix? Yes or NO – john_w Jul 08 '16 at 08:35
  • Orthogonally diagonalizable if and only if symmetric. Symmetric $\ne$ projection. – Gerry Myerson Jul 08 '16 at 09:26
  • Thanks Gerry. But can you also tell me if individually those lamda_i mu_i mu_i^T whether they are projection matrix themselves? – john_w Jul 11 '16 at 09:13
  • So if we add the individuals lambda_i * mu_i * mu_i ^T together, then the sum is not a projection matrix? but individually each one is a projection matrix, is that right? – john_w Jul 11 '16 at 09:14

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