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If $A$ is an $n \times n$ matrix, and $A^{T}A = A$, then $A$ is symmetric and $A=A^2$.

Let $A = [a_{ij}]$

Suppose $A^{T}A = A$.

Then, we want to show that $[A^{T}A]_{ij} = [A^{T}A]_{ji}$ (symmetric $A=A^T$)

$[A^{T}A]_{ij}$

= $[A^{T}]_{ij}\times [A]_{ij}$

= $[A]_{ji}\times [A^{T}]_{ji}$

= $[A^{T}]_{ji}\times [A]_{ji}$ (switched order)

= $[A^{T}A]_{ji}$.

I have showed that it is symmetric.

How do I show $A = A^2$?

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    If $A$ is symmetric then $A=A^t$. – 2017-01-23
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    If $A^T=A \to A^TA=A \to AA=A \\\to A^2=A$ – 2017-01-23
  • 0
    The matrix $A^TA$ is symmetric for every matrix $A$ (even nonsquare), – 2017-01-23

1 Answers 1

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To show that $A$ is symmetric we can also use the identities $(CD)^T = D^T C^T$ and $(C^T)^T = C$. Then

$$ A^T = (A^TA)^T = A^T(A^T)^T = A^TA = A $$

and

$$ A = A^T A = A(A) = A^2. $$

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    hahaha, one second faster, +1 – 2017-01-23
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    Thanks alot, I understand. I think the way you showed symmetry is better – 2017-01-23