If A is any real $m \times n$ matrix, then $A^TA$ is called the Gram matrix of $A$. Show that the Gram matrix of $A$ is always symmetric with nonnegative diagonal elements.

I have tried several matrices to prove that this is true but im not sure how I go about showing for a general matrix that this always holds true. Please help.

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    $\begingroup$ What is the $(i,j)$ entry of $A^T$? What is the $(i,j)$ entry of the matrix product $BA$? What is the $(i,j)$ entry of the matrix product $A^TA$? $\endgroup$ – user108903 Jan 21 '13 at 19:15
  • $\begingroup$ math.stackexchange.com/questions/282045/… $\endgroup$ – Git Gud Jan 21 '13 at 19:47


  1. Prove that the Gram matrix is always symmetric: What does symmetry have to do with transposes? What happens when you transpose $A^TA$?

  2. Prove that the diagonal elements are nonnegative: Let $a_{ij}$ be the $(i,j)$ element of A. Can you find an expression for the $k$-th diagonal element of $A^TA$ in terms of the $a_{ij}$?

  • $\begingroup$ @user56881 You're making the symmetry argument much harder than you need to. What is $(A^TA)^T$? $\endgroup$ – Jonathan Christensen Jan 21 '13 at 19:36

Some hints: for first part, use $(AB)^T=B^TA^T$

for second part, use $a_{ii}=e_i^TA^TAe_i=(Ae_i)^T(Ae_i)=||Ae_i||^2\geq0$. ($e_i$ is $i^{th}$ column of identity matrix)

  • $\begingroup$ @user56881 You sort of got it. We don't need to show that $A^T = A$, we need to show that $G^T = G$, which is what you showed. $A$ may not even be square, so it's certainly not necessarily symmetric. $\endgroup$ – Jonathan Christensen Jan 21 '13 at 20:30

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