Proving Matrix Equality Using Singular Value Decomposition

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This discussion focuses on proving matrix equality using Singular Value Decomposition (SVD). It establishes that for two matrices A and B, the condition $AA^T = BB^T$ holds if and only if $A = BO$, where O is an orthogonal matrix. The forward direction requires SVD to demonstrate that the singular values of A match those of B, while the reverse direction does not necessitate SVD. The conclusion emphasizes the importance of SVD in the forward proof but clarifies its redundancy in the reverse proof.

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  • Understanding of matrix operations, specifically matrix transposition and multiplication.
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  • Knowledge of orthogonal matrices and their characteristics.
  • Basic linear algebra concepts, including eigenvalues and eigenvectors.
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linearishard
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Hi, I have another question, if A and B are mxn matrices, how do I prove that $AA^T = BB^T$ iff $A = BO$ where $O$ is some orthogonal matrix? I think I need to use a singular value decomposition but I am not sure. Thanks!
 
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Can you at least prove the reverse direction, that is, if $A = BO$ for some orthogonal matrix $O$, then $AA^T = BB^T$? You don't need to use SVD for this.
 
Yeah I did that but it seemed too simple, my study guide says I should be using SVD. Is it actually unnecessary?
 
You use SVD for the forward direction, not the reverse direction.
 
what do you mean by that? What is the forward and reverse directions?
 
The forward direction: If $AA^T = BB^T$, then $A = BO$ where $O$ is some orthogonal matrix. The reverse direction: If $A = BO$ where $O$ is an orthogonal matrix, then $AA^T = BB^T$.
 
If $AA^T = BB^T$, then the singular values of $A$ are the singular values of $B$. You can write $A = U\Sigma V^T$ and $B = U\Sigma Q^T$ for some orthogonal matrices $U, V$, and $Q$. Then $A = BO$ where $O = QV^T$. Since the transpose and product of orthogonal matrices are orthogonal, $O$ is orthogonal.
 

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