Prove that for any square matrix A, the eigenvalues of AAᵀ and AᵀA are the same (excluding zero).
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Step-by-Step Solution
Step 1: Define Eigenvalues and Eigenvectors
An eigenvalue λ of a matrix M is a scalar such that there exists a non-zero vector v (called an eigenvector) satisfying the equation Mv=λv. We want to show that the non-zero eigenvalues of AAT and ATA are identical.
Step 2: Consider an Eigenvalue of ATA
Let λ be a non-zero eigenvalue of ATA with a corresponding eigenvector v. This means ATAv=λv, where v=0.
Step 3: Multiply by A
Multiply both sides of the equation ATAv=λv by A from the left. This gives A(ATAv)=A(λv), which simplifies to (AAT)(Av)=λ(Av).
Step 4: Show Av is an Eigenvector of AAT
For Av to be an eigenvector of AAT, it must be non-zero. Since λ=0 and ATAv=λv, if Av=0, then AT(0)=λv, which implies 0=λv. Since λ=0, this would mean v=0, contradicting our assumption that v is a non-zero eigenvector. Therefore, Av=0.
Step 5: Conclusion for one direction
Since Av=0, the equation (AAT)(Av)=λ(Av) shows that λ is an eigenvalue of AAT with Av as its corresponding eigenvector. This proves that every non-zero eigenvalue of ATA is also an eigenvalue of AAT.
Step 6: Symmetry of the Proof
The proof is symmetric. Let μ be a non-zero eigenvalue of AAT with eigenvector w. Then AATw=μw. Multiplying by AT from the left gives AT(AATw)=AT(μw), which simplifies to (ATA)(ATw)=μ(ATw). Similar to the previous steps, we can show that ATw=0. Thus, every non-zero eigenvalue of AAT is also an eigenvalue of ATA.