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Rank-Eigenvalue Lemma

Featured image for the Rank–Eigenvalue Lemma relating matrix rank to the number of nonzero eigenvalues.

In this proof, we will show that if B = ATA and rank(B) = r, then B has exactly r nonzero eigenvalues, counting repeated eigenvalues.

The key idea is to use the Spectral Theorem to diagonalize B. Since multiplying by invertible matrices does not change rank, the rank of B is equal to the rank of its diagonal matrix of eigenvalues.

The rank therefore tells us exactly how many of those eigenvalues are nonzero.

This result will be especially useful in the Singular Value Decomposition, where the nonzero eigenvalues of ATA determine the positive singular values of A.

Handwritten slide stating the Rank–Eigenvalue Lemma for B = AᵀA with rank(B) = r.
Handwritten proof slide relating rank(B) to the rank of its diagonal eigenvalue matrix D.
Handwritten proof slide concluding that rank(B) = r gives exactly r nonzero eigenvalues, counting repeated eigenvalues.
Handwritten proof slide noting that B = AᵀA is real symmetric and positive semidefinite.
Handwritten proof slide applying the Spectral Theorem to orthogonally diagonalize the real symmetric matrix B.
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Angela Morgan

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