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Let $B$ be an $n\times m$ real matrix, $c$ be a positive constant, and $I$ be the identity.

Is $B\cdot B^{\mathrm T} + cI$ guaranteed to be invertible?

pitchounet
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Douglas B. Staple
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2 Answers2

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Yes. You can check that $BB^T+cI$ is positive definite; its eigenvalues are all at least $c>0$. So it is invertible.

kimchi lover
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$BB^T$ for any $B$ is symmetric semi-positive definite (see explanation below), which means that its eigenvalues are all $\geq 0$. Adding $cI$ shifts "to the right" all the spectrum by $c$, thus all eigenvalues become $>0$. And, as this operation preserves symmetry, the resulting matrix est symmetric positive definite.

Explanation : for any (vertical) vector $X \in \mathbb{R^n}$,

$$X^T(BB^T)X= (B^TX)^T(B^TX)=\|B^TX\|^2 \geq 0$$

which is a criterion for semi definite positiveness of matrix $BB^T$.

Jean Marie
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