Can any one help me by following question: Let assume we have two real-valued matrices $A,B\in R^{m\times n}$. Then, let define the matrix $G:=(A+t(B-A))(A+t(B-A))^T$ for a $t\in (0,1)$. Now, I am wondering, which conditions are sufficient to make sure that G is always invertible for every $t\in (0,1)$? BTW: My first thought was that if A and B have full rank, then the line segment between them have also full rank (which concludes that G is invertible) but actually this is not true.
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A square matrix with real entries $M$ is invertible if and only if $MM^T$ so, if you only work with square matrices, then you only need to worry about making sure $A+t(B-A)$ is invertible. If you also assume $A$ is invertible (that's the case $t=0$, by the way), then you can say something reasonably simple:
Claim: Suppose $A$ and $B$ are square matrices of the same size with real entries and $A$ is invertible. Then, the following are equivalent:
- $A+t(B-A)$ is invertible for $0<t<1$.
- The spectrum of $C:=A^{-1}B$ does not contain any negative numbers.
Basically, you just multiply $A+t(B-A)$ by $A^{-1}$ (which doesn't affect its invertibility) to get $I+t(C-I)$. Then, after a bit of algebra, you see that, for $t \neq 0$, $I+t(C-I)$ is invertible if and only if $C-\frac{1-t}{t}I$ is invertible. Since $\left\{\frac{1-t}{t}: 0 < t < 1\right\} = (-\infty,0)$, we get the desired conclusion.
Mike F
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I am also wondering about the cases that M is a non squared matrix. Then, can we say any thing about invertibility of $MM^T$ based on $M$? – Mehr Apr 25 '22 at 07:19
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$MM^T$ is invertible if and only if $M$ is surjective if and only if $M^T$ is injective. – Mike F Apr 26 '22 at 02:35