Aryanna Fisher

2022-11-12

Is it always true that

$det({A}^{T}A)=0$ , for $A=n\times m$ matrix with $n<m$?

$det({A}^{T}A)=0$ , for $A=n\times m$ matrix with $n<m$?

lavarcar2d2

Beginner2022-11-13Added 18 answers

From the way you wrote it, the product is size $m.$. However, the maximum rank is $n.$ which is smaller. The matrix ${A}^{T}A$ being square and of non-maximal rank, it has determinant $0.$.

Rhett Guerrero

Beginner2022-11-14Added 6 answers

hint

$\mathrm{R}\mathrm{a}\mathrm{n}\mathrm{k}(AB)\le min(\mathrm{R}\mathrm{a}\mathrm{n}\mathrm{k}A,\mathrm{R}\mathrm{a}\mathrm{n}\mathrm{k}B)$ Notice that ${A}^{\mathrm{T}}A$ is $m\times m$

$\mathrm{R}\mathrm{a}\mathrm{n}\mathrm{k}(AB)\le min(\mathrm{R}\mathrm{a}\mathrm{n}\mathrm{k}A,\mathrm{R}\mathrm{a}\mathrm{n}\mathrm{k}B)$ Notice that ${A}^{\mathrm{T}}A$ is $m\times m$

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