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I'm quite new in Linear Algebra and I read the question on a book, try to know how to solve that: Let $X$ be a matrix 5 by 3. Does there exist the inverse of $XX^T$?

Chuong Huynh
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    Note that $XX^T$ is a square matrix (specifically 5 by 5 in this case), however there is no guarantee that it is invertible. Consider $X$ with all entries zero. – Hendrix Jul 19 '19 at 02:02
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    It's useful to know that if $A$ is a real $m \times n$ matrix then $A$ and $A^T A$ have the same null space. Proof: $Ax = 0 \implies A^T A x = 0.$ Conversely, $A^T Ax = 0 \implies x^T A^T Ax = 0 \implies | Ax |^2 = 0 \implies Ax = 0$. – littleO Jul 19 '19 at 02:03
  • $X^T$ is $3 \times 5$, so the dimension of its image is at most $3$, hence the dimension of its null space is at least $5-3 = 2$. Thus the dimension of the null space of $XX^T$ is also at least $2$, and in particular is not $0$, so $XX^T$ cannot be invertible. –  Jul 19 '19 at 02:28

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There does not. The rank goes smaller under matrix multiplication. Since $X$ has size $5\times 3$, $r(X)\leq3$, where $r$ denotes the rank. Hence $r(XX^T)\leq r(X)\leq3$. But $XX^T$ is invertible $\iff$ it has rank $5$. Hence it is not invertible.

trisct
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  • Can you explain why $XX^T$ is invertible if it has rank 5? – Chuong Huynh Jul 19 '19 at 02:26
  • @ChuongHuynh See https://math.stackexchange.com/questions/350991/n-by-n-invertible-matrix-a-has-textranka-n – Hendrix Jul 19 '19 at 02:28
  • @ChuongHuynh I think there are more than one way to prove it, and any textbook should have a proof of this. Since I don't know what you have already learned I'd suggest you to first find it in your textbook or check out this link shared by Hendrix. – trisct Jul 19 '19 at 02:44