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Let $A$, $B$ be $n\times n$ complex matrices and $I$ be the $n\times n$ identity matrix.

Is $\left(\begin{array}{cc}A&I\\I&B\end{array}\right)$ being invertible the same as $\det(AB-I)\ne 0$?

If not, what is the right condition for this $2n\times 2n$ matrix to be invertible? Thank you.

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    $\begingroup$ Your "claim" is correct. Hint about proving it: elementary row operations do not change whether a matrix is invertible, so consider what the steps to row reduce your block matrix will lead to... $\endgroup$
    – hardmath
    Sep 10, 2016 at 16:56
  • $\begingroup$ Just for information : It's true if the matrix has coefficient in a field, but it's not true in general ! Take $$\begin{pmatrix}1&0\\0&2\end{pmatrix}.$$ It's determinant is $2$ but it's not invertible over $\mathbb Z$. $\endgroup$
    – Surb
    Sep 10, 2016 at 17:11
  • $\begingroup$ @hardmath Thank you very much! $\endgroup$
    – HLC
    Sep 10, 2016 at 17:30
  • $\begingroup$ @Surb For a matrix $M$ to be invertible over a ring $R$ you only need $\det(M)$ to be a unit in $R$. So here we only need $\det(AB-I)$ to be a unit. $\endgroup$
    – HLC
    Sep 11, 2016 at 5:34
  • $\begingroup$ @HLC: Yes, it's correct ! (I wouldn't use "only" since to be a unit is not evident, but yes, it must be a unit) $\endgroup$
    – Surb
    Sep 11, 2016 at 6:50

1 Answer 1

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This follows from $ \begin{pmatrix} I & 0 \\ - A & I \end{pmatrix} \begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix} \begin{pmatrix} A & I \\ I & B \end{pmatrix} = \begin{pmatrix} I & B \\ 0 & I - AB \end{pmatrix}. $

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  • $\begingroup$ I think your proof is kinda the same as row operations but don't see exactly how that works yet. Can you illustrate what row operations do we essentially perform by multiplying by $\left(\begin{array}{cc}I&0\\-A&I\end{array}\right)$ and $\left(\begin{array}{cc}0&I\\I&0\end{array}\right)$ please? Thank you. $\endgroup$
    – HLC
    Sep 10, 2016 at 17:21
  • $\begingroup$ $\begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix}$ swaps the first and second block of rows, this puts identity matrices in the diagonal. We can then use the identity matrices on the first diagonal block and zero out what lies beneath it and end up with block triangular matrices. Perform the multiplications one by one and you will see what I mean. $\endgroup$
    – Arin Chaudhuri
    Sep 10, 2016 at 17:25
  • $\begingroup$ Of course you are right! I'm just being stupid. I will delete these comments later. Thank you very much! $\endgroup$
    – HLC
    Sep 10, 2016 at 17:30

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