1
$\begingroup$

Let $AB \in K^{n\times n}$, $AB$ is a diagonal matrix and the elements on the diagonal are non-zero.

Is $A$ invertible?

Since $AB$ is a diagonal matrix and the elements on the diagonal are non-zero, $AB$ is invertible. Also neither $A$ nor $B$ can possibly contain zero-vectors or $AB$ would too which it doesn't. However not containing zero-vectors is not sufficient for a matrix to be invertible, right?

Please be aware that I don't know much about linear algebra yet, so a thorough explanation would be much appreciated.

$\endgroup$
  • 2
    $\begingroup$ $A(B(AB)^{-1})=I$, done. $\endgroup$ – user641 Nov 18 '12 at 16:10
2
$\begingroup$

Use the determinant:

$$AB=\begin{pmatrix} a_1&0&...&0\\...&...&...&...\\0&0&...&a_n\end{pmatrix}\,\,,\,a_i\neq 0\Longrightarrow \det A\cdot\det B=\det AB=\prod_{i=1}^na_i\neq 0\Longrightarrow$$

$$\Longrightarrow \det A\neq 0\Longrightarrow \,\,A\;\;\text{ is regular and thus invertible}$$

$\endgroup$
  • $\begingroup$ Thanks for your answer though I'll have to do a bit of reading to understand it because we haven't covered the determinant yet. $\endgroup$ – Christian Nov 18 '12 at 11:27
  • $\begingroup$ Well, then re-take your argument: a matrix is non-invertible( i.e. singular) iff upon reducing it (by rows or columns, it doesn't matter) we get one at some point one row (column) of zeros, If this was the case for $\,A\,$ then in $\,AB\,$ (after reducing $A$) we'd get one row (column) or zeros, which is impossible since, at you pointed out, $\,AB\,$ is invertible. $\endgroup$ – DonAntonio Nov 18 '12 at 11:38
  • $\begingroup$ Makes sense. I wasn't exactly sure if I could assume A to be reduced already. So this also means if $A$ had linearly dependant columns (and could thus be reduced to have zero columns) the columns in $AB$ would also be linearly dependent, right? $\endgroup$ – Christian Nov 18 '12 at 11:41
  • $\begingroup$ Indeed so, @Christian $\endgroup$ – DonAntonio Nov 18 '12 at 11:41
4
$\begingroup$

$$AB = \Lambda$$ Since $\Lambda$ is a diagonal matrix with non-zero elements it's invertible $$\left ( AB \right)^{-1} = \Lambda^{-1}$$ If a product is invertible, then multiplicands are also invertible $$\left( AB \right)^{-1} = B^{-1}A^{-1} = \Lambda^{-1} $$ therefore $A^{-1} = B\Lambda^{-1}$

$\endgroup$
  • $\begingroup$ My prof's script contains this rule with the constraint that $A, B \in GL(n, K)$. Is this really true for all matrices in $K^{n\times n}$? $\endgroup$ – Christian Nov 18 '12 at 11:26
  • $\begingroup$ @Kaster , your proof assumes what must be proved, i.e. that $\,A\,$ is invertible. To write $\,(AB)^{-1}=B^{-1}A^{-1}\,$ is meaningless unless we know a priori that both $\,A,B\,$ are invertible. $\endgroup$ – DonAntonio Nov 18 '12 at 11:41
1
$\begingroup$

Yes, both $A$ and $B$ are invertible provided they are square in the first place. We could however say that $A$ and $B$ both have full-rank since $$\text{rank(AB)} \leq \min \{\text{rank(A), rank(B)}\}$$ Since $AB$ is diagonal with non-zero entries, $\text{rank(AB)} = n$. If $A \in \mathbb{R}^{n \times m}, B \in \mathbb{R}^{m \times n}$, then the previous result means that $m \geq n$. If $m=n$, we could also conclude that $A$ and $B$ are invertible.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.