If $A$ and $B$ are two invertible $5 \times 5$ matrices, does $B^{T}A$ remain invertible?
I cannot find out is there any properties of invertible matrix to my question.
Thank you!
$\begingroup$
$\endgroup$
5
asked Oct 24, 2012 at 16:03
-
$\begingroup$ A common trick to answer the question "Is foo invertible?" is to actually write down the inverse of foo. In many cases this is pretty easy to do, as seen in the answers. $\endgroup$– user14972Oct 24, 2012 at 16:10
-
$\begingroup$ @Hurkyl What does foo mean? $\endgroup$– Jill CloverOct 24, 2012 at 16:13
-
$\begingroup$ @PENGTENG It is a nonsense word used as a generic placeholder in math and computer science. It's being used as a variable for an object here, but it's more frequently used for properties rather than objects. $\endgroup$– rschwiebOct 24, 2012 at 16:14
-
$\begingroup$ @rschwieb I get it. Thanks! $\endgroup$– Jill CloverOct 24, 2012 at 16:17
-
$\begingroup$ How is this well researched? C'mon guys $\endgroup$– Alec TealAug 3, 2015 at 6:41
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
3
Of course: $B$ invertible implies $B^T$ invertible, and the product of two invertible matrices is clearly invertible.
This is easily seen from these equations: $$BB^{-1}=I\implies (BB^{-1})^T=I\implies (B^{-1})^TB^T=1,$$ and the fact that if $X$ and $Y$ are invertible, $(XY)^{-1}=Y^{-1}X^{-1}$.
Perhaps the general properties you should take away are these:
$(XY)^T=Y^TX^T$ and $(XY)^{-1}=Y^{-1}X^{-1}$.
answered Oct 24, 2012 at 16:06
-
3$\begingroup$ And this is of course true for $n \times n$-matrices and not just $5 \times 5$-matrices. $\endgroup$– N.U.Oct 24, 2012 at 16:11
-
$\begingroup$ "...the product of two matrices is clearly invertible." Only if the two matrices are themselves invertible: math.stackexchange.com/a/1026628 $\endgroup$– nacnudusAug 3, 2015 at 2:16
-
$\begingroup$ @nacnudus of course, a reasonable person can tell this is a typo of the form of an omitted word from context. Thanks for indirectly alerting me to it. $\endgroup$– rschwiebAug 3, 2015 at 3:27
$\begingroup$
$\endgroup$
Yes. $$ \det(B^T\,A)=\det(B^T)\det(A)=\det(B)\det(A)\ne0. $$ Moreover $$ (B^T\,A)^{-1}=A^{-1}(B^{-1})^T. $$
answered Oct 24, 2012 at 16:06