# Math Help - orthogonal matrices question

1. ## orthogonal matrices question

there is an orthogonal matrices nxn,in which all of its members are
whole numbers (like -1 ,-7,4,8,..).
also there is vector b $b\epsilon R^{n}$ its members are whole too.

prove that for the system Ax=b we have a single solution "c"

where c=(c1,..,cn) and c1 .. cn are whole numbers

2. ## Re: orthogonal matrices question

Originally Posted by transgalactic
there is an orthogonal matrices nxn,in which all of its members are
whole numbers (like -1 ,-7,4,8,..).
also there is vector b $b\epsilon R^{n}$ its members are whole too.

prove that for the system Ax=b we have a single solution "c"

where c=(c1,..,cn) and c1 .. cn are whole numbers

In such a square system and since an orthogonal matrix is always invertible, the solution's given by $x=A^{-1}b$ , and this is an integer

vector because $A^{-1}$ is an integer matrix (why? What's the determinant of an orthogonal matrix?)

Tonio

3. ## Re: orthogonal matrices question

tonio, i think that invoking the determinant is misleading. orthogonal matrices exist that have non-integral entries, example:

[-1/√2 -1/√2]
[ 1/√2 -1/√2].

more pertinent is the fact that for an orthogonal matrix A, A^-1 = A^T, and clearly A^T has integer entries iff A does.

4. ## Re: orthogonal matrices question

Originally Posted by Deveno
tonio, i think that invoking the determinant is misleading. orthogonal matrices exist that have non-integral entries

Whoever implied otherwise, and what does this have to do with what I wrote? The important fact for the OP is that the determinant

of an orthogonal matrix is always $\pm 1$ , and this means the inverse matrix is also an integer one...

Tonio

, example:

[-1/√2 -1/√2]
[ 1/√2 -1/√2].

more pertinent is the fact that for an orthogonal matrix A, A^-1 = A^T, and clearly A^T has integer entries iff A does.
I don't think the above is more pertinent at all for the supposed level of the OP: if he's asking what he's

asking, relations between the transpose of a matrix and its inverse and , which usually are studied

within the frame of vector spaces with inner product, could still be far away from him.

Tonio

5. ## Re: orthogonal matrices question

sure, i understand, but how do you go from that fact that det(A) = ±1, to conclude that A^-1 has integral entries?

i exhibited an orthogonal matrix for which this is NOT true. in other words, i am wondering what YOU were thinking as the answer

to the question: "Why (does A^-1 have integer entries)?. What's the determinant of an orthogonal matrix?".

because if your answer is "because det(A) = ±1", that's not correct. having a determinant of ±1, does NOT imply integral entries.

6. ## Re: orthogonal matrices question

Originally Posted by Deveno
sure, i understand, but how do you go from that fact that det(A) = ±1, to conclude that A^-1 has integral entries?

By his question he already must know how to calculate the inverse of a matrix: $A^{-1}=\frac{1}{\det A}\,Adj(A)$, so it is now

obvious that the inverse is going to be an integer matrix (in fact, having determinant $\pm 1$ is an iff condition for an integer matrix

to have an integer inverse)...

i exhibited an orthogonal matrix for which this is NOT true.

Oh dear! The matrix we've been working with is an INTEGER one: all I've done has taken this into account all through! I never talked

about general matrices or whatever but focused on the OP where we're given an integer matrix.

in other words, i am wondering what YOU were thinking as the answer

to the question: "Why (does A^-1 have integer entries)?. What's the determinant of an orthogonal matrix?".

because if your answer is "because det(A) = ±1", that's not correct. having a determinant of ±1, does NOT imply integral entries.

Of course not! Please do read my above observation in blue to you, and let's hope the OP also

reads this so that she/he won't confused by your bizarre remarks.

Tonio