Question

I know that if a matrix is symmetric, then the eigenvectors corresponding to the distinct eigenvalues are orthogonal to each other. I also know that a matrix itself is orthogonal if A^-1 = A^T. Does this imply that a matrix must be symmetric in order for it to be orthogonal?

Answer 1

6 1
2 -3

is othogonal right?

Answer 2

if so, is it symmetric?

Answer 3

Hm something seems wrong there.
\[\left[\begin{matrix}6 & 1 \\ 2 & -3 \end{matrix}\right]\] is not orthogonal because
\[A ^{-1} = \left[\begin{matrix}3/20 & 1/20 \\ 1/10 & -3/10\end{matrix}\right] \neq \left[\begin{matrix}6 & 2 \\ 1 & -3\end{matrix}\right] = A ^{T}\]

Also the vectors
\[\left(\begin{matrix}6 \\ 2\end{matrix}\right) and \left(\begin{matrix}1 \\ -3\end{matrix}\right) \]
have to be othonormal for the matrix to be orthogonal. The 2 vectors above are orthogonal but not orthonormal therefore the matrix you proposed can't be orthogonal.

Answer 4

you are confusing terminology

Answer 5

orthogonal basis just have to have perp vectors; once you have perp vectors; and orthonormal basis units them

Answer 6

<6,2>.<1,-3> = 6-6=0 which defines orthogonal vectors

Answer 7

The definition in my book states An nxn matrix is orthogonal if and only if the colums(or rows) form an orthoNORMAL set of vectors, not simply orthogonal.

An alternative definition in the book is that a non-singular matrix A is called orthogonal if \[A ^{-1} = A ^{T}\]
which i showed is not true either. So I dont understand what you are trying to say, Im going strictly by definitions.

Answer 8

http://mathworld.wolfram.com/OrthogonalBasis.html

there is something buried in here that says the Cs need not be equal to 1; yet if they are equal to 1 then the orthoG is an orthoN.

Answer 9

ive been trying to find a better definition tho :)

Answer 10

If I normalize the vectors \[\left(\begin{matrix}6 \\ 2\end{matrix}\right) and \left(\begin{matrix}1 \\ -3\end{matrix}\right)\] then they are orthonormal and therefore the matrix
\[\left[\begin{matrix}6 & 2\\ 1& -3\end{matrix}\right]\]
is orthogonal by definition.

Using the normalized equation:

\[\left[\begin{matrix}6/\sqrt{40} & 1/\sqrt{10}\\ 2/\sqrt{40}& -3/\sqrt{10}\end{matrix}\right]\]

also satisfies

\[A ^{-1} = A ^{T}\]

But the new normalized vector is no longer symmetric

Answer 11

http://books.google.com/books?id=1ebLTz_MtUcC&pg=PA21&dq=orthogonal+basis&hl=en&sa=X&ei=VdyhT_rACpSc8QTfsL3DCQ&ved=0CDMQ6AEwAQ#v=onepage&q=orthogonal%20basis&f=false

Answer 12

all orthoN are orthoG basis; but not all orthoG are orthoN basis ...

Answer 13

but it looks like you answered it appropriately with my example nonetheless :)

Answer 14

yes you're right i did answer my question digging into your example :) thanks

Answer 15

youre welcome :)