A Matrix with Orthonormal Columns

[e(ho0n3]

Homework Statement
Let M be an n x n matrix. Prove that the columns of M form an orthonormal set if and only if M^-1 = M^T.

The attempt at a solution
Let's consider the following 2 x 2 matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

If the inverse equals its transpose, i.e.

\frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} a & c \\ b & d \end{pmatrix}

then

\begin{align*}<br /> a &amp; = \frac{d}{ad - bc} \\<br /> b &amp; = \frac{-c}{ad - bc} \\<br /> c &amp; = \frac{-b}{ad - bc} \\<br /> d &amp; = \frac{a}{ad - bc}<br /> \end{align*}

or rather that ad - bc = 1. Does this mean that ab + cd = 0? Not necessarily: Consider a = 2 and b = c = d = 1 so that ad - bc = 2 - 1 = 1 but ab + cd = 2 + 1 = 3. What gives?

And doeas ab + cd = 0 imply the equations above for a, b, c and d? I think not.

I'm stumped.

 

[n_bourbaki]

Just multiply out the matrices and think.

 

[xalvyn]

\left| ad - bc \right|=1, but ad - bc may not necessarily be 1. For example, the columns of \begin{pmatrix} -1 &amp; 0 \\ 0 &amp; 1 \end{pmatrix} are orthonormal, but det \begin{pmatrix} -1 &amp; 0 \\ 0 &amp; 1 \end{pmatrix}=-1. I guess it would be easier to work from the relation \begin{pmatrix} a &amp; b \\ c &amp; d \end{pmatrix} \begin{pmatrix} a &amp; c \\ b &amp; d \end{pmatrix} = \begin{pmatrix} 1 &amp; 0 \\ 0 &amp; 1 \end{pmatrix}, which follows from the hypothesis that the transpose of M is its inverse. The equations should be simpler to work with. In general, the product of any 2 x 2 matrix M and its transpose M^T may be expressed as : \begin{pmatrix} a &amp; b \\ c &amp; d \end{pmatrix} \times \begin{pmatrix} a &amp; c \\ b &amp; d \end{pmatrix} = \begin{pmatrix} (r_1,r_1) &amp; (r_1,r_2) \\ (r_2,r_1) &amp; (r_2,r_2) \end{pmatrix}, where r_i denotes the ith row of M, and (r_i,r_j) denotes the scalar product of r_i and r_j. Now equate the latter expression with the identity matrix. This shows that the rows of M are orthonormal, and the argument can be extended to any (n x n) matrix. To show that the columns are also orthonormal, we could use the fact that if MM^T = I, then M^TM=I, and thence express the product M^TM as we did above.

 

[tiny-tim]

Hi e(ho0n3!

Why make it so complicated?

In problems like this, just write out the definition!

Let M be an n x n matrix. Prove that the columns of M form an orthonormal set if and only if M^-1 = M^T.

Hint: MM^T = I means, for any i and j, ∑a_ika_jk = I_ij.

Put i = j: then … ?

Put i ≠ j: then … ?

 

[HallsofIvy]

Homework Statement
Let M be an n x n matrix. Prove that the columns of M form an orthonormal set if and only if M^-1 = M^T.

The attempt at a solution
Let's consider the following 2 x 2 matrix

\begin{pmatrix} a &amp; b \\ c &amp; d \end{pmatrix}

If the inverse equals its transpose, i.e.

\frac{1}{ad - bc} \begin{pmatrix} d &amp; -b \\ -c &amp; a \end{pmatrix} = \begin{pmatrix} a &amp; c \\ b &amp; d \end{pmatrix}

then

\begin{align*}<br /> a &amp; = \frac{d}{ad - bc} \\<br /> b &amp; = \frac{-c}{ad - bc} \\<br /> c &amp; = \frac{-b}{ad - bc} \\<br /> d &amp; = \frac{a}{ad - bc}<br /> \end{align*}

or rather that ad - bc = 1. Does this mean that ab + cd = 0? Not necessarily: Consider a = 2 and b = c = d = 1 so that ad - bc = 2 - 1 = 1 but ab + cd = 2 + 1 = 3. What gives?

And doeas ab + cd = 0 imply the equations above for a, b, c and d? I think not.

I'm stumped.

I don't understand your point with this example. The inverse of
A= \left[\begin{array}{cc}2 &amp; 1 \\ 1 &amp; 1\end{array}\right]
is
A^{-1}= \left[\begin{array}{cc}2 &amp; -1 \\ -1 &amp; 1\end{array}\right]
NOT the transpose of A and so has nothing to do with this problem.

 

[e(ho0n3]

OK. Looks like I really messed up on this one. Thank you all for the pointers.