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

$$\begin{align*} a & = \frac{d}{ad - bc} \\ b & = \frac{-c}{ad - bc} \\ c & = \frac{-b}{ad - bc} \\ d & = \frac{a}{ad - bc} \end{align*}$$

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 $$\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$$ are orthonormal, but $$det \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}=-1.$$ I guess it would be easier to work from the relation $$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} a & c \\ b & d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$, 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 : $$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \times \begin{pmatrix} a & c \\ b & d \end{pmatrix} = \begin{pmatrix} (r_1,r_1) & (r_1,r_2) \\ (r_2,r_1) & (r_2,r_2) \end{pmatrix}$$, 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

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

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

$$\begin{align*} a & = \frac{d}{ad - bc} \\ b & = \frac{-c}{ad - bc} \\ c & = \frac{-b}{ad - bc} \\ d & = \frac{a}{ad - bc} \end{align*}$$

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
$$A= \left[\begin{array}{cc}2 & 1 \\ 1 & 1\end{array}\right]$$
is
$$A^{-1}= \left[\begin{array}{cc}2 & -1 \\ -1 & 1\end{array}\right]$$
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.