Show that if x is a nonzero column vector in R^n, then the nxn matrix:

A = I - 2/||x||^2 * xx^T
is orthogonal.

Notation key:
||x|| = norm of x
x^T = transpose of x
I = identity matrix.

Let me try to convince a math student to use "physics" notations that many mathematicians don't like.

Let's work with the elements of matrices and use a notation for that instead of the matrix itself. What we do in physics is we attach indices to the matrix. We write:


to denote the matrix element in the i-th row and j-th column (starting from the upper left corner go i steps down and j steps to the right).

The unit matrix is denoted by the Kronecker delta symbol:


which is zero if i is not equal to j and is one if they are equal.

Finally we use the Einstein summation convention for repeated indices.If
C = A B, you can write this as:

C_{i,j} = A_{i,k}B_{k,j}

the repeated index k is summed over.

You can write the square of the norm of x as the inner product with itself, so:

||x||^2 = x_{k}x_{k}

The matrix A in this problem can be written in the "index notation" as:

A_{i,j} =
delta_{i,j}-2/||x||^2 x_{i}x_{j}

You see that we now don't need that stupid T symbol on the x. We don't need to indicate that the index j runs from left to right so that you would have to rotate the column vector x so that it lies on his side.

For an orthogonal matrix the matrix product with its transpose is the unit matrix. The transpose of A is A itself in this case, as you easily see by interchanging i with j:
A_{i,j} = A_{j,i}.

In the general case you need to show that:

A_{i,k}A_{j,k} = delta_{i,j}

In our case A_{j,k} = A_{k,j} so this reduces to an ordinary matrix product, but that doesn't really matter.

If you work out the product (and sum over the repeated index) you obtain four terms.
The product of the deltas gives you:


The three other terms cancel. You obtain twice a product with a delta and the second term in A_{i,j} which after summation of the repeated index gives you:


The product of the two seciond terms in A gives you


Summation over the repeated index k gives a factor ||x||^2 and this then cancels the other term.

Thanks Count. I finally had time to go through and understand this proof. I now understand, but this wasn't simple. I really appreciate the help.

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