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Find an orthonormal basis for the subspace U of M2,2(R) spanned by −1 0 1 1 ,  0 1 1 0 .

Question

Find an orthonormal basis for the subspace U of M2,2(R) spanned by

(10 11),(01 10).\begin{pmatrix}-1 & 0 \\\ 1 & 1\end{pmatrix}, \begin{pmatrix}0 & 1 \\\ 1 & 0\end{pmatrix}.

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Solution

The given vectors are:

A = [ -1 0 1 1 ]

B = [ 0 1 1 0 ]

Step 1: Check if the vectors are orthogonal

Two vectors are orthogonal if their dot product is zero. In the case of matrices, we use the Frobenius inner product, which is the sum of the products of the corresponding entries.

The Frobenius inner product of A and B is:

(-10 + 01 + 11 + 10) = 1

Since the inner product is not zero, the vectors are not orthogonal.

Step 2: Gram-Schmidt Process

We can use the Gram-Schmidt process to orthogonalize the vectors and then normalize them to get an orthonormal basis.

First, let's orthogonalize:

Let u1 = A = [ -1 0 1 1 ]

To find u2, we subtract the projection of B onto A from B:

u2 = B - ((B dot u1) / (u1 dot u1)) * u1

Calculating the dot products:

B dot u1 = (0*-1 + 10 + 11 + 01) = 1 u1 dot u1 = (-1-1 + 00 + 11 + 1*1) = 3

So,

u2 = B - (1/3)*u1 = [ 0 1 1 0 ]

              - (1/3)*[ -1 0
                         1 1 ]

              = [ 1/3 1
                  2/3 -1/3 ]

Step 3: Normalize the vectors

To normalize, we divide each vector by its magnitude:

v1 = u1 / ||u1|| = [ -1/√3 0 1/√3 1/√3 ]

v2 = u2 / ||u2|| = [ 1/√(1/9 + 1 + 4/9 + 1/9) 1/√(1/9 + 1 + 4/9 + 1/9) 2/√(1/9 + 1 + 4/9 + 1/9) -1/3√(1/9 + 1 + 4/9 + 1/9) ]

Simplifying, we get:

v1 = [ -1/√3 0 1/√3 1/√3 ]

v2 = [ 1/√2 1/√2 2/√2 -1/6√2 ]

So, the orthonormal basis for the subspace U is {v1, v2}.

This problem has been solved

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