Let M1 = eginbmatrix -1 -1 1 1 endbmatrix and M2 =...

Let $M_1 = egin{bmatrix} -1 & -1 \ 1 & 1 end{bmatrix}$ and $M_2 = egin{bmatrix} -6 & -2 \ 2 & 2 end{bmatrix}$. \ Consider the inner product $<A, B> = trace(A^TB)$ in the vector space $mathbb{R}^{2 imes 2}$ of $2 imes 2$ matrices. Use the Gram-Schmidt process to determine an orthonormal basis for the subspace of $mathbb{R}^{2 imes 2}$ spanned by the matrices $M_1$ and $M_2$.

Transcript

00:01 For this question we are given two 2x2 matrices.

00:05 We have m1 and m2.

00:07 We are also given an inner product and our role with this inner product is to find an orthonormal basis for the span of m1 and m2.

00:17 So what we're going to do is first we're going to find an orthogonal basis and this will be done using the gram -schmidt orthogonalization process.

00:31 Let's call this orthogonal basis v1 and v2.

00:38 So the way that this works is we first choose v1 to equal one of our matrices.

00:47 In this case we'll let v1 equal m1 and then we use orthogonal projections to get the other matrix.

00:58 So in this case the v2 that we're looking for is going to be m2 minus and then we have the orthogonal projection from v1 which uses the inner product.

01:13 It will be m2 with v1 over v1 with itself and this is all times v1.

01:28 So v1 is already figured out.

01:31 Let's go ahead and figure out these two inner products.

01:34 So if we do m2 and v1 this is going to equal by definition the trace of m2 transpose times v1.

01:53 So let's go ahead and put these together.

02:00 So let's see trace of m2 transpose we should get for the first column negative six negative two and then two two for the second column.

02:21 Okay and then we also have v1 which is equal to m1 and there we go.

02:39 All right next we would like to compute the matrix multiplication.

02:44 So let's see when we do that you should get eight eight four and four.

02:55 So there's your matrix multiplication.

02:57 You take the trace which is the sum of the diagonal elements which in this case is 12.

03:03 Okay now we do it again this time for v1 and itself.

03:11 So using the definition of the inner product this is the trace of v1 transpose times v1.

03:21 So we have trace of we have v1 transpose so we get negative one negative one and then one one times and then we have v1.

03:47 Perfect.

03:48 So there are the two matrices...

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