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Chapter 6: Problem 11

What happens if the Gram-Schmidt procedure is applied to a list of vectors that is not linearly independent?

Short Answer

Expert verified
When the Gram-Schmidt procedure is applied to a list of vectors that are not linearly independent, it results in an orthogonal component later in the process becoming the zero vector. This occurs because one of the vectors becomes a linear combination of previous vectors, and it's impossible to normalize the zero vector. To create an orthonormal basis for the subspace spanned by the linearly dependent vectors, we can modify the procedure to skip the normalization step when encountering a zero vector, thus removing linearly dependent vectors while preserving the subspace spanned by the basis.

Step by step solution

01

Understand the Gram-Schmidt procedure and linear independence

The Gram-Schmidt procedure is a method to orthogonalize a set of vectors and create an orthonormal basis for the subspace spanned by the original set of vectors. Linearly independent vectors are vectors that cannot be expressed as linear combinations of one another. A set of vectors is linearly independent if none of the vectors in the set is a linear combination of the remaining vectors in the set.
02

Applying the Gram-Schmidt procedure

To apply the Gram-Schmidt procedure on a set of vectors \(v_1, v_2, ..., v_n\) follow these steps: 1. Normalize the first vector: \(u_1 = \frac{v_1}{||v_1||}\) 2. For each subsequent vector, subtract the projection of the current vector onto the previous orthogonalized vectors and normalize the result: For \(i = 2\) to \(n\): a. Calculate the orthogonal component: \(w_i = v_i - \sum_{j=1}^{i-1} \langle v_i, u_j\rangle u_j\) b. Normalize the orthogonal component: \(u_i = \frac{w_i}{||w_i||}\) The resulting set of vectors \(u_1, u_2, ..., u_n\) is an orthonormal basis for the subspace spanned by the original set of vectors.
03

Applying the Gram-Schmidt procedure on linearly dependent vectors

If the Gram-Schmidt procedure is applied on a set of vectors that are not linearly independent, then at some step, the orthogonal component \(w_i\) will be the zero vector. This is because \(v_i\) will be a linear combination of previous vectors, causing the subtraction part to result in the zero vector. It's impossible to normalize the zero vector, and therefore, the Gram-Schmidt procedure cannot be applied directly to create an orthonormal basis for a set of linearly dependent vectors.
04

Handling linearly dependent vectors in the Gram-Schmidt procedure

To create an orthonormal basis that spans the subspace of a set of linearly dependent vectors, we can modify the Gram-Schmidt procedure to skip the normalization step when encountering a zero vector. This will essentially remove linearly dependent vectors from our basis without affecting the subspace spanned by the basis. The resulting set of vectors will be linearly independent and form an orthonormal basis for the subspace spanned by the original set of linearly dependent vectors.

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Most popular questions from this chapter

Prove or disprove: there is an inner product on \(\mathbf{R}^{2}\) such that the associated norm is given by $$\left\|\left(x_{1}, x_{2}\right)\right\|=\left|x_{1}\right|+\left|x_{2}\right|$$ for all \(\left(x_{1}, x_{2}\right) \in \mathbf{R}^{2}\)

Find a polynomial \(q \in \mathcal{P}_{2}(\mathbf{R})\) such that $$p\left(\frac{1}{2}\right)=\int_{0}^{1} p(x) q(x) d x$$ for every \(p \in \mathcal{P}_{2}(\mathbf{R})\)

Prove that if \(P \in \mathcal{L}(V)\) is such that \(P^{2}=P\) and every vector in null \(P\) is orthogonal to every vector in range \(P,\) then \(P\) is an orthogonal projection.

Prove that if \(V\) is a complex inner-product space, then $$\langle u, v\rangle=\frac{\|u+v\|^{2}-\|u-v\|^{2}+\|u+i v\|^{2} i-\|u-i v\|^{2} i}{4}$$ for all \(u, v \in V\)

Prove that if \(x, y\) are nonzero vectors in \(\mathbf{R}^{2}\), then $$\langle x, y\rangle=\|x\|\|y\| \cos \theta$$ where \(\theta\) is the angle between \(x\) and \(y\) (thinking of \(x\) and \(y\) as arrows with initial point at the origin). Hint: draw the triangle formed by \(x, y,\) and \(x-y ;\) then use the law of cosines.
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