24. (T/F) Not every orthogonal set in $\mathbb{R}^{n}$ is linearly independent.
25. (T/F) If $\mathbf{y}$ is a linear combination of nonzero vectors from an orthogonal set, then the weights in the linear combination can be computed without row operations on a matrix.
26. (T/F) If a set $S=\left\{\mathbf{u}_{1}, \ldots, \mathbf{u}_{p} ight\}$ has the property that $\mathbf{u}_{i} \cdot \mathbf{u}_{j}=0$ whenever $i
eq j$, then $S$ is an orthonormal set.

Question

24. (T/F) Not every orthogonal set in $\mathbb{R}^{n}$ is linearly independent.
25. (T/F) If $\mathbf{y}$ is a linear combination of nonzero vectors from an orthogonal set, then the weights in the linear combination can be computed without row operations on a matrix.
26. (T/F) If a set $S=\left\{\mathbf{u}_{1}, \ldots, \mathbf{u}_{p}ight\}$ has the property that $\mathbf{u}_{i} \cdot \mathbf{u}_{j}=0$ whenever $i 
eq j$, then $S$ is an orthonormal set.

Accepted Answer

∻Solution by Steps∻ ### Question 24 ‖ step 1 ⋮ Identify the statement: "Not every orthogonal set in $\mathbb{R}^{n}$ is linearly independent." ‖ ‖ step 2 ⋮ Recall the definition of an orthogonal set: A set of vectors $\{\mathbf{u}_1, \mathbf{u}_2, \ldots, \mathbf{u}_p\}$ is orthogonal if $\mathbf{u}_i \cdot \mathbf{u}_j = 0$ for all $i eq j$. ‖ ‖ step 3 ⋮ Recall the property of orthogonal sets: In $\mathbb{R}^{n}$, any orthogonal set of nonzero vectors is linearly independent. ‖ ‖ step 4 ⋮ Conclude that the statement is False because every orthogonal set of nonzero vectors in $\mathbb{R}^{n}$ is linearly independent. ‖ ∻Answer∻ ⚹ False ⚹ ∻Key Concept∻ ⚹Orthogonal sets and linear independence⚹ ∻Explanation∻ ⚹In $\mathbb{R}^{n}$, any orthogonal set of nonzero vectors is always linearly independent.⚹ ### Question 25 ‖ step 1 ⋮ Identify the statement: "If $\mathbf{y}$ is a linear combination of nonzero vectors from an orthogonal set, then the weights in the linear combination can be computed without row operations on a matrix." ‖ ‖ step 2 ⋮ Recall the property of orthogonal sets: If $\{\mathbf{u}_1, \mathbf{u}_2, \ldots, \mathbf{u}_p\}$ is an orthogonal set, then any vector $\mathbf{y}$ that is a linear combination of these vectors can be written as $\mathbf{y} = c_1 \mathbf{u}_1 + c_2 \mathbf{u}_2 + \ldots + c_p \mathbf{u}_p$. ‖ ‖ step 3 ⋮ The weights $c_i$ can be computed using the formula $c_i = \frac{\mathbf{y} \cdot \mathbf{u}_i}{\mathbf{u}_i \cdot \mathbf{u}_i}$, which does not require row operations on a matrix. ‖ ‖ step 4 ⋮ Conclude that the statement is True because the weights can be computed directly using the dot product. ‖ ∻Answer∻ ⚹ True ⚹ ∻Key Concept∻ ⚹Weights in linear combinations of orthogonal sets⚹ ∻Explanation∻ ⚹The weights in the linear combination of vectors from an orthogonal set can be computed directly using dot products, without the need for row operations on a matrix.⚹ ### Question 26 ‖ step 1 ⋮ Identify the statement: "If a set $S=\left\{\mathbf{u}_1, \ldots, \mathbf{u}_p ight\}$ has the property that $\mathbf{u}_i \cdot \mathbf{u}_j=0$ whenever $i eq j$, then $S$ is an orthonormal set." ‖ ‖ step 2 ⋮ Recall the definition of an orthonormal set: A set of vectors $\{\mathbf{u}_1, \mathbf{u}_2, \ldots, \mathbf{u}_p\}$ is orthonormal if $\mathbf{u}_i \cdot \mathbf{u}_j = 0$ for all $i eq j$ and $\|\mathbf{u}_i\| = 1$ for all $i$. ‖ ‖ step 3 ⋮ Note that the given property $\mathbf{u}_i \cdot \mathbf{u}_j=0$ for $i eq j$ only ensures orthogonality, not orthonormality. ‖ ‖ step 4 ⋮ Conclude that the statement is False because orthonormality also requires each vector to have a unit norm, which is not guaranteed by orthogonality alone. ‖ ∻Answer∻ ⚹ False ⚹ ∻Key Concept∻ ⚹Orthogonal vs. Orthonormal sets⚹ ∻Explanation∻ ⚹An orthonormal set requires both orthogonality and unit norm for each vector, whereas the given property only ensures orthogonality.⚹◊Is it possible for a set to be orthogonal but not orthonormal in $\mathbb{R}^{n}$?⍭ Generate me a similar question◊

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