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24.07.2022 - 21:34

Is B an orthogonal basis for R4 ?

Question:

Let {eq}B=left{\left[ begin{array}{c} -1\ 1\ -1\ 1 end{array}right],\left[ begin{array}{c} 1\ 1\ 1\ 1 end{array}right]\left[ begin{array}{c} 1\ 1\ -1\ -1 end{array}right]\left[ begin{array}{c} 1\ -1\ -1\ 1 end{array}right]right} {/eq}

{eq}y=\left[ begin{array}{c} -3\ 4\ 2\ 1 end{array}right] {/eq} is {eq}B {/eq} an orthogonal basis for {eq}mathbb{R}^4 {/eq} ?

Answers (1)

Esther April 17, 2023 в 19:37

No, {eq}B {/eq} is not an orthogonal basis for {eq}mathbb{R}^4 {/eq}. To check for orthogonality, we need to check if the dot products between any two vectors in {eq}B {/eq} are zero. However, since {eq}B {/eq} is not a set of perpendicular vectors, the dot products between some pairs of vectors are nonzero. For example, {eq}left[begin{array}{c} -1\ 1\ -1\ 1 end{array}right]\cdot\left[ begin{array}{c} 1\ 1\ 1\ 1 end{array}right] = -1 + 1 -1 + 1 = 0{/eq} {eq}left[begin{array}{c} 1\ 1\ 1\ 1 end{array}right]\cdot\left[ begin{array}{c} 1\ 1\ -1\ -1 end{array}right] = 1 + 1 - 1 - 1 = 0{/eq} {eq}left[begin{array}{c} 1\ -1\ -1\ 1 end{array}right]\cdot\left[ begin{array}{c} 1\ 1\ -1\ -1 end{array}right] = 1 - 1 + 1 - 1 = 0{/eq} However, the dot product between {eq}left[begin{array}{c} -1\ 1\ -1\ 1 end{array}right] {/eq} and {eq}left[begin{array}{c} 1\ -1\ -1\ 1 end{array}right] {/eq} is not zero: {eq}left[begin{array}{c} -1\ 1\ -1\ 1 end{array}right]cdot\left[ begin{array}{c} 1\ -1\ -1\ 1 end{array}right] = -1 -1 + 1 + 1 = 0 {/eq} Therefore, {eq}B {/eq} is not an orthogonal basis for {eq}mathbb{R}^4 {/eq}.

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