Field Theory. Eigenvalues of Orthogonal Matrices Have Length 1. Leave a Reply Cancel reply Your email address will not be published. This website is no longer maintained by Yu. ST is the new administrator. Linear Algebra Problems by Topics The list of linear algebra problems is available here. Subscribe to Blog via Email Enter your email address to subscribe to this blog and receive notifications of new posts by email.
Sponsored Links. Although it spans R 2 , it is not linearly independent. No collection of 3 or more vectors from R 2 can be independent. Although it is linearly independent, it does not span all of R 3. A space may have many different bases. In fact, any collection containing exactly two linearly independent vectors from R 2 is a basis for R 2.
Similarly, any collection containing exactly three linearly independent vectors from R 3 is a basis for R 3 , and so on. Although no nontrivial subspace of R n has a unique basis, there is something that all bases for a given space must have in common. If V has a basis containing exactly r vectors, then every basis for V contains exactly r vectors.
That is, the choice of basis vectors for a given space is not unique, but the number of basis vectors is unique. Example 6 : In R 3 , the vectors i and k span a subspace of dimension 2. Figure 1. See Figure. The subspaces of R 1 , R 2 , and R 3 , some of which have been illustrated in the preceding examples, can be summarized as follows:. Example 9 : Find the dimension of the subspace V of R 4 spanned by the vectors.
Since these vectors are in R 5 , their span, S , is a subspace of R 5. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Connect and share knowledge within a single location that is structured and easy to search. My text says a basis B for a vector space V is a linearly independent subset of V that generates V. OK then. I need to see if these vectors are linearly independent, yes? Something tells me that this is wrong. But I am having a hell of a time figuring this stuff out. Please someone help, and I ask: pretend I am the dumbest student you ever met. So you have, in fact, shown linear independence.
Your confusion stems from the fact that you showed that the homogeneous system had only the trivial solution 0,0,0 , and indeed homogeneous systems will always have this solution.
The criteria for linear dependence is that there exist other, nontrivial solutions. Another way to check for linear independence is simply to stack the vectors into a square matrix and find its determinant - if it is 0, they are dependent, otherwise they are independent. This method saves a bit of work if you are so inclined. If the determinant is zero then the set is linearly dependent else i. Sign up to join this community.
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